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Introduction
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection0
Tue, 12 Apr 2016 23:00:00 GMT
<p>Motion is vital to life, and to science. In many ways it was the investigation of motion, initiated by Galileo Galilei in the late sixteenth century, and brought to a head by Isaac Newton in the seventeenth, that inaugurated the modern era of physics. Progress since that time has been so great that describing motion is now regarded as a fundamental part of science rather than one of its frontiers. Nonetheless, the description of motion played a central role in Einstein's formulation of the special theory of relativity in 1905, and it continues to provide an excellent starting point for the quantitative investigation of nature.</p><p>This OpenLearn course provides a sample of level 2 study in <span class="oucontentlinkwithtip"><a class="oucontenthyperlink" href="http://www.open.ac.uk/courses/find/science?utm_source=openlearn&utm_campaign=ou&utm_medium=ebook">Science</a></span></p>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection0
IntroductionS207_2<p>Motion is vital to life, and to science. In many ways it was the investigation of motion, initiated by Galileo Galilei in the late sixteenth century, and brought to a head by Isaac Newton in the seventeenth, that inaugurated the modern era of physics. Progress since that time has been so great that describing motion is now regarded as a fundamental part of science rather than one of its frontiers. Nonetheless, the description of motion played a central role in Einstein's formulation of the special theory of relativity in 1905, and it continues to provide an excellent starting point for the quantitative investigation of nature.</p><p>This OpenLearn course provides a sample of level 2 study in <span class="oucontentlinkwithtip"><a class="oucontenthyperlink" href="http://www.open.ac.uk/courses/find/science?utm_source=openlearn&utm_campaign=ou&utm_medium=ebook">Science</a></span></p>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

Learning outcomes
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsectionlearningoutcomes
Tue, 12 Apr 2016 23:00:00 GMT
<p>After studying this course, you should be able to:</p><ul><li><p>explain the meaning of all the newly defined (emboldened) terms introduced in this course</p></li><li><p>draw, analyse and interpret positiontime, displacementtime, velocitytime and accelerationtime graphs. Where appropriate, you should also be able to relate those graphs one to another and to the functions or equations that describe them, particularly in the case of straightline graphs</p></li><li><p>find the derivatives of simple polynomial functions, express physical rates of change as derivatives, and relate derivatives to the gradients of appropriate graphs</p></li><li><p>solve simple problems involving uniform motion and uniformly accelerated motion by using appropriate equations. You should also be able to rearrange simple equations, to change the subject of an equation, and to eliminate variables between sets of equations</p></li><li><p>describe the nature and purpose of droptowers and dropshafts, with particular reference to their role in simulating the near weightless conditions of space.</p></li></ul>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsectionlearningoutcomes
Learning outcomesS207_2<p>After studying this course, you should be able to:</p><ul><li><p>explain the meaning of all the newly defined (emboldened) terms introduced in this course</p></li><li><p>draw, analyse and interpret positiontime, displacementtime, velocitytime and accelerationtime graphs. Where appropriate, you should also be able to relate those graphs one to another and to the functions or equations that describe them, particularly in the case of straightline graphs</p></li><li><p>find the derivatives of simple polynomial functions, express physical rates of change as derivatives, and relate derivatives to the gradients of appropriate graphs</p></li><li><p>solve simple problems involving uniform motion and uniformly accelerated motion by using appropriate equations. You should also be able to rearrange simple equations, to change the subject of an equation, and to eliminate variables between sets of equations</p></li><li><p>describe the nature and purpose of droptowers and dropshafts, with particular reference to their role in simulating the near weightless conditions of space.</p></li></ul>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

1 The description of motion
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection1
Tue, 12 Apr 2016 23:00:00 GMT
<p>The concepts that have been developed to allow the description of motion  concepts such as <i>speed</i>, <i>velocity</i> and <i>acceleration</i>  are now so much a part of everyday language that we rarely think about them. Just consider the number of times each day you have to describe some aspect of motion or understand an instruction about motion; obey a speed limit or work out a journey time. We may take the description of motion for granted, but the concepts involved are so fundamental and so much depends upon them that they really deserve careful consideration. This was clearly understood by Einstein, but it was also well known long before his time.</p><p>To the ancient Greek philosopher Zeno, motion seemed such a selfcontradictory feature of the world that he and his followers became convinced that the apparent existence of motion only served to indicate the fundamental unreliability of our senses. For Zeno the description of motion was not only a fundamental problem, it was, perhaps, <i>the</i> fundamental problem.</p><div class="oucontentfigure" style="width:488px;" id="fig000_001"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/eb0f8a87/s207_2_001i.jpg" alt="" width="488" height="279" style="maxwidth:488px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 1:</b> The paradox of Achilles and the tortoise. Achilles moves faster than the tortoise, but each time he reaches the previous position of the tortoise the tortoise has moved on. The gap gets smaller and smaller but, according to Zeno, never completely vanishes. Does this really mean that Achilles can never quite catch up with the tortoise? Such a conclusion would clearly conflict with our everyday experience of how things move</span></div></div></div><p>Zeno's arguments against the reality of motion are still remembered because of three paradoxes he based upon them. The best known of these paradoxes concerns a race between the ancient hero Achilles and a tortoise (Figure 1). Naturally, Achilles can run much faster than a tortoise, so the outcome of the race seems obvious. Even if the tortoise is given a head start, Achilles will quickly overtake it and go on to win the race. However, Zeno argued, this cannot really be the case. According to Zeno, whatever head start is given to the tortoise, Achilles will take some time to reach the starting position of the tortoise and during that time the tortoise, no matter how slowly it moves, will have reached some new position, still ahead of Achilles. Now, starting from the tortoise's original position, Achilles will take a much shorter time to reach the new position of the tortoise, but by the time he does so the tortoise will again have moved on a little, so the reptile will still be ahead of the athlete. In Zeno's view this process can go on forever with the tortoise always moving on, at least a little, in the time that Achilles takes to reach its previous position. Achilles, according to Zeno, will never quite manage to close the gap. Since this conclusion disagrees with everyday experience, Zeno concluded that everyday experience was misleading. Unlike most modern scientists, Zeno preferred to trust his reason rather than his experience of the world.</p><p>Modern science is able to resolve Zeno's paradox. Motion is not in conflict with reason, but the resolution relies on mathematical concepts that were not known to the ancient Greeks, nor even to Galileo. This course deals with motion along a line and with the ways in which such motion can be represented. It will show you how <i>graphs</i> can be used to depict motion and how <i>equations</i> can provide even more powerful summaries of such graphical information. Crucially, it will also introduce you to some of the basic ideas of <i>differential calculus</i>, the branch of mathematics that concerns small changes and their influence.</p><p>Click to view part 1 of Newton's Revolution</p><div id="vid001" class="oucontentmedia oucontentaudiovideo" style="width:400px;"><div class="oucontentdefaultfilter "><span class="oumediafilter"><a href="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/0155c615/8c0a1f9c/s207_2_001v.mp4?forcedownload=1" class="oumedialinknoscript ompspacer">Download this video clip.</a><span class="accesshide">Video player: Video 1</span><a href="#" class="ompentermedia ompaccesshide" tabindex="1">
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</span></div><div class="filter_transcript" id="transcript_09c2923d1"><div><a href="#skip_transcript_09c2923d1" class="accesshide">Skip transcript: Video 1</a><h4 class="accesshide">Transcript: Video 1</h4></div><div class="filter_transcript_box" tabindex="0" id="content_transcript_09c2923d1"><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOHN</div><div class="oucontentdialogueremark">Here's an intriguing experiment. The cannon on this train is pointing vertically upwards. When a ball is fired from it, it goes straight up and straight down. But what would happen if the cannon was fired when the train was in motion? Will the ball land... near the cannon...? or behind the train...? or where? What do you think?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">Well, it lands more or less back on the cannon.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">Maybe that experiment surprised you. Many of our intuitive ideas about motion are wrong and yet they were embodied in the theories of one of the greatest thinkers in history  Aristotle. He lived in Greece in the fourth century BC. He was a pupil of Plato and tutor to Alexander the Great...</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Do you still consider the Universe illogical? Does not nature enjoy simplicity and go about its business by means of laws every bit as rigorous as those which govern the affairs of men?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">Then what law, master, decrees that a stone, hurled into the air, must always fall back down to Earth?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">An excellent question! And here we have a stone to provide you with an answer. See, it sits, on the floor, immobile. Now why is it motionless?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">Because it has nothing to make it move.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Correct. It is in its natural state, at rest. Only if I force it to move, so...</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">... is its natural resistance to movement overcome. When I cease to exert an influence, it ceases to move. In other words, for a stone to be in motion, there must be a reason, a cause  a force  to persuade it to move.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">For the stone to rise into the air, I must provide a force to lift it.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Just so.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">And yet, when I let it go, it falls, without any prompting from me.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">I applied no force to make it move downwards. Did I?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Look all around you, Alexander. Do you see any heavy object that is not resting on the ground?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">No, master.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">That is because all heavy objects naturally exist on the ground. Though you might perform a violent action in raising that stone, it will always try to return to the ground where it belongs. As soldiers, sent to battle, yearn for home, so a stone raised in the air will want to return to its proper place. That tendency, in itself, is a force of nature, whereby all heavy objects will take the most direct route to the ground, by falling vertically downwards.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">I comprehend. Master, might I ask  If I were in a chariot, drawn by four strong horses, and I raised a stone above my head, and let it fall  would it land at my feet in the chariot?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Is your chariot moving at speed?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">Oh, yes! Like Phoebus' chariot, charging through the heavens!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Well, your stone would fall vertically, taking the straightest route to the ground, while your chariot moves forward, away from the direct path of the stone. In consequence, the stone would fall behind your chariot. Why, in the time it takes for your stone to return to its natural place, your chariot would have sped on many paces! It is the same as if you let fall a stone from the mast of a ship. If the ship were moving fast over the sea, the stone would plunge into he waves behind it.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">Yees</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">There is no need for experiment, when the logic is impeccable.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">So then, master  what is there to stop the moon from falling to the ground?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">The moon?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">Yes, and the same goes for the sun and the stars, why do these heavenly bodies not fall towards us?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">I have heard it said that the Celtic tribes believe the sky will fall on their heads. Have you been speaking to these savages?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">No master, but I am unsure of how to think of these matters. There are philosophers who say that the Earth travels in a great circle, and that the ground we stand upon is moving.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Alexander, who is your tutor? Me, or those charlatans?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">You, master.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Yes, I! I, Aristotle, standing upon my solid ground! While you Alexander, standing on your moving ground, do you experience any sense of giddiness? Do you fancy the whole world is flying around in a great circle?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">Well... no.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">"Well, no"  if the whole Earth were in motion, what would happen to the birds of the air? They would get left behind as soon as they left their perches! And if I were to throw that stone into the air, with the Earth moving and me upon it, why, the stone would come down on the far side of the Parthenon! Which is why this whole notion of a moving Earth is so patently silly! I forbid you to entertain such freakish ideas! The Earth is in its proper place at the centre of the cosmos, and the ground we stand upon is still!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">I believe you... </div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">At last!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">... but, what about the moon?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Oh, the moon, the moon ... !</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">By all your laws of motion, the moon must fall to the Earth if there is no cause or force to hold it in the sky!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Is it heavy?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">The moon? I cannot say.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">I think you can. Logic, Alexander, logic! If the heavenly bodies were as heavy as Earthly matter, they would fall. Yet they do not. Conclusion? The moon is made of a substance vastly different to that of a stone. I have called this substance aether. Aether, such a good word.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">So, master, according to your laws, the natural place for such bodies is in the heavens.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Of course. The moon has no tendency to fall to the ground. The same applies to the sun and the stars, and everything is in its place.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">Then, if a man could sprout wings, and fly to the Moon, he could not stand upon it?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">I do not think that standing upon the Moon would be very good for the Moon or for the Man. Let us put these speculations aside, Alexander. You have worlds enough to conquer here on Earth, without dreaming of the Moon.</div><div class="clearer"></div></div></div><span class="accesshide" id="skip_transcript_09c2923d1">End transcript: Video 1</span></div><div class="filter_transcript_output" id="output_transcript_09c2923d1"><div class="filter_transcript_copy"><a href="#" id="action_link5be568dfab2677" class="actionicon" ><img class="icon iconsmall" alt="Copy this transcript to the clipboard" title="Copy this transcript to the clipboard" src="https://www.open.edu/openlearn/ocw/theme/image.php/_s/openlearnng/core/1541687917/t/copy" /></a></div><div class="filter_transcript_print"><a href="#" id="action_link5be568dfab2678" class="actionicon" ><img class="icon iconsmall" alt="Print this transcript" title="Print this transcript" src="https://www.open.edu/openlearn/ocw/theme/image.php/_s/openlearnng/core/1541687917/t/print" /></a></div></div><div class="oucontentfiguretext"><div class="oucontenttranscriptlink"><span class="filter_transcript_button" id="button_transcript_09c2923d1">Show transcriptHide transcript</span></div><div class="oucontentmediadownload"><a href="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/0155c615/8c0a1f9c/s207_2_001v.mp4?forcedownload=1" title="Download this video clip">Download</a></div><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption">Video 1</span></div></div></div><div class="oucontentinteractionprint"><div class="oucontentinteractionunavailable">Interactive feature not available in single page view (<a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection1#vid001">see it in standard view</a>).</div></div><p>Click to view part 2 of Newton's Revolution</p><div id="vid002" class="oucontentmedia oucontentaudiovideo" style="width:400px;"><div class="oucontentdefaultfilter "><span class="oumediafilter"><a href="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/0155c615/8093cbbc/s207_2_002v.mp4?forcedownload=1" class="oumedialinknoscript ompspacer">Download this video clip.</a><span class="accesshide">Video player: Video 2</span><a href="#" class="ompentermedia ompaccesshide" tabindex="1">
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</span></div><div class="filter_transcript" id="transcript_59cae9a12"><div><a href="#skip_transcript_59cae9a12" class="accesshide">Skip transcript: Video 2</a><h4 class="accesshide">Transcript: Video 2</h4></div><div class="filter_transcript_box" tabindex="0" id="content_transcript_59cae9a12"><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">Aristotle was convinced that the Earth is stationary. It feels stationary and things don't get left behind when they are thrown up in the air. But then, neither did the ball get left behind on the moving train. So Aristotle's ideas were not correct. Yet, for centuries his theories were generally accepted. It wasn't until the Middle Ages that people began to seriously challenge his world system.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">Galileo Galilei, in about 1600, made one of the earliest studies of the moon and the stars, using the newly invented telescope. This gave him a radical new perspective on the Universe, but it got him into terrible trouble with the authorities.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Your thoughts, Signor  are heresy!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">The truth is the truth! Is the Mother Church so fearful of science? Our Earth behaves as do all the wandering stars! I have observed that there are four moons that circle Jupiter in orbits of their own. So does our Moon circle the Earth, as the Earth describes her orbit round the Sun.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Oh, you are deranged! If the Earth were truly spinning around the Sun  surely we would feel this ceaseless motion! The very air we breathe would be snatched away from our lips! And a ball thrown upwards would vanish over the horizon!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">No, it would not! I have explained, and I have proven.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Proven, Signor?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Where do I begin? A ball, falling from a great height, will fall faster and faster till it reaches the ground.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Why faster and faster?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">There is a gravity which pulls the ball downwards. The speed of its fall increases in proportion to the duration of the fall. In my experiments I have shown that all bodies fall at the same increasing rate. That is vertical motion. But I can also make a ball move horizontally?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Sideways motion.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Sideways motion. Give me a plank of infinite length, smooth and horizontal, and could I but remove all impediments, this marble could roll along that plank at a uniform speed, forever.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Without an external force to keep it in motion? Oh come now! Nothing is eternal, but the Maker of all things.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">You have been sent to trap me.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Galileo, no.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">I have suffered enough! I have recanted what I know to be the truth! The Earth moves round the Sun!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">A dangerous contention. Senor, I am Simplicio, a simple man. Enlighten me.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Simplicio? I have stared too long at the Moon  their instruments of torture  I do not see too well...!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Signor, how can a marble roll along a plank forever? Enlighten me.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Simplicio, imagine this incline were as smooth as polished silver. I place the marble at the top, and release it. What would happen?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">It would roll downwards, faster and faster, until it reached the bottom.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Faster and faster? You learn quickly! Now, starting at the base of the board  will the marble roll upwards?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Only if it were pushed.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Then I give it a thrust, to send it rolling up the incline. Its motion?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">It would slow down constantly, given that the marble's natural tendency is to roll downwards, not upwards, which is contrary to nature.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">You declare that on a downward inclining plane, a body continuously speeds up. And on an upward incline it continuously slows down. But what would happen to the selfsame body, placed on a surface with no upward or downward tilt.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Indeed, it would go nowhere.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Not if it were propelled?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Ah, then it would travel in the direction in which it was pushed.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Would it accelerate?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">No.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Would it decelerate?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">No.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Then, at that constant speed, how far should the body continue to travel?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">For as far as the surface remained flat.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">On a limitless surface, perfectly straight and flat, on the Earth's surface the body's motion would be perpetual?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Well ... yes.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Then we agree! In the absence of friction or resistance, the motion of a body on a horizontal plane is constant. Now, when a body has both horizontal and vertical motion, it's path will be a parabola.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">But I don't see...</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">I have proved this. I propel the marble, so that it moves across the board this way, while at the same time accelerating downwards  its curve.... is a parabola. Now  take note.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">If I, at one and the same time, propel this marble horizontally and simply release this second one, they reach the bottom at the exact same instant. Thus, the two parts of the motion  the vertical and the horizontal  are independent, and distinct from one another.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">This is all very well. But how does it reflect on the question of whether the Earth moves or not, and if it moves, why the air is not snatched away, and the birds left behind?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">I have answered your question!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Signor, I am a simple man, merely helping you to write your Dialogues...</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Imagine a ship  a Man o' War ploughing through the seas. You stand in the crow's nest, and you let fall a brass ball. Where would it land?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Behind the Mast!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">No, no, no, no, no, no, no!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Where then?!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">It would fall directly beneath you! Motion in two directions, remember, the horizontal and the vertical! Whilst the ball is in your hand at the top of the mast, the ball is moving with horizontal velocity, the same speed as the ship! And that is its motion horizontally when you let it drop. Its vertical motion accelerates as it falls, but it retains its original horizontal motion, the same speed as the ship, so it lands at the foot of the mast!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">That's as much as to say that the fall of an object on a moving ship is the same as if the ship were not moving!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Simplicio, you have seen the light! Here we stand on a floor that does not appear to move, but we are on the surface of a revolving Earth. We have horizontal motion. And the air stays with us because its horizontal motion is the same as ours, and the birds of the air do not get left behind for the selfsame reason! Now, please, do you see?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Yet...</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Yet?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">... the heavenly bodies, Jupiter, and his "moons". And why does our moon not fall downwards? Hmm?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">If I understood the nature of gravity and the role of forces, I have drawn her in detail, I have measured her mountains, but what keeps her there, and the stars in the heavens, forever out of reach, I cannot say, but gravity must be the key!</div><div class="clearer"></div></div></div><span class="accesshide" id="skip_transcript_59cae9a12">End transcript: Video 2</span></div><div class="filter_transcript_output" id="output_transcript_59cae9a12"><div class="filter_transcript_copy"><a href="#" id="action_link5be568dfab2679" class="actionicon" ><img class="icon iconsmall" alt="Copy this transcript to the clipboard" title="Copy this transcript to the clipboard" src="https://www.open.edu/openlearn/ocw/theme/image.php/_s/openlearnng/core/1541687917/t/copy" /></a></div><div class="filter_transcript_print"><a href="#" id="action_link5be568dfab26710" class="actionicon" ><img class="icon iconsmall" alt="Print this transcript" title="Print this transcript" src="https://www.open.edu/openlearn/ocw/theme/image.php/_s/openlearnng/core/1541687917/t/print" /></a></div></div><div class="oucontentfiguretext"><div class="oucontenttranscriptlink"><span class="filter_transcript_button" id="button_transcript_59cae9a12">Show transcriptHide transcript</span></div><div class="oucontentmediadownload"><a href="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/0155c615/8093cbbc/s207_2_002v.mp4?forcedownload=1" title="Download this video clip">Download</a></div><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption">Video 2</span></div></div></div><div class="oucontentinteractionprint"><div class="oucontentinteractionunavailable">Interactive feature not available in single page view (<a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection1#vid002">see it in standard view</a>).</div></div><p>Click to view part 3 of Newton's Revolution</p><div id="vid003" class="oucontentmedia oucontentaudiovideo" style="width:400px;"><div class="oucontentdefaultfilter "><span class="oumediafilter"><a href="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/0155c615/e687ac5b/s207_2_003v.mp4?forcedownload=1" class="oumedialinknoscript ompspacer">Download this video clip.</a><span class="accesshide">Video player: Video 3</span><a href="#" class="ompentermedia ompaccesshide" tabindex="1">
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</span></div><div class="filter_transcript" id="transcript_68771a793"><div><a href="#skip_transcript_68771a793" class="accesshide">Skip transcript: Video 3</a><h4 class="accesshide">Transcript: Video 3</h4></div><div class="filter_transcript_box" tabindex="0" id="content_transcript_68771a793"><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOHN</div><div class="oucontentdialogueremark">So Galileo had solved the problem of the moving Earth. It's very similar towhat happened with the moving train. Galileo would have said that when the ball is fired it pursues its upward and downward motion whilst maintaining the horizontal motion that it had on the train  in fact, taking a parabolic path.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">But the story is not yet complete. Galileo has rightly separated motion into its horizontal and vertical components. But this was his difficulty. The concepts of horizontal and vertical are intimately tied up with a fact of standing on the surface of the Earth. So Galileo was unable to extend his understanding to the motion of the planets. What was needed was a synthesis to understand motion in any circumstances. The man who knitted these ideas together into a coherent theory was born in 1642, the year that Galileo died. His name was Isaac Newton.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">NEWTON</div><div class="oucontentdialogueremark">It was in the year of our Lord, 1684, that my friend Edmund Halley approached me with a question: "what would be the shape of the orbits of the planets around the Sun if the force of their attraction to the Sun diminished as the square of the distance to the planet?" I was pleased to give him the answer instantly, 'it is an elipse' I cried. I have proved it.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">It must have been twenty years since I had performed the calculations, and when I retraced my steps on paper, I saw to my satisfaction that I had already written down the Universal Laws governing the motion of all bodies. I have Halley to thank for the publication of my Principia. And I acknowledge my debt to that former giant, Galileo, upon whose shoulders I admit I stood. Of course, the Greeks came two thousand years before him, but the trouble with the likes of Aristotle was that he assumed that the natural state of motion....is rest! Whereas my first law of motion states 'that with no forces acting upon it, a body can be at rest, but it can also move in a straight line at constant speed'.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">This was Galileo's understanding of horizontal motion, but my law is universal. You see Galileo believed that an object travelling horizontally, without friction, would go on forever. But for him this entailed that the object must newtons revolution transcript remain at the surface of the Earth, where it would travel, not in a straight line, but in a great curve around the Earth's circumference.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">Effectively Galileo was saying that an object could travel thus, but with no forces acting on it. That can not be correct. According to my law, as it does not travel in a straight line, there must be a force acting upon it. And, indeed, there is a force  the force of gravity.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">We all know the legend of the apple falling on Newton's head. It was more likely that Newton was observing the moon through the branches of an apple tree, and wondering 'well, if an apple can fall to the ground, why can't the moon?'. Then he realised that the moon is falling towards the Earth. I'll go and ask him about it...</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">Sir Issac?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">NEWTON</div><div class="oucontentdialogueremark">Mmm? Oh, I had forgotten I was expecting a visitor. Do forgive me dear lady  my study is the bane of my housekeeper's existence.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">I'm sure you have other things on your mind, Sir Isaac. I was wondering if perhaps we might talk a little bit about gravity?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">NEWTON</div><div class="oucontentdialogueremark">Yes, thank you. Well, in that case, I would need to expound my laws of motion. Pray, be seated  oh, I beg your pardon  clutter, clutter, clutter!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">Motion itself does not require a force: that is the basis for my First Law, which goes on: for there to be any deviation from motion at a constant speed in a straight line, there must be a force to cause the deviation. My second Law states that the force causes the body to accelerate, but only in the direction of the force. And the acceleration is equal to the force divided by the mass. So, in the case of an object thrown horizontally, it accelerates only downwards, the direction of gravity: the horizontal motion is unchanged, as Galileo showed.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">But what about the Moon orbiting the Earth, or the planets orbiting the Sun? How do you explain those?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">NEWTON</div><div class="oucontentdialogueremark">The Moon is travelling here in its orbit round the Earth. Now, suppose that at this point there were no gravitational force acting on it  it would continue in this straight line at the same speed. But, observe, the pull of gravity tugs it round, towards the Earth and causes it to deviate, and here it does again, and again, and the motion curves into a closed orbit  an ellipse. </div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">Of course, and that shape, the ellipse, that's due to the nature of gravity  your Law of Universal Gravitation...</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">NEWTON</div><div class="oucontentdialogueremark">Indeed. My law of Universal Gravitation states simply that any two particles in the Universe attract one another with the force of gravity. This force is proportional to the product of their masses and inversely proportional to the square of the distance between them.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">Ah yes, the inverse square law. Sir Edmund Halley must have been pleased with that discovery.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">NEWTON</div><div class="oucontentdialogueremark">Halley, why Halley? Certes, it allows me to predict the motions of planets and comets, as much as the tides of the ocean, the path of a stone through the air, newtons revolution transcript the fall of an apple. It enabled me to predict the true shape of the Earth, whose constant rotation requires it to bulge slightly at the Equator, and to be flattened somewhat at the Poles. Did Halley tell us the Earth is not round after all?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">Well, it would seem then that your theory is the last word on motion in the Universe?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">NEWTON</div><div class="oucontentdialogueremark">I thought God always had the last word? Have not our laws of motion always arisen to fit the current concept of our place in God's cosmos?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">But surely, Sir Isaac, your theory is so successful  it predicts so much, and so accurately  it will be difficult for anyone to challenge it.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">NEWTON</div><div class="oucontentdialogueremark">Dear lady, I do not know what I may seem to the world. But to myself I seem to have been only a boy playing on the seashore, diverting myself now and then finding another smoother pebble or prettier shell than the ordinary, whilst the great ocean of truth lay all undiscovered before me...</div><div class="clearer"></div></div></div><span class="accesshide" id="skip_transcript_68771a793">End transcript: Video 3</span></div><div class="filter_transcript_output" id="output_transcript_68771a793"><div class="filter_transcript_copy"><a href="#" id="action_link5be568dfab26711" class="actionicon" ><img class="icon iconsmall" alt="Copy this transcript to the clipboard" title="Copy this transcript to the clipboard" src="https://www.open.edu/openlearn/ocw/theme/image.php/_s/openlearnng/core/1541687917/t/copy" /></a></div><div class="filter_transcript_print"><a href="#" id="action_link5be568dfab26712" class="actionicon" ><img class="icon iconsmall" alt="Print this transcript" title="Print this transcript" src="https://www.open.edu/openlearn/ocw/theme/image.php/_s/openlearnng/core/1541687917/t/print" /></a></div></div><div class="oucontentfiguretext"><div class="oucontenttranscriptlink"><span class="filter_transcript_button" id="button_transcript_68771a793">Show transcriptHide transcript</span></div><div class="oucontentmediadownload"><a href="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/0155c615/e687ac5b/s207_2_003v.mp4?forcedownload=1" title="Download this video clip">Download</a></div><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption">Video 3</span></div></div></div><div class="oucontentinteractionprint"><div class="oucontentinteractionunavailable">Interactive feature not available in single page view (<a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection1#vid003">see it in standard view</a>).</div></div> <script>
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https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection1
1 The description of motionS207_2<p>The concepts that have been developed to allow the description of motion  concepts such as <i>speed</i>, <i>velocity</i> and <i>acceleration</i>  are now so much a part of everyday language that we rarely think about them. Just consider the number of times each day you have to describe some aspect of motion or understand an instruction about motion; obey a speed limit or work out a journey time. We may take the description of motion for granted, but the concepts involved are so fundamental and so much depends upon them that they really deserve careful consideration. This was clearly understood by Einstein, but it was also well known long before his time.</p><p>To the ancient Greek philosopher Zeno, motion seemed such a selfcontradictory feature of the world that he and his followers became convinced that the apparent existence of motion only served to indicate the fundamental unreliability of our senses. For Zeno the description of motion was not only a fundamental problem, it was, perhaps, <i>the</i> fundamental problem.</p><div class="oucontentfigure" style="width:488px;" id="fig000_001"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/eb0f8a87/s207_2_001i.jpg" alt="" width="488" height="279" style="maxwidth:488px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 1:</b> The paradox of Achilles and the tortoise. Achilles moves faster than the tortoise, but each time he reaches the previous position of the tortoise the tortoise has moved on. The gap gets smaller and smaller but, according to Zeno, never completely vanishes. Does this really mean that Achilles can never quite catch up with the tortoise? Such a conclusion would clearly conflict with our everyday experience of how things move</span></div></div></div><p>Zeno's arguments against the reality of motion are still remembered because of three paradoxes he based upon them. The best known of these paradoxes concerns a race between the ancient hero Achilles and a tortoise (Figure 1). Naturally, Achilles can run much faster than a tortoise, so the outcome of the race seems obvious. Even if the tortoise is given a head start, Achilles will quickly overtake it and go on to win the race. However, Zeno argued, this cannot really be the case. According to Zeno, whatever head start is given to the tortoise, Achilles will take some time to reach the starting position of the tortoise and during that time the tortoise, no matter how slowly it moves, will have reached some new position, still ahead of Achilles. Now, starting from the tortoise's original position, Achilles will take a much shorter time to reach the new position of the tortoise, but by the time he does so the tortoise will again have moved on a little, so the reptile will still be ahead of the athlete. In Zeno's view this process can go on forever with the tortoise always moving on, at least a little, in the time that Achilles takes to reach its previous position. Achilles, according to Zeno, will never quite manage to close the gap. Since this conclusion disagrees with everyday experience, Zeno concluded that everyday experience was misleading. Unlike most modern scientists, Zeno preferred to trust his reason rather than his experience of the world.</p><p>Modern science is able to resolve Zeno's paradox. Motion is not in conflict with reason, but the resolution relies on mathematical concepts that were not known to the ancient Greeks, nor even to Galileo. This course deals with motion along a line and with the ways in which such motion can be represented. It will show you how <i>graphs</i> can be used to depict motion and how <i>equations</i> can provide even more powerful summaries of such graphical information. Crucially, it will also introduce you to some of the basic ideas of <i>differential calculus</i>, the branch of mathematics that concerns small changes and their influence.</p><p>Click to view part 1 of Newton's Revolution</p><div id="vid001" class="oucontentmedia oucontentaudiovideo" style="width:400px;"><div class="oucontentdefaultfilter "><span class="oumediafilter"><a href="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/0155c615/8c0a1f9c/s207_2_001v.mp4?forcedownload=1" class="oumedialinknoscript ompspacer">Download this video clip.</a><span class="accesshide">Video player: Video 1</span><a href="#" class="ompentermedia ompaccesshide" tabindex="1">
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</span></div><div class="filter_transcript" id="transcript_09c2923d1"><div><a href="#skip_transcript_09c2923d1" class="accesshide">Skip transcript: Video 1</a><h4 class="accesshide">Transcript: Video 1</h4></div><div class="filter_transcript_box" tabindex="0" id="content_transcript_09c2923d1"><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOHN</div><div class="oucontentdialogueremark">Here's an intriguing experiment. The cannon on this train is pointing vertically upwards. When a ball is fired from it, it goes straight up and straight down. But what would happen if the cannon was fired when the train was in motion? Will the ball land... near the cannon...? or behind the train...? or where? What do you think?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">Well, it lands more or less back on the cannon.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">Maybe that experiment surprised you. Many of our intuitive ideas about motion are wrong and yet they were embodied in the theories of one of the greatest thinkers in history  Aristotle. He lived in Greece in the fourth century BC. He was a pupil of Plato and tutor to Alexander the Great...</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Do you still consider the Universe illogical? Does not nature enjoy simplicity and go about its business by means of laws every bit as rigorous as those which govern the affairs of men?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">Then what law, master, decrees that a stone, hurled into the air, must always fall back down to Earth?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">An excellent question! And here we have a stone to provide you with an answer. See, it sits, on the floor, immobile. Now why is it motionless?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">Because it has nothing to make it move.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Correct. It is in its natural state, at rest. Only if I force it to move, so...</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">... is its natural resistance to movement overcome. When I cease to exert an influence, it ceases to move. In other words, for a stone to be in motion, there must be a reason, a cause  a force  to persuade it to move.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">For the stone to rise into the air, I must provide a force to lift it.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Just so.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">And yet, when I let it go, it falls, without any prompting from me.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">I applied no force to make it move downwards. Did I?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Look all around you, Alexander. Do you see any heavy object that is not resting on the ground?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">No, master.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">That is because all heavy objects naturally exist on the ground. Though you might perform a violent action in raising that stone, it will always try to return to the ground where it belongs. As soldiers, sent to battle, yearn for home, so a stone raised in the air will want to return to its proper place. That tendency, in itself, is a force of nature, whereby all heavy objects will take the most direct route to the ground, by falling vertically downwards.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">I comprehend. Master, might I ask  If I were in a chariot, drawn by four strong horses, and I raised a stone above my head, and let it fall  would it land at my feet in the chariot?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Is your chariot moving at speed?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">Oh, yes! Like Phoebus' chariot, charging through the heavens!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Well, your stone would fall vertically, taking the straightest route to the ground, while your chariot moves forward, away from the direct path of the stone. In consequence, the stone would fall behind your chariot. Why, in the time it takes for your stone to return to its natural place, your chariot would have sped on many paces! It is the same as if you let fall a stone from the mast of a ship. If the ship were moving fast over the sea, the stone would plunge into he waves behind it.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">Yees</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">There is no need for experiment, when the logic is impeccable.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">So then, master  what is there to stop the moon from falling to the ground?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">The moon?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">Yes, and the same goes for the sun and the stars, why do these heavenly bodies not fall towards us?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">I have heard it said that the Celtic tribes believe the sky will fall on their heads. Have you been speaking to these savages?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">No master, but I am unsure of how to think of these matters. There are philosophers who say that the Earth travels in a great circle, and that the ground we stand upon is moving.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Alexander, who is your tutor? Me, or those charlatans?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">You, master.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Yes, I! I, Aristotle, standing upon my solid ground! While you Alexander, standing on your moving ground, do you experience any sense of giddiness? Do you fancy the whole world is flying around in a great circle?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">Well... no.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">"Well, no"  if the whole Earth were in motion, what would happen to the birds of the air? They would get left behind as soon as they left their perches! And if I were to throw that stone into the air, with the Earth moving and me upon it, why, the stone would come down on the far side of the Parthenon! Which is why this whole notion of a moving Earth is so patently silly! I forbid you to entertain such freakish ideas! The Earth is in its proper place at the centre of the cosmos, and the ground we stand upon is still!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">I believe you... </div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">At last!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">... but, what about the moon?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Oh, the moon, the moon ... !</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">By all your laws of motion, the moon must fall to the Earth if there is no cause or force to hold it in the sky!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Is it heavy?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">The moon? I cannot say.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">I think you can. Logic, Alexander, logic! If the heavenly bodies were as heavy as Earthly matter, they would fall. Yet they do not. Conclusion? The moon is made of a substance vastly different to that of a stone. I have called this substance aether. Aether, such a good word.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">So, master, according to your laws, the natural place for such bodies is in the heavens.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">Of course. The moon has no tendency to fall to the ground. The same applies to the sun and the stars, and everything is in its place.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ALEXANDER</div><div class="oucontentdialogueremark">Then, if a man could sprout wings, and fly to the Moon, he could not stand upon it?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">ARISTOTLE</div><div class="oucontentdialogueremark">I do not think that standing upon the Moon would be very good for the Moon or for the Man. Let us put these speculations aside, Alexander. You have worlds enough to conquer here on Earth, without dreaming of the Moon.</div><div class="clearer"></div></div></div><span class="accesshide" id="skip_transcript_09c2923d1">End transcript: Video 1</span></div><div class="filter_transcript_output" id="output_transcript_09c2923d1"><div class="filter_transcript_copy"><a href="#" id="action_link5be568dfab2677" class="actionicon" ><img class="icon iconsmall" alt="Copy this transcript to the clipboard" title="Copy this transcript to the clipboard" src="https://www.open.edu/openlearn/ocw/theme/image.php/_s/openlearnng/core/1541687917/t/copy" /></a></div><div class="filter_transcript_print"><a href="#" id="action_link5be568dfab2678" class="actionicon" ><img class="icon iconsmall" alt="Print this transcript" title="Print this transcript" src="https://www.open.edu/openlearn/ocw/theme/image.php/_s/openlearnng/core/1541687917/t/print" /></a></div></div><div class="oucontentfiguretext"><div class="oucontenttranscriptlink"><span class="filter_transcript_button" id="button_transcript_09c2923d1">Show transcriptHide transcript</span></div><div class="oucontentmediadownload"><a href="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/0155c615/8c0a1f9c/s207_2_001v.mp4?forcedownload=1" title="Download this video clip">Download</a></div><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption">Video 1</span></div></div></div><div class="oucontentinteractionprint"><div class="oucontentinteractionunavailable">Interactive feature not available in single page view (<a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection1#vid001">see it in standard view</a>).</div></div><p>Click to view part 2 of Newton's Revolution</p><div id="vid002" class="oucontentmedia oucontentaudiovideo" style="width:400px;"><div class="oucontentdefaultfilter "><span class="oumediafilter"><a href="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/0155c615/8093cbbc/s207_2_002v.mp4?forcedownload=1" class="oumedialinknoscript ompspacer">Download this video clip.</a><span class="accesshide">Video player: Video 2</span><a href="#" class="ompentermedia ompaccesshide" tabindex="1">
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</span></div><div class="filter_transcript" id="transcript_59cae9a12"><div><a href="#skip_transcript_59cae9a12" class="accesshide">Skip transcript: Video 2</a><h4 class="accesshide">Transcript: Video 2</h4></div><div class="filter_transcript_box" tabindex="0" id="content_transcript_59cae9a12"><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">Aristotle was convinced that the Earth is stationary. It feels stationary and things don't get left behind when they are thrown up in the air. But then, neither did the ball get left behind on the moving train. So Aristotle's ideas were not correct. Yet, for centuries his theories were generally accepted. It wasn't until the Middle Ages that people began to seriously challenge his world system.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">Galileo Galilei, in about 1600, made one of the earliest studies of the moon and the stars, using the newly invented telescope. This gave him a radical new perspective on the Universe, but it got him into terrible trouble with the authorities.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Your thoughts, Signor  are heresy!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">The truth is the truth! Is the Mother Church so fearful of science? Our Earth behaves as do all the wandering stars! I have observed that there are four moons that circle Jupiter in orbits of their own. So does our Moon circle the Earth, as the Earth describes her orbit round the Sun.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Oh, you are deranged! If the Earth were truly spinning around the Sun  surely we would feel this ceaseless motion! The very air we breathe would be snatched away from our lips! And a ball thrown upwards would vanish over the horizon!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">No, it would not! I have explained, and I have proven.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Proven, Signor?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Where do I begin? A ball, falling from a great height, will fall faster and faster till it reaches the ground.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Why faster and faster?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">There is a gravity which pulls the ball downwards. The speed of its fall increases in proportion to the duration of the fall. In my experiments I have shown that all bodies fall at the same increasing rate. That is vertical motion. But I can also make a ball move horizontally?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Sideways motion.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Sideways motion. Give me a plank of infinite length, smooth and horizontal, and could I but remove all impediments, this marble could roll along that plank at a uniform speed, forever.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Without an external force to keep it in motion? Oh come now! Nothing is eternal, but the Maker of all things.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">You have been sent to trap me.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Galileo, no.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">I have suffered enough! I have recanted what I know to be the truth! The Earth moves round the Sun!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">A dangerous contention. Senor, I am Simplicio, a simple man. Enlighten me.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Simplicio? I have stared too long at the Moon  their instruments of torture  I do not see too well...!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Signor, how can a marble roll along a plank forever? Enlighten me.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Simplicio, imagine this incline were as smooth as polished silver. I place the marble at the top, and release it. What would happen?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">It would roll downwards, faster and faster, until it reached the bottom.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Faster and faster? You learn quickly! Now, starting at the base of the board  will the marble roll upwards?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Only if it were pushed.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Then I give it a thrust, to send it rolling up the incline. Its motion?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">It would slow down constantly, given that the marble's natural tendency is to roll downwards, not upwards, which is contrary to nature.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">You declare that on a downward inclining plane, a body continuously speeds up. And on an upward incline it continuously slows down. But what would happen to the selfsame body, placed on a surface with no upward or downward tilt.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Indeed, it would go nowhere.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Not if it were propelled?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Ah, then it would travel in the direction in which it was pushed.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Would it accelerate?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">No.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Would it decelerate?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">No.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Then, at that constant speed, how far should the body continue to travel?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">For as far as the surface remained flat.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">On a limitless surface, perfectly straight and flat, on the Earth's surface the body's motion would be perpetual?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Well ... yes.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Then we agree! In the absence of friction or resistance, the motion of a body on a horizontal plane is constant. Now, when a body has both horizontal and vertical motion, it's path will be a parabola.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">But I don't see...</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">I have proved this. I propel the marble, so that it moves across the board this way, while at the same time accelerating downwards  its curve.... is a parabola. Now  take note.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">If I, at one and the same time, propel this marble horizontally and simply release this second one, they reach the bottom at the exact same instant. Thus, the two parts of the motion  the vertical and the horizontal  are independent, and distinct from one another.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">This is all very well. But how does it reflect on the question of whether the Earth moves or not, and if it moves, why the air is not snatched away, and the birds left behind?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">I have answered your question!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Signor, I am a simple man, merely helping you to write your Dialogues...</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Imagine a ship  a Man o' War ploughing through the seas. You stand in the crow's nest, and you let fall a brass ball. Where would it land?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Behind the Mast!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">No, no, no, no, no, no, no!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Where then?!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">It would fall directly beneath you! Motion in two directions, remember, the horizontal and the vertical! Whilst the ball is in your hand at the top of the mast, the ball is moving with horizontal velocity, the same speed as the ship! And that is its motion horizontally when you let it drop. Its vertical motion accelerates as it falls, but it retains its original horizontal motion, the same speed as the ship, so it lands at the foot of the mast!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">That's as much as to say that the fall of an object on a moving ship is the same as if the ship were not moving!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Simplicio, you have seen the light! Here we stand on a floor that does not appear to move, but we are on the surface of a revolving Earth. We have horizontal motion. And the air stays with us because its horizontal motion is the same as ours, and the birds of the air do not get left behind for the selfsame reason! Now, please, do you see?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">Yet...</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">Yet?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">SIMPLICIO</div><div class="oucontentdialogueremark">... the heavenly bodies, Jupiter, and his "moons". And why does our moon not fall downwards? Hmm?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">GALILEO</div><div class="oucontentdialogueremark">If I understood the nature of gravity and the role of forces, I have drawn her in detail, I have measured her mountains, but what keeps her there, and the stars in the heavens, forever out of reach, I cannot say, but gravity must be the key!</div><div class="clearer"></div></div></div><span class="accesshide" id="skip_transcript_59cae9a12">End transcript: Video 2</span></div><div class="filter_transcript_output" id="output_transcript_59cae9a12"><div class="filter_transcript_copy"><a href="#" id="action_link5be568dfab2679" class="actionicon" ><img class="icon iconsmall" alt="Copy this transcript to the clipboard" title="Copy this transcript to the clipboard" src="https://www.open.edu/openlearn/ocw/theme/image.php/_s/openlearnng/core/1541687917/t/copy" /></a></div><div class="filter_transcript_print"><a href="#" id="action_link5be568dfab26710" class="actionicon" ><img class="icon iconsmall" alt="Print this transcript" title="Print this transcript" src="https://www.open.edu/openlearn/ocw/theme/image.php/_s/openlearnng/core/1541687917/t/print" /></a></div></div><div class="oucontentfiguretext"><div class="oucontenttranscriptlink"><span class="filter_transcript_button" id="button_transcript_59cae9a12">Show transcriptHide transcript</span></div><div class="oucontentmediadownload"><a href="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/0155c615/8093cbbc/s207_2_002v.mp4?forcedownload=1" title="Download this video clip">Download</a></div><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption">Video 2</span></div></div></div><div class="oucontentinteractionprint"><div class="oucontentinteractionunavailable">Interactive feature not available in single page view (<a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection1#vid002">see it in standard view</a>).</div></div><p>Click to view part 3 of Newton's Revolution</p><div id="vid003" class="oucontentmedia oucontentaudiovideo" style="width:400px;"><div class="oucontentdefaultfilter "><span class="oumediafilter"><a href="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/0155c615/e687ac5b/s207_2_003v.mp4?forcedownload=1" class="oumedialinknoscript ompspacer">Download this video clip.</a><span class="accesshide">Video player: Video 3</span><a href="#" class="ompentermedia ompaccesshide" tabindex="1">
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</span></div><div class="filter_transcript" id="transcript_68771a793"><div><a href="#skip_transcript_68771a793" class="accesshide">Skip transcript: Video 3</a><h4 class="accesshide">Transcript: Video 3</h4></div><div class="filter_transcript_box" tabindex="0" id="content_transcript_68771a793"><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOHN</div><div class="oucontentdialogueremark">So Galileo had solved the problem of the moving Earth. It's very similar towhat happened with the moving train. Galileo would have said that when the ball is fired it pursues its upward and downward motion whilst maintaining the horizontal motion that it had on the train  in fact, taking a parabolic path.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">But the story is not yet complete. Galileo has rightly separated motion into its horizontal and vertical components. But this was his difficulty. The concepts of horizontal and vertical are intimately tied up with a fact of standing on the surface of the Earth. So Galileo was unable to extend his understanding to the motion of the planets. What was needed was a synthesis to understand motion in any circumstances. The man who knitted these ideas together into a coherent theory was born in 1642, the year that Galileo died. His name was Isaac Newton.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">NEWTON</div><div class="oucontentdialogueremark">It was in the year of our Lord, 1684, that my friend Edmund Halley approached me with a question: "what would be the shape of the orbits of the planets around the Sun if the force of their attraction to the Sun diminished as the square of the distance to the planet?" I was pleased to give him the answer instantly, 'it is an elipse' I cried. I have proved it.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">It must have been twenty years since I had performed the calculations, and when I retraced my steps on paper, I saw to my satisfaction that I had already written down the Universal Laws governing the motion of all bodies. I have Halley to thank for the publication of my Principia. And I acknowledge my debt to that former giant, Galileo, upon whose shoulders I admit I stood. Of course, the Greeks came two thousand years before him, but the trouble with the likes of Aristotle was that he assumed that the natural state of motion....is rest! Whereas my first law of motion states 'that with no forces acting upon it, a body can be at rest, but it can also move in a straight line at constant speed'.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">This was Galileo's understanding of horizontal motion, but my law is universal. You see Galileo believed that an object travelling horizontally, without friction, would go on forever. But for him this entailed that the object must newtons revolution transcript remain at the surface of the Earth, where it would travel, not in a straight line, but in a great curve around the Earth's circumference.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">Effectively Galileo was saying that an object could travel thus, but with no forces acting on it. That can not be correct. According to my law, as it does not travel in a straight line, there must be a force acting upon it. And, indeed, there is a force  the force of gravity.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">We all know the legend of the apple falling on Newton's head. It was more likely that Newton was observing the moon through the branches of an apple tree, and wondering 'well, if an apple can fall to the ground, why can't the moon?'. Then he realised that the moon is falling towards the Earth. I'll go and ask him about it...</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">Sir Issac?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">NEWTON</div><div class="oucontentdialogueremark">Mmm? Oh, I had forgotten I was expecting a visitor. Do forgive me dear lady  my study is the bane of my housekeeper's existence.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">I'm sure you have other things on your mind, Sir Isaac. I was wondering if perhaps we might talk a little bit about gravity?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">NEWTON</div><div class="oucontentdialogueremark">Yes, thank you. Well, in that case, I would need to expound my laws of motion. Pray, be seated  oh, I beg your pardon  clutter, clutter, clutter!</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialogueremark">Motion itself does not require a force: that is the basis for my First Law, which goes on: for there to be any deviation from motion at a constant speed in a straight line, there must be a force to cause the deviation. My second Law states that the force causes the body to accelerate, but only in the direction of the force. And the acceleration is equal to the force divided by the mass. So, in the case of an object thrown horizontally, it accelerates only downwards, the direction of gravity: the horizontal motion is unchanged, as Galileo showed.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">But what about the Moon orbiting the Earth, or the planets orbiting the Sun? How do you explain those?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">NEWTON</div><div class="oucontentdialogueremark">The Moon is travelling here in its orbit round the Earth. Now, suppose that at this point there were no gravitational force acting on it  it would continue in this straight line at the same speed. But, observe, the pull of gravity tugs it round, towards the Earth and causes it to deviate, and here it does again, and again, and the motion curves into a closed orbit  an ellipse. </div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">Of course, and that shape, the ellipse, that's due to the nature of gravity  your Law of Universal Gravitation...</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">NEWTON</div><div class="oucontentdialogueremark">Indeed. My law of Universal Gravitation states simply that any two particles in the Universe attract one another with the force of gravity. This force is proportional to the product of their masses and inversely proportional to the square of the distance between them.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">Ah yes, the inverse square law. Sir Edmund Halley must have been pleased with that discovery.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">NEWTON</div><div class="oucontentdialogueremark">Halley, why Halley? Certes, it allows me to predict the motions of planets and comets, as much as the tides of the ocean, the path of a stone through the air, newtons revolution transcript the fall of an apple. It enabled me to predict the true shape of the Earth, whose constant rotation requires it to bulge slightly at the Equator, and to be flattened somewhat at the Poles. Did Halley tell us the Earth is not round after all?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">Well, it would seem then that your theory is the last word on motion in the Universe?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">NEWTON</div><div class="oucontentdialogueremark">I thought God always had the last word? Have not our laws of motion always arisen to fit the current concept of our place in God's cosmos?</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">JOY</div><div class="oucontentdialogueremark">But surely, Sir Isaac, your theory is so successful  it predicts so much, and so accurately  it will be difficult for anyone to challenge it.</div><div class="clearer"></div></div><div class="oucontentdialogueline"><div class="oucontentdialoguespeaker">NEWTON</div><div class="oucontentdialogueremark">Dear lady, I do not know what I may seem to the world. But to myself I seem to have been only a boy playing on the seashore, diverting myself now and then finding another smoother pebble or prettier shell than the ordinary, whilst the great ocean of truth lay all undiscovered before me...</div><div class="clearer"></div></div></div><span class="accesshide" id="skip_transcript_68771a793">End transcript: Video 3</span></div><div class="filter_transcript_output" id="output_transcript_68771a793"><div class="filter_transcript_copy"><a href="#" id="action_link5be568dfab26711" class="actionicon" ><img class="icon iconsmall" alt="Copy this transcript to the clipboard" title="Copy this transcript to the clipboard" src="https://www.open.edu/openlearn/ocw/theme/image.php/_s/openlearnng/core/1541687917/t/copy" /></a></div><div class="filter_transcript_print"><a href="#" id="action_link5be568dfab26712" class="actionicon" ><img class="icon iconsmall" alt="Print this transcript" title="Print this transcript" src="https://www.open.edu/openlearn/ocw/theme/image.php/_s/openlearnng/core/1541687917/t/print" /></a></div></div><div class="oucontentfiguretext"><div class="oucontenttranscriptlink"><span class="filter_transcript_button" id="button_transcript_68771a793">Show transcriptHide transcript</span></div><div class="oucontentmediadownload"><a href="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/0155c615/e687ac5b/s207_2_003v.mp4?forcedownload=1" title="Download this video clip">Download</a></div><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption">Video 3</span></div></div></div><div class="oucontentinteractionprint"><div class="oucontentinteractionunavailable">Interactive feature not available in single page view (<a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection1#vid003">see it in standard view</a>).</div></div> <script>
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</script>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

1.1 From droptowers to Oblivion  some applications of linear motion
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection1.1
Tue, 12 Apr 2016 23:00:00 GMT
<p>We have all experienced that momentary feeling of lightness when an elevator begins its downward motion. It is almost as if our weight had suddenly been reduced or, conceivably, that the pull of the Earth's gravity had decreased for a moment. But imagine what it would be like if the lift cable had suddenly snapped and the lift, with you in it, had plummeted downward. Apart from stark terror, what else do you think you would experience during your fall? What would the <i>physical experience</i> of such a disaster be like?</p><p>Well, it would be just like jumping from a high tower. If your descent was unimpeded by the resistance of the air, almost all sense of weight would vanish while you were falling. You would feel weightless, just as though you were an astronaut in outer space.</p><p>Not surprisingly, scientists who want to know how equipment will behave under the conditions found in spacecraft are keen to simulate the same conditions here on Earth. One way in which they can do this is by dropping their equipment from the top of a tower, or down a vertical shaft. There are a number of research centres around the world where drop facilities of this kind are available. These are specialised facilities where steps are taken to avoid or overcome the effects of air resistance: simply dropping an object in the Earth's atmosphere is not a satisfactory way of simulating the environment of outer space.</p><p>Figure 2 shows the 140 m droptower in Bremen, Germany. The tower is airtight, so all the air can be pumped out. Equipment under test is placed inside a specially constructed dropvehicle and monitored by closedcircuit TV as it falls from the top to the bottom of the tower. About five seconds of free fall can be achieved in this way. During those few seconds, within the falling dropvehicle, the effects of gravity are reduced to a tiny fraction of their usual value, a condition known as 'microgravity'.</p><div class="oucontentfigure oucontentmediamini" id="fig001_0018"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/8ccdfb2a/s207_2_002i.jpg" alt="" width="261" height="269" style="maxwidth:261px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 2:</b> The 140 m droptower in Bremen, Germany</span></div></div></div><p>In the USA, at the Lewis Research Center in Ohio, NASA operates a 143 m dropshaft, as part of its Zero Gravity Research Facility. Microgravity investigations conducted at the research facility have concerned the spread of fire, the flow of liquids, and the feasibility of spacebased industrial processes that would be impossible under normal terrestrial conditions. Figure 3 shows the facility's bulletshaped dropvehicle being given a soft landing at the end of a drop, to avoid destroying the expensive equipment that it contains.</p><p>At the time of writing, the world's longest dropshaft is in Japan. The Japan Microgravity Center (JAMIC) has a 700 m drop housed in a disused mine shaft. It would be impossible to evacuate the air from such a big shaft, so in this case the rocketshaped test capsule is propelled down the shaft by gasjets with a thrust that is designed to compensate for air resistance. Inside this capsule, there is a second capsule and the space between the capsules is a vacuum. The experiments are carried out in the inner capsule which, to a very good approximation, is in free fall. The two capsules decelerate during the final 200 m of the fall.</p><p>By the time you finish this course you should be able to work out the duration of the fall in the JAMIC facility, and the highest speed attained by the capsule. You should also be able to work out the length of shaft that would be required to produce any given duration of microgravity.</p><div class="oucontentfigure oucontentmediamini" id="fig001_002"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/4b1a2986/s207_2_003i.jpg" alt="" width="300" height="445" style="maxwidth:300px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 3:</b> The linear motion of a falling test vehicle is stopped safely at the NASA Lewis Zero Gravity Research Facility in Ohio, USA</span></div></div></div><p>If all this sounds a bit esoteric you might prefer to consider a different kind of dropfacility. Figure 4 shows <i>Oblivion</i>, a ride at the Alton Towers Adventure Park, UK. <i>Oblivion</i> is described as 'the world's first verticaldrop rollercoaster'. It will not simulate the space environment, but it will produce a few seconds of terror from a simple application of linear motion.</p><div class="oucontentfigure oucontentmediamini" id="fig001_003"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/f19ec698/s207_2_004i.jpg" alt="" width="261" height="415" style="maxwidth:261px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 4:</b> <i>Oblivion</i>, the verticaldrop rollercoaster at the Alton Towers Adventure Park, UK</span></div></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection1.1
1.1 From droptowers to Oblivion  some applications of linear motionS207_2<p>We have all experienced that momentary feeling of lightness when an elevator begins its downward motion. It is almost as if our weight had suddenly been reduced or, conceivably, that the pull of the Earth's gravity had decreased for a moment. But imagine what it would be like if the lift cable had suddenly snapped and the lift, with you in it, had plummeted downward. Apart from stark terror, what else do you think you would experience during your fall? What would the <i>physical experience</i> of such a disaster be like?</p><p>Well, it would be just like jumping from a high tower. If your descent was unimpeded by the resistance of the air, almost all sense of weight would vanish while you were falling. You would feel weightless, just as though you were an astronaut in outer space.</p><p>Not surprisingly, scientists who want to know how equipment will behave under the conditions found in spacecraft are keen to simulate the same conditions here on Earth. One way in which they can do this is by dropping their equipment from the top of a tower, or down a vertical shaft. There are a number of research centres around the world where drop facilities of this kind are available. These are specialised facilities where steps are taken to avoid or overcome the effects of air resistance: simply dropping an object in the Earth's atmosphere is not a satisfactory way of simulating the environment of outer space.</p><p>Figure 2 shows the 140 m droptower in Bremen, Germany. The tower is airtight, so all the air can be pumped out. Equipment under test is placed inside a specially constructed dropvehicle and monitored by closedcircuit TV as it falls from the top to the bottom of the tower. About five seconds of free fall can be achieved in this way. During those few seconds, within the falling dropvehicle, the effects of gravity are reduced to a tiny fraction of their usual value, a condition known as 'microgravity'.</p><div class="oucontentfigure oucontentmediamini" id="fig001_0018"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/8ccdfb2a/s207_2_002i.jpg" alt="" width="261" height="269" style="maxwidth:261px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 2:</b> The 140 m droptower in Bremen, Germany</span></div></div></div><p>In the USA, at the Lewis Research Center in Ohio, NASA operates a 143 m dropshaft, as part of its Zero Gravity Research Facility. Microgravity investigations conducted at the research facility have concerned the spread of fire, the flow of liquids, and the feasibility of spacebased industrial processes that would be impossible under normal terrestrial conditions. Figure 3 shows the facility's bulletshaped dropvehicle being given a soft landing at the end of a drop, to avoid destroying the expensive equipment that it contains.</p><p>At the time of writing, the world's longest dropshaft is in Japan. The Japan Microgravity Center (JAMIC) has a 700 m drop housed in a disused mine shaft. It would be impossible to evacuate the air from such a big shaft, so in this case the rocketshaped test capsule is propelled down the shaft by gasjets with a thrust that is designed to compensate for air resistance. Inside this capsule, there is a second capsule and the space between the capsules is a vacuum. The experiments are carried out in the inner capsule which, to a very good approximation, is in free fall. The two capsules decelerate during the final 200 m of the fall.</p><p>By the time you finish this course you should be able to work out the duration of the fall in the JAMIC facility, and the highest speed attained by the capsule. You should also be able to work out the length of shaft that would be required to produce any given duration of microgravity.</p><div class="oucontentfigure oucontentmediamini" id="fig001_002"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/4b1a2986/s207_2_003i.jpg" alt="" width="300" height="445" style="maxwidth:300px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 3:</b> The linear motion of a falling test vehicle is stopped safely at the NASA Lewis Zero Gravity Research Facility in Ohio, USA</span></div></div></div><p>If all this sounds a bit esoteric you might prefer to consider a different kind of dropfacility. Figure 4 shows <i>Oblivion</i>, a ride at the Alton Towers Adventure Park, UK. <i>Oblivion</i> is described as 'the world's first verticaldrop rollercoaster'. It will not simulate the space environment, but it will produce a few seconds of terror from a simple application of linear motion.</p><div class="oucontentfigure oucontentmediamini" id="fig001_003"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/f19ec698/s207_2_004i.jpg" alt="" width="261" height="415" style="maxwidth:261px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 4:</b> <i>Oblivion</i>, the verticaldrop rollercoaster at the Alton Towers Adventure Park, UK</span></div></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

2.1 Simplification and modelling
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection2.1
Tue, 12 Apr 2016 23:00:00 GMT
<p>Everyday experience teaches us that unconfined objects are free to move in three independent directions. I can move my hand up or down, left or right, backwards or forwards. By combining movements in these three directions I can, at least in principle, move my hand to any point in space. The fact that there are just three independent directions, and that these suffice to reach any point, shows that the space in which my hand moves is <b>threedimensional</b>.</p><p>The motion of a large object, such as an aeroplane, moving in threedimensional space is very difficult to describe exactly. The aeroplane may flex, rotate and vibrate as it moves, and there may be complicated changes taking place within it. To avoid such complexities at the start of our investigation of motion we shall initially restrict our attention to objects that move in just one dimension along a line.</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox oucontentsnoheading " id="box001_002"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>We shall treat the object concerned as a <b>particle</b>, that is, a pointlike concentration of matter that has no size, no shape and no internal structure.</p></div></div></div><p>Treating a real object, such as an aeroplane, as though it is a particle is clearly a simplification. Real objects certainly do have size, shape and internal structure, but such details can often be neglected in specific contexts. Making simplifications of this kind is an important part of the skill of scientific modelling in physics. A good <b>model</b> uses the welldefined concepts of physics to represent the essential features of a problem while omitting the irrelevant details. The trick is not to oversimplify. The model should be as simple as it can be, but no simpler. Just what this entails will depend on the problem being analysed. For example, the use of the year as a unit of time is a result of the orbital motion of the Earth around the Sun. This orbital motion is described quite easily while treating the Earth as a particle. The Earth's diameter is about 10 000 times smaller than the distance between the Earth and the Sun, so a particle model is a very good approximation in this case. However, a particle model of the Earth cannot account for the distinction between day and night since that depends on the rotation of the Earth.</p><p>In this course we shall only consider problems that can be adequately modelled by particles moving in one dimension, that is, along a straight line.</p><p>Describing the motion of a particle moving along a line may sound like a fairly simple undertaking, but, as you will see, it will present plenty of challenges and will allow us to gain significant insights into the operation of systems such as zero gravity droptowers and verticaldrop rollercoasters.</p><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_001"><div class="oucontentouterbox"><h3 class="oucontenth3 oucontentheading oucontentnonumber">Question 1</h3><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>List some more examples of real motions that might, in your opinion, be reasonably well modelled by particles moving along a line.</p></div>
<div class="oucontentsaqanswer"><h4 class="oucontenth4">Answer</h4><p>Your list might well include items such as: the motion of a passenger on a train, or in a plane or in any other vehicle, as long as it is the passenger's overall position that is important, and not their posture or internal movement. You might also have listed the vehicles themselves, provided the same conditions apply. Indeed, you might list almost anything, including the Earth or the Sun, provided you are considering a context in which the moving object can be treated as pointlike.</p></div></div></div></div><p>The branch of physics that is concerned with the description of motion is known as <b>kinematics</b>. Kinematics is not concerned with <i>forces</i>, nor with the causes of motion; those topics are central to the study of <i>dynamics</i>. Typical questions that we might ask about the kinematics of a particle are:</p><ul class="oucontentbulleted"><li><p>Where is the particle?</p></li><li><p>How fast is it moving, and in what direction?</p></li><li><p>How rapidly is it speeding up or slowing down?</p></li></ul><p>How such questions are to be answered is the main concern of the rest of this course.</p>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection2.1
2.1 Simplification and modellingS207_2<p>Everyday experience teaches us that unconfined objects are free to move in three independent directions. I can move my hand up or down, left or right, backwards or forwards. By combining movements in these three directions I can, at least in principle, move my hand to any point in space. The fact that there are just three independent directions, and that these suffice to reach any point, shows that the space in which my hand moves is <b>threedimensional</b>.</p><p>The motion of a large object, such as an aeroplane, moving in threedimensional space is very difficult to describe exactly. The aeroplane may flex, rotate and vibrate as it moves, and there may be complicated changes taking place within it. To avoid such complexities at the start of our investigation of motion we shall initially restrict our attention to objects that move in just one dimension along a line.</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox
oucontentsnoheading
" id="box001_002"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>We shall treat the object concerned as a <b>particle</b>, that is, a pointlike concentration of matter that has no size, no shape and no internal structure.</p></div></div></div><p>Treating a real object, such as an aeroplane, as though it is a particle is clearly a simplification. Real objects certainly do have size, shape and internal structure, but such details can often be neglected in specific contexts. Making simplifications of this kind is an important part of the skill of scientific modelling in physics. A good <b>model</b> uses the welldefined concepts of physics to represent the essential features of a problem while omitting the irrelevant details. The trick is not to oversimplify. The model should be as simple as it can be, but no simpler. Just what this entails will depend on the problem being analysed. For example, the use of the year as a unit of time is a result of the orbital motion of the Earth around the Sun. This orbital motion is described quite easily while treating the Earth as a particle. The Earth's diameter is about 10 000 times smaller than the distance between the Earth and the Sun, so a particle model is a very good approximation in this case. However, a particle model of the Earth cannot account for the distinction between day and night since that depends on the rotation of the Earth.</p><p>In this course we shall only consider problems that can be adequately modelled by particles moving in one dimension, that is, along a straight line.</p><p>Describing the motion of a particle moving along a line may sound like a fairly simple undertaking, but, as you will see, it will present plenty of challenges and will allow us to gain significant insights into the operation of systems such as zero gravity droptowers and verticaldrop rollercoasters.</p><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_001"><div class="oucontentouterbox"><h3 class="oucontenth3 oucontentheading oucontentnonumber">Question 1</h3><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>List some more examples of real motions that might, in your opinion, be reasonably well modelled by particles moving along a line.</p></div>
<div class="oucontentsaqanswer"><h4 class="oucontenth4">Answer</h4><p>Your list might well include items such as: the motion of a passenger on a train, or in a plane or in any other vehicle, as long as it is the passenger's overall position that is important, and not their posture or internal movement. You might also have listed the vehicles themselves, provided the same conditions apply. Indeed, you might list almost anything, including the Earth or the Sun, provided you are considering a context in which the moving object can be treated as pointlike.</p></div></div></div></div><p>The branch of physics that is concerned with the description of motion is known as <b>kinematics</b>. Kinematics is not concerned with <i>forces</i>, nor with the causes of motion; those topics are central to the study of <i>dynamics</i>. Typical questions that we might ask about the kinematics of a particle are:</p><ul class="oucontentbulleted"><li><p>Where is the particle?</p></li><li><p>How fast is it moving, and in what direction?</p></li><li><p>How rapidly is it speeding up or slowing down?</p></li></ul><p>How such questions are to be answered is the main concern of the rest of this course.</p>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

2.2 Describing positions along a line
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection2.2
Tue, 12 Apr 2016 23:00:00 GMT
<p>To take a definite case, consider a car moving along a straight horizontal road. The car can be modelled as a particle by supposing the particle to be located at, say, the midpoint of the car. It is clearly convenient to measure the progress of the car with respect to the road, and for this purpose you might use the set of uniformly spaced redtopped posts along the righthand side of the road (see Figure 5). The posts provide a way of assigning a unique <b>position coordinate</b> to the car (regarded as a particle) at any instant. The instantaneous position coordinate of the car is simply the number of the nearest post at that moment.</p><div class="oucontentfigure" style="width:462px;" id="fig001_004"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/24aec4da/s207_2_005i.jpg" alt="" width="462" height="305" style="maxwidth:462px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 5:</b> A long straight road and a set of uniformly spaced posts along it</span></div></div></div><p>When used in this way the posts provide the basis of a onedimensional <b>coordinate system</b>  a systematic means of assigning position coordinates along a line. Taken together, the posts constitute an <b>axis</b> of the system; a straight line along which distances can be measured. One point on the axis must be chosen as the <b>origin</b> and assigned the value 0. Points on one side of the origin can then be labelled by their distance from the origin (10 m or 20 m say), while points on the other side are labelled by <i>minus</i> their distance from the origin (−10 m or −20 m for example). Conventionally, we might represent any of these values by the algebraic symbol <i>x</i>, in which case it would be called the <i>x</i>coordinate of the corresponding point on the axis, and the axis itself would be called the <i>x</i>axis.</p><p>Figure 6 shows an example, corresponding to a particular choice of origin, and with <i>x</i> increasing smoothly from left to right, as indicated by the arrow. Note that the numbers <i>x</i> increase from left to right everywhere on the <i>x</i>axis, not only on the positive segment. Thus for example −﻿10 is larger than −﻿20, and so if you subtract −﻿20 from −﻿10 you get −﻿10 − (﻿−﻿20﻿) = 10, a positive number. At the time illustrated the car is at <i>x</i> = 30 m, and a pedestrian (also modelled as a particle) is at <i>x</i> = −20 m. Notice that it is essential to include the units in the specification of <i>x</i>. It makes no sense to refer to the position of the car as being '30', with no mention of the units, i.e. metres. In one dimension, the specification of a physical <b>position</b> <i>x</i> consists of a positive or negative number multiplied by an appropriate unit, in this case the metre.</p><div class="oucontentfigure" style="width:511px;" id="fig001_005"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/fb0476ab/s207_2_006i.jpg" alt="" width="511" height="106" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 6:</b> A choice of <i>x</i>axis. Setting up the axis involves choosing an origin and a direction of increasing <i>x</i> (from left to right). Distances are measured in metres (m)</span></div></div></div><p>It is worth emphasising that setting up an <i>x</i>axis involves some degree of choice. The origin and the direction of increasing <i>x</i> are both chosen in an arbitrary way, usually so as to simplify the problem. The choice of units of measurement is also arbitrary, though usually guided by convention. Even the decision to call the axis an <i>x</i>axis is arbitrary; <i>x</i> is conventional, but not compulsory. It does not really matter <i>what</i> choices are made, but it is essential to stick to the <i>same</i> choices throughout the description of a given motion.</p><p>The conventions about units deserve special attention, so they have been set apart from the main text in Box 1.</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox " id="box001_003"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Box 1 Introducing SI units</h2><div class="oucontentinnerbox"><p>The standard way for scientists to measure lengths, distances and positions is in metric units (metres, millimetres, kilometres, etc., as opposed to miles, feet or inches). This is part of an internationally agreed system of units known as <b>SI</b> (which stands for the <i>Système Internationale</i>). This course uses SI units throughout.</p><p>The standard SI abbreviation for the <b>metre</b> is m. Since 1983 the speed of light travelling in a vacuum has been defined to be exactly</p><p>2.99792458 × 10<sup>8</sup> metres per second.</p><p>This is an internationally agreed <i>definition</i>, not the result of a measurement, so the metre may be similarly defined as the distance that light travels through a vacuum in (1 / 299 792 458) second. Typical lengths of interest to physicists range from the diameter of the atomic nucleus, which is about 10<sup>−15</sup> m, to the diameter of the visible Universe, which is about 10<sup>27</sup> m.</p><p>In view of the wide range of lengths that are of interest, it would be inconvenient to use only the metre for their measurement. To avoid this there are standard <b>SI multiples</b> and <b>SI submultiples</b> that may also be used. You will already be familiar with some of these; the prefix <b>kilo</b> means 10<sup>3</sup> as in kilometre and the prefix <b>milli</b> stands for 10<sup>−3</sup> as in millimetre. Table 1 gives the standard SI multiples; you are not expected to remember all of them but it is certainly worth learning the more common ones.</p><p>Another example of an SI unit is the <b>second</b> which is the unit of time. The standard abbreviation for the second is s. (Notice that, like m, <i>this</i> symbol is always lower case. The distinction <i>does</i> matter.) The time values of interest to experimental physicists range from the 10<sup>−24</sup> s duration of certain events in subatomic physics, to the present age of the Universe, which is about 10<sup>18</sup> s, though speculations about the birth and death of the Universe have involved times ranging from 10<sup>−45</sup> s to 10<sup>140</sup> s, or more.</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_001"><h3 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 1:</b> Standard SI multiples and submultiples</h3><div class="oucontenttablewrapper"><table><tr><th scope="col">Multiple</th><th scope="col">Prefix</th><th scope="col">Symbol for prefix</th><th scope="col">Submultiple</th><th scope="col">Prefix</th><th scope="col">Symbol for prefix</th></tr><tr><td>10<sup>12</sup></td><td>tera</td><td>T</td><td>10<sup>−3</sup></td><td>milli</td><td>m</td></tr><tr><td>10<sup>9</sup></td><td>giga</td><td>G</td><td>10<sup>−6</sup></td><td>micro</td><td>μ</td></tr><tr><td>10<sup>6</sup></td><td>mega</td><td>M</td><td>10<sup>−9</sup></td><td>nano</td><td>n</td></tr><tr><td>10<sup>3</sup></td><td>kilo</td><td>k</td><td>10<sup>−12</sup></td><td>pico</td><td>p</td></tr><tr><td>10<sup>0</sup></td><td/><td/><td>10<sup>−15</sup></td><td>femto</td><td>f</td></tr></table></div><div class="oucontentsourcereference"></div></div></div></div></div><p>Having chosen a convenient <i>x</i>axis and selected units of length and time, we are well placed to describe the position of a particle as it moves along a straight line. All we need to do is to equip ourselves with a clock, choose an origin of time (that is, an instant at which the time <i>t</i> = 0 s) and note the position of the particle at a series of closely spaced intervals. Table 2 shows typical results for the kind of car in Figure 6. The use of an oblique slash (/) or <i>solidus</i> in the column headings in Table 2 is another convention. It reminds us that the quantity to the left of the solidus is being measured in the units listed on the right of the solidus.</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_002"><h2 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 2:</b> The position coordinate <i>x</i> of the car in Figure 6 at various times <i>t</i></h2><div class="oucontenttablewrapper"><table><tr><th scope="col"><i>t</i>/s</th><th scope="col"><i>x</i>/m</th></tr><tr><td>0</td><td class="oucontenttablemiddle ">0</td></tr><tr><td>5</td><td class="oucontenttablemiddle ">1.7</td></tr><tr><td>10</td><td class="oucontenttablemiddle ">6.8</td></tr><tr><td>15</td><td class="oucontenttablemiddle ">15</td></tr><tr><td>20</td><td class="oucontenttablemiddle ">26</td></tr><tr><td>25</td><td class="oucontenttablemiddle ">39</td></tr><tr><td>30</td><td class="oucontenttablemiddle ">53</td></tr><tr><td>35</td><td class="oucontenttablemiddle ">68</td></tr><tr><td>40</td><td class="oucontenttablemiddle ">84</td></tr><tr><td>45</td><td class="oucontenttablemiddle ">99</td></tr><tr><td>50</td><td class="oucontenttablemiddle ">115</td></tr><tr><td>55</td><td class="oucontenttablemiddle ">131</td></tr><tr><td>60</td><td class="oucontenttablemiddle ">146</td></tr></table></div><div class="oucontentsourcereference"></div></div><p>In fact, if you remember that a physical quantity such as <i>x</i> represents the <i>product</i> of a number and a unit of measurement, as in 20 m, for example, you can see that <i>x</i>/m may be thought of as indicating 'the value of <i>x</i> divided by 1 metre', which would just be the number 20 if <i>x</i> was 20 m. It therefore makes good sense to see that the entries in the <i>x</i>/m column of Table 2 are indeed just numbers. In effect, <i>x</i>/m indicates that the units have been divided out. Take care always to remember that symbols such as <i>x</i>, that are used to represent physical quantities, conventionally include the relevant units while quantities such as <i>x</i>/m are purely numerical.</p>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection2.2
2.2 Describing positions along a lineS207_2<p>To take a definite case, consider a car moving along a straight horizontal road. The car can be modelled as a particle by supposing the particle to be located at, say, the midpoint of the car. It is clearly convenient to measure the progress of the car with respect to the road, and for this purpose you might use the set of uniformly spaced redtopped posts along the righthand side of the road (see Figure 5). The posts provide a way of assigning a unique <b>position coordinate</b> to the car (regarded as a particle) at any instant. The instantaneous position coordinate of the car is simply the number of the nearest post at that moment.</p><div class="oucontentfigure" style="width:462px;" id="fig001_004"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/24aec4da/s207_2_005i.jpg" alt="" width="462" height="305" style="maxwidth:462px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 5:</b> A long straight road and a set of uniformly spaced posts along it</span></div></div></div><p>When used in this way the posts provide the basis of a onedimensional <b>coordinate system</b>  a systematic means of assigning position coordinates along a line. Taken together, the posts constitute an <b>axis</b> of the system; a straight line along which distances can be measured. One point on the axis must be chosen as the <b>origin</b> and assigned the value 0. Points on one side of the origin can then be labelled by their distance from the origin (10 m or 20 m say), while points on the other side are labelled by <i>minus</i> their distance from the origin (−10 m or −20 m for example). Conventionally, we might represent any of these values by the algebraic symbol <i>x</i>, in which case it would be called the <i>x</i>coordinate of the corresponding point on the axis, and the axis itself would be called the <i>x</i>axis.</p><p>Figure 6 shows an example, corresponding to a particular choice of origin, and with <i>x</i> increasing smoothly from left to right, as indicated by the arrow. Note that the numbers <i>x</i> increase from left to right everywhere on the <i>x</i>axis, not only on the positive segment. Thus for example −10 is larger than −20, and so if you subtract −20 from −10 you get −10 − (−20) = 10, a positive number. At the time illustrated the car is at <i>x</i> = 30 m, and a pedestrian (also modelled as a particle) is at <i>x</i> = −20 m. Notice that it is essential to include the units in the specification of <i>x</i>. It makes no sense to refer to the position of the car as being '30', with no mention of the units, i.e. metres. In one dimension, the specification of a physical <b>position</b> <i>x</i> consists of a positive or negative number multiplied by an appropriate unit, in this case the metre.</p><div class="oucontentfigure" style="width:511px;" id="fig001_005"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/fb0476ab/s207_2_006i.jpg" alt="" width="511" height="106" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 6:</b> A choice of <i>x</i>axis. Setting up the axis involves choosing an origin and a direction of increasing <i>x</i> (from left to right). Distances are measured in metres (m)</span></div></div></div><p>It is worth emphasising that setting up an <i>x</i>axis involves some degree of choice. The origin and the direction of increasing <i>x</i> are both chosen in an arbitrary way, usually so as to simplify the problem. The choice of units of measurement is also arbitrary, though usually guided by convention. Even the decision to call the axis an <i>x</i>axis is arbitrary; <i>x</i> is conventional, but not compulsory. It does not really matter <i>what</i> choices are made, but it is essential to stick to the <i>same</i> choices throughout the description of a given motion.</p><p>The conventions about units deserve special attention, so they have been set apart from the main text in Box 1.</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox " id="box001_003"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Box 1 Introducing SI units</h2><div class="oucontentinnerbox"><p>The standard way for scientists to measure lengths, distances and positions is in metric units (metres, millimetres, kilometres, etc., as opposed to miles, feet or inches). This is part of an internationally agreed system of units known as <b>SI</b> (which stands for the <i>Système Internationale</i>). This course uses SI units throughout.</p><p>The standard SI abbreviation for the <b>metre</b> is m. Since 1983 the speed of light travelling in a vacuum has been defined to be exactly</p><p>2.99792458 × 10<sup>8</sup> metres per second.</p><p>This is an internationally agreed <i>definition</i>, not the result of a measurement, so the metre may be similarly defined as the distance that light travels through a vacuum in (1 / 299 792 458) second. Typical lengths of interest to physicists range from the diameter of the atomic nucleus, which is about 10<sup>−15</sup> m, to the diameter of the visible Universe, which is about 10<sup>27</sup> m.</p><p>In view of the wide range of lengths that are of interest, it would be inconvenient to use only the metre for their measurement. To avoid this there are standard <b>SI multiples</b> and <b>SI submultiples</b> that may also be used. You will already be familiar with some of these; the prefix <b>kilo</b> means 10<sup>3</sup> as in kilometre and the prefix <b>milli</b> stands for 10<sup>−3</sup> as in millimetre. Table 1 gives the standard SI multiples; you are not expected to remember all of them but it is certainly worth learning the more common ones.</p><p>Another example of an SI unit is the <b>second</b> which is the unit of time. The standard abbreviation for the second is s. (Notice that, like m, <i>this</i> symbol is always lower case. The distinction <i>does</i> matter.) The time values of interest to experimental physicists range from the 10<sup>−24</sup> s duration of certain events in subatomic physics, to the present age of the Universe, which is about 10<sup>18</sup> s, though speculations about the birth and death of the Universe have involved times ranging from 10<sup>−45</sup> s to 10<sup>140</sup> s, or more.</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_001"><h3 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 1:</b> Standard SI multiples and submultiples</h3><div class="oucontenttablewrapper"><table><tr><th scope="col">Multiple</th><th scope="col">Prefix</th><th scope="col">Symbol for prefix</th><th scope="col">Submultiple</th><th scope="col">Prefix</th><th scope="col">Symbol for prefix</th></tr><tr><td>10<sup>12</sup></td><td>tera</td><td>T</td><td>10<sup>−3</sup></td><td>milli</td><td>m</td></tr><tr><td>10<sup>9</sup></td><td>giga</td><td>G</td><td>10<sup>−6</sup></td><td>micro</td><td>μ</td></tr><tr><td>10<sup>6</sup></td><td>mega</td><td>M</td><td>10<sup>−9</sup></td><td>nano</td><td>n</td></tr><tr><td>10<sup>3</sup></td><td>kilo</td><td>k</td><td>10<sup>−12</sup></td><td>pico</td><td>p</td></tr><tr><td>10<sup>0</sup></td><td/><td/><td>10<sup>−15</sup></td><td>femto</td><td>f</td></tr></table></div><div class="oucontentsourcereference"></div></div></div></div></div><p>Having chosen a convenient <i>x</i>axis and selected units of length and time, we are well placed to describe the position of a particle as it moves along a straight line. All we need to do is to equip ourselves with a clock, choose an origin of time (that is, an instant at which the time <i>t</i> = 0 s) and note the position of the particle at a series of closely spaced intervals. Table 2 shows typical results for the kind of car in Figure 6. The use of an oblique slash (/) or <i>solidus</i> in the column headings in Table 2 is another convention. It reminds us that the quantity to the left of the solidus is being measured in the units listed on the right of the solidus.</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_002"><h2 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 2:</b> The position coordinate <i>x</i> of the car in Figure 6 at various times <i>t</i></h2><div class="oucontenttablewrapper"><table><tr><th scope="col"><i>t</i>/s</th><th scope="col"><i>x</i>/m</th></tr><tr><td>0</td><td class="oucontenttablemiddle ">0</td></tr><tr><td>5</td><td class="oucontenttablemiddle ">1.7</td></tr><tr><td>10</td><td class="oucontenttablemiddle ">6.8</td></tr><tr><td>15</td><td class="oucontenttablemiddle ">15</td></tr><tr><td>20</td><td class="oucontenttablemiddle ">26</td></tr><tr><td>25</td><td class="oucontenttablemiddle ">39</td></tr><tr><td>30</td><td class="oucontenttablemiddle ">53</td></tr><tr><td>35</td><td class="oucontenttablemiddle ">68</td></tr><tr><td>40</td><td class="oucontenttablemiddle ">84</td></tr><tr><td>45</td><td class="oucontenttablemiddle ">99</td></tr><tr><td>50</td><td class="oucontenttablemiddle ">115</td></tr><tr><td>55</td><td class="oucontenttablemiddle ">131</td></tr><tr><td>60</td><td class="oucontenttablemiddle ">146</td></tr></table></div><div class="oucontentsourcereference"></div></div><p>In fact, if you remember that a physical quantity such as <i>x</i> represents the <i>product</i> of a number and a unit of measurement, as in 20 m, for example, you can see that <i>x</i>/m may be thought of as indicating 'the value of <i>x</i> divided by 1 metre', which would just be the number 20 if <i>x</i> was 20 m. It therefore makes good sense to see that the entries in the <i>x</i>/m column of Table 2 are indeed just numbers. In effect, <i>x</i>/m indicates that the units have been divided out. Take care always to remember that symbols such as <i>x</i>, that are used to represent physical quantities, conventionally include the relevant units while quantities such as <i>x</i>/m are purely numerical.</p>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

2.3 Positiontime graphs
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection2.3
Tue, 12 Apr 2016 23:00:00 GMT
<p>Tables do not give a very striking impression of how one thing varies with respect to another. A visual form of presentation, such as a graph, is usually much more effective. This is evident from Figure 7, which shows the graph obtained by plotting the data in Table 2 and then drawing a smooth curve through the resulting points.</p><div class="oucontentfigure" style="width:511px;" id="fig001_006"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/f0990c86/s207_2_007i.jpg" alt="" width="511" height="502" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 7:</b> A positiontime graph based on the results in Table 2</span></div></div></div><p>The smooth curve drawn in Figure 7 is called the <b>positiontime graph</b> of the car's motion. It can be used to read off the position of the car at any instant of time, or to find when the car passes a certain point. In graphs such as this, it is conventional to plot the time <i>t</i> along the horizontal axis. The vertical axis is used for the position <i>x</i>. This is just a standard way of displaying information about quantities that depend on time; it does <i>not</i> imply that the <i>x</i>axis of Figure 6 is vertical!</p><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_002"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 2</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>(a) Use Figure 7 to estimate the position of the car at <i>t</i> = 32 s.</p><p>(b) Estimate the time at which the car reaches the position shown in Figure 6.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) From Figure 7, the position at <i>t</i> = 32 s is 59 m.</p><p>(b) In Figure 6 the car is at <i>x</i> = 30 m. According to Figure 7, the time corresponding to <i>x</i> = 30 m is 21 s.</p></div></div></div></div><p>A positiontime graph provides a very straightforward way of describing motion along a line. It is easy to construct from a table of measurements, and easy to use to determine details of the motion. However, you should realise that the appearance of the graph depends on the precise choice made for the <i>x</i>axis. Figure 8 shows a new coordinate system in which the origin is 20 m to the right of the origin in Figure 6. As a result, the position coordinate of the car, measured in this new system, at any of the times listed in Table 2, will be 20 m <i>less</i> than the value given in the table, and the corresponding positiontime graph will look like Figure 9. The car now starts at <i>x</i> = −﻿20 m, and the steadily increasing <i>x</i>coordinate only becomes positive after <i>t</i> = 8 s. Different choices of origin simply shift the positiontime graph upwards or downwards, without changing its shape.</p><div class="oucontentfigure" style="width:511px;" id="fig001_007"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/b60704c0/s207_2_008i.jpg" alt="" width="511" height="108" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 8:</b> An alternative choice of <i>x</i>axis with the origin moved 20 m to the right</span></div></div></div><div class="oucontentfigure oucontentmediamini" id="fig001_008"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/de5f033d/s207_2_009i.jpg" alt="" width="234" height="254" style="maxwidth:234px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 9:</b> The positiontime graph when the position of the car is measured relative to the <i>x</i>axis of Figure 8</span></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_003"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 3</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>What change in the description of the motion would shift the positiontime graph to the right or the left, without changing its shape?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>The whole graph could be shifted to the right or the left (without altering its shape) by choosing the origin of time (<i>t</i> = 0 s) to be earlier or later than that used in Table 2.</p></div></div></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection2.3
2.3 Positiontime graphsS207_2<p>Tables do not give a very striking impression of how one thing varies with respect to another. A visual form of presentation, such as a graph, is usually much more effective. This is evident from Figure 7, which shows the graph obtained by plotting the data in Table 2 and then drawing a smooth curve through the resulting points.</p><div class="oucontentfigure" style="width:511px;" id="fig001_006"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/f0990c86/s207_2_007i.jpg" alt="" width="511" height="502" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 7:</b> A positiontime graph based on the results in Table 2</span></div></div></div><p>The smooth curve drawn in Figure 7 is called the <b>positiontime graph</b> of the car's motion. It can be used to read off the position of the car at any instant of time, or to find when the car passes a certain point. In graphs such as this, it is conventional to plot the time <i>t</i> along the horizontal axis. The vertical axis is used for the position <i>x</i>. This is just a standard way of displaying information about quantities that depend on time; it does <i>not</i> imply that the <i>x</i>axis of Figure 6 is vertical!</p><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_002"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 2</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>(a) Use Figure 7 to estimate the position of the car at <i>t</i> = 32 s.</p><p>(b) Estimate the time at which the car reaches the position shown in Figure 6.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) From Figure 7, the position at <i>t</i> = 32 s is 59 m.</p><p>(b) In Figure 6 the car is at <i>x</i> = 30 m. According to Figure 7, the time corresponding to <i>x</i> = 30 m is 21 s.</p></div></div></div></div><p>A positiontime graph provides a very straightforward way of describing motion along a line. It is easy to construct from a table of measurements, and easy to use to determine details of the motion. However, you should realise that the appearance of the graph depends on the precise choice made for the <i>x</i>axis. Figure 8 shows a new coordinate system in which the origin is 20 m to the right of the origin in Figure 6. As a result, the position coordinate of the car, measured in this new system, at any of the times listed in Table 2, will be 20 m <i>less</i> than the value given in the table, and the corresponding positiontime graph will look like Figure 9. The car now starts at <i>x</i> = −20 m, and the steadily increasing <i>x</i>coordinate only becomes positive after <i>t</i> = 8 s. Different choices of origin simply shift the positiontime graph upwards or downwards, without changing its shape.</p><div class="oucontentfigure" style="width:511px;" id="fig001_007"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/b60704c0/s207_2_008i.jpg" alt="" width="511" height="108" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 8:</b> An alternative choice of <i>x</i>axis with the origin moved 20 m to the right</span></div></div></div><div class="oucontentfigure oucontentmediamini" id="fig001_008"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/de5f033d/s207_2_009i.jpg" alt="" width="234" height="254" style="maxwidth:234px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 9:</b> The positiontime graph when the position of the car is measured relative to the <i>x</i>axis of Figure 8</span></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_003"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 3</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>What change in the description of the motion would shift the positiontime graph to the right or the left, without changing its shape?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>The whole graph could be shifted to the right or the left (without altering its shape) by choosing the origin of time (<i>t</i> = 0 s) to be earlier or later than that used in Table 2.</p></div></div></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

2.4 Displacementtime graphs
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection2.4
Tue, 12 Apr 2016 23:00:00 GMT
<div class="oucontentfigure" style="width:511px;" id="fig001_009"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/90613466/s207_2_010i.small.jpg" alt="" width="511" height="78" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 10:</b> A police car in pursuit of a getaway car</span></div></div></div><p>A particle's position, <i>x</i>, is always measured from the origin of the coordinate system. However, in describing real motions it is often important to know where something is located relative to a point other than the origin. Figure 10 shows a case in point; in a highspeed pursuit neither the police nor the robbers are likely to be very interested in their location relative to the origin, but both will be interested in the location of their own vehicle relative to the other. The physical quantity used to describe the location of one point relative to another is called <b>displacement</b>. In the case of Figure 10, the displacement of the getaway car from the police car is 400 m and the displacement of the police car from the getaway car is −400 m. In each case the displacement of a body is determined by subtracting the position coordinate of the reference body from the position coordinate of the body of interest. Thus the displacement (measured along the <i>x</i>axis) of a particle with position <i>x</i> from a chosen reference point with position <i>x</i><sub>ref</sub> is given by</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="eqn001_001"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/878328ae/s207_1_ue001i.gif" alt=""/></div><p>Notice that displacements, like positions, may be positive or negative depending on the direction of the displacement. Thus <i>s<sub>x</sub></i> is always positive when <i>x</i> is to the right of <i>x</i><sub>ref</sub> and is negative when <i>x</i> is to the left of <i>x</i><sub>ref</sub>. Also note that since displacements are measured along a definite axis it makes sense to represent them by a symbol, <i>s<sub>x</sub></i>, that includes a reference to that axis. This symbol may be read as '<i>s</i> subscript <i>x</i>' or, more simply, as '﻿<i>s</i> sub <i>x</i>﻿'.</p><p>A special case of Equation 1 is when the reference point <i>x</i><sub>ref</sub> is the origin. Then <i>x</i><sub>ref</sub> = 0 and <i>s<sub>x</sub> </i> = <i>x</i>. Thus the displacement of a point from the origin is the position coordinate of the point. Once we know the displacement <i>s<sub>x</sub> </i> of one object from another, it is easy to work out the <b>distance</b> <i>s</i> between those two objects. The distance is just the numerical value of the displacement, including the unit of measurement but ignoring any overall negative sign. So, when the displacement of the police car from the getaway car is <i>s<sub>x</sub></i> = −400 m, the distance between them is <i>s</i> = 400 m. We describe this simple relationship between distance and displacement by saying that the distance between two objects is given by the <b>magnitude</b> of the displacement of one from the other, and we indicate it mathematically by writing</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="eqn001_002"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2821db48/s207_1_ue002i.gif" alt=""/></div><p>The two vertical bars   constitute a <b>modulus sign</b> and indicate that you should work out the value of whatever they enclose and then take its magnitude (i.e. ignore any overall minus sign).</p><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="ique001_001"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 4</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>In Figure 8, what is the displacement <i>s<sub>x</sub> </i> of the car from the pedestrian, and what is the distance <i>s</i> between them?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p><i>s<sub>x</sub> </i> = 50 m, and <i>s</i> = 50 m.</p></div></div></div></div><p>In many circumstances it is more valuable to plot a <b>displacementtime graph</b> rather than a positiontime graph. In order to do this you either have to know the relevant displacements at various times, or you need to know enough about the positions of both the bodies involved to work out the displacements. Table 3 contains some plausible data about the positions of the two cars in Figure 10; use it to answer the following question.</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_003"><h2 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 3:</b> The position coordinates of the police car <i>x</i><sub>pol</sub> and getaway car <i>x</i><sub>get</sub> at various times <i>t</i></h2><div class="oucontenttablewrapper"><table><tr><th scope="col"><i>t</i>/s</th><th scope="col"><i>x</i><sub>pol</sub>/m</th><th scope="col"><i>x</i><sub>get</sub>/m</th></tr><tr><td class="oucontenttablemiddle ">0</td><td class="oucontenttablemiddle ">−200</td><td class="oucontenttablemiddle ">200</td></tr><tr><td class="oucontenttablemiddle ">5</td><td class="oucontenttablemiddle ">−115</td><td class="oucontenttablemiddle ">237</td></tr><tr><td class="oucontenttablemiddle ">10</td><td class="oucontenttablemiddle ">−45</td><td class="oucontenttablemiddle ">265</td></tr><tr><td class="oucontenttablemiddle ">15</td><td class="oucontenttablemiddle ">10</td><td class="oucontenttablemiddle ">293</td></tr><tr><td class="oucontenttablemiddle ">20</td><td class="oucontenttablemiddle ">62</td><td class="oucontenttablemiddle ">317</td></tr><tr><td class="oucontenttablemiddle ">25</td><td class="oucontenttablemiddle ">108</td><td class="oucontenttablemiddle ">337</td></tr><tr><td class="oucontenttablemiddle ">30</td><td class="oucontenttablemiddle ">149</td><td class="oucontenttablemiddle ">351</td></tr><tr><td class="oucontenttablemiddle ">35</td><td class="oucontenttablemiddle ">184</td><td class="oucontenttablemiddle ">364</td></tr><tr><td class="oucontenttablemiddle ">40</td><td class="oucontenttablemiddle ">213</td><td class="oucontenttablemiddle ">377</td></tr><tr><td class="oucontenttablemiddle ">45</td><td class="oucontenttablemiddle ">239</td><td class="oucontenttablemiddle ">388</td></tr><tr><td class="oucontenttablemiddle ">50</td><td class="oucontenttablemiddle ">263</td><td class="oucontenttablemiddle ">394</td></tr><tr><td class="oucontenttablemiddle ">55</td><td class="oucontenttablemiddle ">283</td><td class="oucontenttablemiddle ">399</td></tr><tr><td class="oucontenttablemiddle ">60</td><td class="oucontenttablemiddle ">300</td><td class="oucontenttablemiddle ">400</td></tr></table></div><div class="oucontentsourcereference"></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_004"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 5</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Plot a graph to show how the displacement of the getaway car from the police car depends on time.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Since we are interested in the displacement of the getaway car from the police car, we need to plot <i>x</i><sub>get</sub> − <i>x</i><sub>pol</sub>. You may find it useful to tabulate these values by adding another column (headed <i>x</i>﻿<sub>get</sub> − <i>x</i>﻿<sub>pol</sub>) to Table 3. The result of the plot is given in Figure 11.</p><div class="oucontentfigure" style="width:392px;" id="fig001_037"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7d41f5fc/s207_2_038i.jpg" alt="" width="392" height="452" style="maxwidth:392px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 11:</b> A displacementtime graph based on Table 3</span></div></div></div></div></div></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection2.4
2.4 Displacementtime graphsS207_2<div class="oucontentfigure" style="width:511px;" id="fig001_009"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/90613466/s207_2_010i.small.jpg" alt="" width="511" height="78" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 10:</b> A police car in pursuit of a getaway car</span></div></div></div><p>A particle's position, <i>x</i>, is always measured from the origin of the coordinate system. However, in describing real motions it is often important to know where something is located relative to a point other than the origin. Figure 10 shows a case in point; in a highspeed pursuit neither the police nor the robbers are likely to be very interested in their location relative to the origin, but both will be interested in the location of their own vehicle relative to the other. The physical quantity used to describe the location of one point relative to another is called <b>displacement</b>. In the case of Figure 10, the displacement of the getaway car from the police car is 400 m and the displacement of the police car from the getaway car is −400 m. In each case the displacement of a body is determined by subtracting the position coordinate of the reference body from the position coordinate of the body of interest. Thus the displacement (measured along the <i>x</i>axis) of a particle with position <i>x</i> from a chosen reference point with position <i>x</i><sub>ref</sub> is given by</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="eqn001_001"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/878328ae/s207_1_ue001i.gif" alt=""/></div><p>Notice that displacements, like positions, may be positive or negative depending on the direction of the displacement. Thus <i>s<sub>x</sub></i> is always positive when <i>x</i> is to the right of <i>x</i><sub>ref</sub> and is negative when <i>x</i> is to the left of <i>x</i><sub>ref</sub>. Also note that since displacements are measured along a definite axis it makes sense to represent them by a symbol, <i>s<sub>x</sub></i>, that includes a reference to that axis. This symbol may be read as '<i>s</i> subscript <i>x</i>' or, more simply, as '<i>s</i> sub <i>x</i>'.</p><p>A special case of Equation 1 is when the reference point <i>x</i><sub>ref</sub> is the origin. Then <i>x</i><sub>ref</sub> = 0 and <i>s<sub>x</sub> </i> = <i>x</i>. Thus the displacement of a point from the origin is the position coordinate of the point. Once we know the displacement <i>s<sub>x</sub> </i> of one object from another, it is easy to work out the <b>distance</b> <i>s</i> between those two objects. The distance is just the numerical value of the displacement, including the unit of measurement but ignoring any overall negative sign. So, when the displacement of the police car from the getaway car is <i>s<sub>x</sub></i> = −400 m, the distance between them is <i>s</i> = 400 m. We describe this simple relationship between distance and displacement by saying that the distance between two objects is given by the <b>magnitude</b> of the displacement of one from the other, and we indicate it mathematically by writing</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="eqn001_002"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2821db48/s207_1_ue002i.gif" alt=""/></div><p>The two vertical bars   constitute a <b>modulus sign</b> and indicate that you should work out the value of whatever they enclose and then take its magnitude (i.e. ignore any overall minus sign).</p><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="ique001_001"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 4</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>In Figure 8, what is the displacement <i>s<sub>x</sub> </i> of the car from the pedestrian, and what is the distance <i>s</i> between them?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p><i>s<sub>x</sub> </i> = 50 m, and <i>s</i> = 50 m.</p></div></div></div></div><p>In many circumstances it is more valuable to plot a <b>displacementtime graph</b> rather than a positiontime graph. In order to do this you either have to know the relevant displacements at various times, or you need to know enough about the positions of both the bodies involved to work out the displacements. Table 3 contains some plausible data about the positions of the two cars in Figure 10; use it to answer the following question.</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_003"><h2 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 3:</b> The position coordinates of the police car <i>x</i><sub>pol</sub> and getaway car <i>x</i><sub>get</sub> at various times <i>t</i></h2><div class="oucontenttablewrapper"><table><tr><th scope="col"><i>t</i>/s</th><th scope="col"><i>x</i><sub>pol</sub>/m</th><th scope="col"><i>x</i><sub>get</sub>/m</th></tr><tr><td class="oucontenttablemiddle ">0</td><td class="oucontenttablemiddle ">−200</td><td class="oucontenttablemiddle ">200</td></tr><tr><td class="oucontenttablemiddle ">5</td><td class="oucontenttablemiddle ">−115</td><td class="oucontenttablemiddle ">237</td></tr><tr><td class="oucontenttablemiddle ">10</td><td class="oucontenttablemiddle ">−45</td><td class="oucontenttablemiddle ">265</td></tr><tr><td class="oucontenttablemiddle ">15</td><td class="oucontenttablemiddle ">10</td><td class="oucontenttablemiddle ">293</td></tr><tr><td class="oucontenttablemiddle ">20</td><td class="oucontenttablemiddle ">62</td><td class="oucontenttablemiddle ">317</td></tr><tr><td class="oucontenttablemiddle ">25</td><td class="oucontenttablemiddle ">108</td><td class="oucontenttablemiddle ">337</td></tr><tr><td class="oucontenttablemiddle ">30</td><td class="oucontenttablemiddle ">149</td><td class="oucontenttablemiddle ">351</td></tr><tr><td class="oucontenttablemiddle ">35</td><td class="oucontenttablemiddle ">184</td><td class="oucontenttablemiddle ">364</td></tr><tr><td class="oucontenttablemiddle ">40</td><td class="oucontenttablemiddle ">213</td><td class="oucontenttablemiddle ">377</td></tr><tr><td class="oucontenttablemiddle ">45</td><td class="oucontenttablemiddle ">239</td><td class="oucontenttablemiddle ">388</td></tr><tr><td class="oucontenttablemiddle ">50</td><td class="oucontenttablemiddle ">263</td><td class="oucontenttablemiddle ">394</td></tr><tr><td class="oucontenttablemiddle ">55</td><td class="oucontenttablemiddle ">283</td><td class="oucontenttablemiddle ">399</td></tr><tr><td class="oucontenttablemiddle ">60</td><td class="oucontenttablemiddle ">300</td><td class="oucontenttablemiddle ">400</td></tr></table></div><div class="oucontentsourcereference"></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_004"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 5</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Plot a graph to show how the displacement of the getaway car from the police car depends on time.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Since we are interested in the displacement of the getaway car from the police car, we need to plot <i>x</i><sub>get</sub> − <i>x</i><sub>pol</sub>. You may find it useful to tabulate these values by adding another column (headed <i>x</i><sub>get</sub> − <i>x</i><sub>pol</sub>) to Table 3. The result of the plot is given in Figure 11.</p><div class="oucontentfigure" style="width:392px;" id="fig001_037"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7d41f5fc/s207_2_038i.jpg" alt="" width="392" height="452" style="maxwidth:392px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 11:</b> A displacementtime graph based on Table 3</span></div></div></div></div></div></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

2.5 A note on graph drawing
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection2.5
Tue, 12 Apr 2016 23:00:00 GMT
<p>There will be many occasions throughout your study of physics when you will need to draw graphs. This subsection gives some important guidelines for this activity.</p><ol class="oucontentnumbered"><li><p><i>Decide</i> which is the <b>independent variable</b> and which the <b>dependent variable</b>. <i>Plot</i> the independent variable along the horizontal axis and the dependent variable along the vertical axis. This is purely a convention but is why, for instance, we usually plot the time along the horizontal axis of a positiontime graph. It is the position that varies with time rather than the time that varies with position. Time is the independent variable since we can choose to make a measurement at any time. Position is the dependent variable.</p></li><li><p><i>Give</i> the graph a title, e.g. distance versus time.</p></li><li><p><i>Arrange</i> the axes so that the vertical axis increases in an upward direction and values along the horizontal axis increase to the right. This is simply a convention.</p></li><li><p><i>Label</i> both axes to show which quantities are being plotted and include the units. By convention only pure numbers are plotted. The physical quantity must be divided by its units before being plotted. This means that each axis should be labelled as quantity/units. This is why in all the graphs we have drawn so far the axes have been labelled by time/s and position/m (or displacement/m).</p></li><li><p><i>Fill</i> as much of the graph paper as reasonably possible. You will obtain greater accuracy if the graph is as big as possible. However, take care to use the graph paper sensibly. Graph paper usually has centimetre and millimetre squares, so it is straightforward to use 2 or 5 or 10 divisions on the paper to one physical unit. What you should avoid are multiples such as 3, 6, 7 ….</p></li><li><p><i>Scale</i> the axes appropriately, especially if the numbers involved are either very large or very small. For example, if the values of time <i>t</i> range from 0 s to 1.0 × 10<sup>﻿−5</sup> s, then, rather than plotting <i>t</i>/s and inserting values such as 1.0 × 10<sup>−6</sup>, 2.0 × 10<sup>−6</sup>, etc. along the axis, it is usually more convenient to change the units to microseconds and plot <i>t</i>/μs; the values along the axis will then simply be 1, 2, 3, etc. It is also acceptable to label the axis <i>t</i>/10<sup>−6</sup> s rather than <i>t</i>/μs if you prefer.</p></li><li><p><i>Plot</i> the points clearly. If you use very small dots they may be confused with other marks on the paper. However, using very big dots is not a good idea since it is hard to tell the position of the centre. Some authors put the dots within small circles. In this course we simply use dots since it is easy to show them clearly in professionally drawn graphs.</p></li><li><p><i>Draw</i> a straight line or smooth curve through the points plotted. The graphs that you draw will generally represent the smooth variation of one quantity with respect to another so a smooth curve is usually appropriate.</p></li></ol><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_005"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 6</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>How many of the above guidelines did you violate in answering Question 5?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Only you will know the answer to this, but it is common to see graphs in which the axes have not been labelled, or the units have been omitted. This is especially true when automated graphplotting packages are used; such packages often require special instructions if they are to show labels and units, and these are easily overlooked. If you are using such a package (or a graphical calculator), don't forget that the line you have to plot is far from being the whole graph: axes and labels are also important.</p></div></div></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection2.5
2.5 A note on graph drawingS207_2<p>There will be many occasions throughout your study of physics when you will need to draw graphs. This subsection gives some important guidelines for this activity.</p><ol class="oucontentnumbered"><li><p><i>Decide</i> which is the <b>independent variable</b> and which the <b>dependent variable</b>. <i>Plot</i> the independent variable along the horizontal axis and the dependent variable along the vertical axis. This is purely a convention but is why, for instance, we usually plot the time along the horizontal axis of a positiontime graph. It is the position that varies with time rather than the time that varies with position. Time is the independent variable since we can choose to make a measurement at any time. Position is the dependent variable.</p></li><li><p><i>Give</i> the graph a title, e.g. distance versus time.</p></li><li><p><i>Arrange</i> the axes so that the vertical axis increases in an upward direction and values along the horizontal axis increase to the right. This is simply a convention.</p></li><li><p><i>Label</i> both axes to show which quantities are being plotted and include the units. By convention only pure numbers are plotted. The physical quantity must be divided by its units before being plotted. This means that each axis should be labelled as quantity/units. This is why in all the graphs we have drawn so far the axes have been labelled by time/s and position/m (or displacement/m).</p></li><li><p><i>Fill</i> as much of the graph paper as reasonably possible. You will obtain greater accuracy if the graph is as big as possible. However, take care to use the graph paper sensibly. Graph paper usually has centimetre and millimetre squares, so it is straightforward to use 2 or 5 or 10 divisions on the paper to one physical unit. What you should avoid are multiples such as 3, 6, 7 ….</p></li><li><p><i>Scale</i> the axes appropriately, especially if the numbers involved are either very large or very small. For example, if the values of time <i>t</i> range from 0 s to 1.0 × 10<sup>−5</sup> s, then, rather than plotting <i>t</i>/s and inserting values such as 1.0 × 10<sup>−6</sup>, 2.0 × 10<sup>−6</sup>, etc. along the axis, it is usually more convenient to change the units to microseconds and plot <i>t</i>/μs; the values along the axis will then simply be 1, 2, 3, etc. It is also acceptable to label the axis <i>t</i>/10<sup>−6</sup> s rather than <i>t</i>/μs if you prefer.</p></li><li><p><i>Plot</i> the points clearly. If you use very small dots they may be confused with other marks on the paper. However, using very big dots is not a good idea since it is hard to tell the position of the centre. Some authors put the dots within small circles. In this course we simply use dots since it is easy to show them clearly in professionally drawn graphs.</p></li><li><p><i>Draw</i> a straight line or smooth curve through the points plotted. The graphs that you draw will generally represent the smooth variation of one quantity with respect to another so a smooth curve is usually appropriate.</p></li></ol><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_005"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 6</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>How many of the above guidelines did you violate in answering Question 5?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Only you will know the answer to this, but it is common to see graphs in which the axes have not been labelled, or the units have been omitted. This is especially true when automated graphplotting packages are used; such packages often require special instructions if they are to show labels and units, and these are easily overlooked. If you are using such a package (or a graphical calculator), don't forget that the line you have to plot is far from being the whole graph: axes and labels are also important.</p></div></div></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

3.1 Describing uniform motion
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection3.1
Tue, 12 Apr 2016 23:00:00 GMT
<p><b>Uniform motion</b> along a line is the very special kind of motion that occurs when an object moves with unvarying speed in a fixed direction. During a fixed period of time, such as one second, an object in uniform motion will always cover the same distance, no matter when the period begins. This is the kind of motion associated with trafficfree motoring along straight roads, with uninterrupted train journeys along straight tracks, and with unhindered straight and level flying (provided the distances involved are sufficiently small that the curvature of the Earth can be ignored).</p><p>Table 4 shows some typical values of position and time for a pedestrian and a car that are both in a state of uniform motion. The corresponding positiontime graphs are shown in Figure 12a and b.</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_004"><h3 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 4:</b> Positions for the pedestrian, <i>x</i><sub>ped</sub>, and the car, <i>x</i><sub>car</sub>, at times <i>t</i> </h3><div class="oucontenttablewrapper"><table><tr><th scope="col"><i>t</i>/s</th><th scope="col"><i>x</i><sub>ped</sub>/m</th><th scope="col"><i>x</i><sub>car</sub>/m</th></tr><tr><td>0</td><td>−20</td><td>30</td></tr><tr><td>10</td><td>−10</td><td>60</td></tr><tr><td>20</td><td>0</td><td>90</td></tr><tr><td>30</td><td>10</td><td>120</td></tr><tr><td>40</td><td>20</td><td>150</td></tr><tr><td>50</td><td>30</td><td>180</td></tr><tr><td>60</td><td>40</td><td>210</td></tr></table></div><div class="oucontentsourcereference"></div></div><div class="oucontentfigure oucontentmediamini" id="fig001_010"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/8c38f2ec/s207_2_011i.jpg" alt="" width="234" height="560" style="maxwidth:234px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 12:</b> Positiontime graphs for the positions of (a) a pedestrian and (b) a car, based on Table 4</span></div></div></div><p>As you can see, both are <i>straightline graphs</i>, as is characteristic of uniform motion. The graphs indicate that in each case the quantities being plotted are related by an equation of the general form</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="eqn001_003"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2e3eb78b/s207_1_ue003i.gif" alt=""/></div><p>where <i>A</i> and <i>B</i> are constants. The two parts of Figure 12 simply correspond to different values of the constants <i>A</i> and <i>B</i>. In fact, as you will see later, by choosing appropriate values for <i>A</i> and <i>B</i>, Equation 3 can be used to represent any straight line that can be drawn on the positiontime graph, except one that runs parallel to the vertical axis. In view of this you will not be surprised to learn that any equation of the general form <i>x</i> = <i>At</i> + <i>B</i> may be referred to as the <b>equation of a straight line</b>, irrespective of the particular constants and variables that it involves.</p><p>Understanding the link between equations (such as Equation 3) and graphs (such as those in Figure 12) is of vital importance throughout physics. Broadly speaking, any equation that relates two variables can be represented as a graph, and any graph showing how one quantity varies with another can be represented by an equation. (There are exceptions to this broad statement, but they tend to be rather unphysical, and won't be considered here.) The skill of looking at a simple equation and visualising it as a graph is one that is well worth developing. Many physicists find that visualisation helps to bring equations to life and makes it possible to look 'into' equations rather than merely looking 'at' them. The equation of a straight line that describes uniform motion is a good place to start developing this skill.</p><p>In the next two subsections you will learn how the constants <i>A</i> and <i>B</i> in the equation of a straight line determine the slope and positioning of the corresponding line, and what those features represent physically in the case of a positiontime graph.</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_005"><h3 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 5:</b> Displacement of the car from the pedestrian at time <i>t</i>, according to Table 4</h3><div class="oucontenttablewrapper"><table><tr><th scope="col"><i>t</i>/s</th><th scope="col">(<i>x</i><sub>car</sub> − <i>x</i><sub>ped</sub>)/m</th></tr><tr><td>0</td><td class="oucontenttablemiddle ">50</td></tr><tr><td>10</td><td class="oucontenttablemiddle ">70</td></tr><tr><td>20</td><td class="oucontenttablemiddle ">90</td></tr><tr><td>30</td><td class="oucontenttablemiddle ">110</td></tr><tr><td>40</td><td class="oucontenttablemiddle ">130</td></tr><tr><td>50</td><td class="oucontenttablemiddle ">150</td></tr><tr><td>60</td><td class="oucontenttablemiddle ">170</td></tr></table></div><div class="oucontentsourcereference"></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_006"><div class="oucontentouterbox"><h3 class="oucontenth3 oucontentheading oucontentnonumber">Question 7</h3><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Use Table 5 to plot a graph showing how the displacement of the car from the pedestrian varies with time. Describe your graph in words and write down the general form of the equation that describes your graph.</p></div>
<div class="oucontentsaqanswer"><h4 class="oucontenth4">Answer</h4><p>The displacement of the car from the pedestrian is plotted in Figure 13. The significant feature is that we get a straight line and consequently the relative motion is uniform, i.e. at constant speed. The straight line does <i>not</i> run through the origin, so this curve is of the general form <i>x</i> = <i>At</i> + <i>B</i>, where <i>x</i> stands for <i>x</i>﻿<sub>car</sub> − <i>x</i>﻿<sub>ped</sub> in this case.</p><div class="oucontentfigure" style="width:394px;" id="fig004_038"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/d9c7c459/s207_2_039i.jpg" alt="" width="394" height="261" style="maxwidth:394px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 13:</b> The result of plotting the displacements in Table 5</span></div></div></div></div></div></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection3.1
3.1 Describing uniform motionS207_2<p><b>Uniform motion</b> along a line is the very special kind of motion that occurs when an object moves with unvarying speed in a fixed direction. During a fixed period of time, such as one second, an object in uniform motion will always cover the same distance, no matter when the period begins. This is the kind of motion associated with trafficfree motoring along straight roads, with uninterrupted train journeys along straight tracks, and with unhindered straight and level flying (provided the distances involved are sufficiently small that the curvature of the Earth can be ignored).</p><p>Table 4 shows some typical values of position and time for a pedestrian and a car that are both in a state of uniform motion. The corresponding positiontime graphs are shown in Figure 12a and b.</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_004"><h3 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 4:</b> Positions for the pedestrian, <i>x</i><sub>ped</sub>, and the car, <i>x</i><sub>car</sub>, at times <i>t</i> </h3><div class="oucontenttablewrapper"><table><tr><th scope="col"><i>t</i>/s</th><th scope="col"><i>x</i><sub>ped</sub>/m</th><th scope="col"><i>x</i><sub>car</sub>/m</th></tr><tr><td>0</td><td>−20</td><td>30</td></tr><tr><td>10</td><td>−10</td><td>60</td></tr><tr><td>20</td><td>0</td><td>90</td></tr><tr><td>30</td><td>10</td><td>120</td></tr><tr><td>40</td><td>20</td><td>150</td></tr><tr><td>50</td><td>30</td><td>180</td></tr><tr><td>60</td><td>40</td><td>210</td></tr></table></div><div class="oucontentsourcereference"></div></div><div class="oucontentfigure oucontentmediamini" id="fig001_010"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/8c38f2ec/s207_2_011i.jpg" alt="" width="234" height="560" style="maxwidth:234px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 12:</b> Positiontime graphs for the positions of (a) a pedestrian and (b) a car, based on Table 4</span></div></div></div><p>As you can see, both are <i>straightline graphs</i>, as is characteristic of uniform motion. The graphs indicate that in each case the quantities being plotted are related by an equation of the general form</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="eqn001_003"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2e3eb78b/s207_1_ue003i.gif" alt=""/></div><p>where <i>A</i> and <i>B</i> are constants. The two parts of Figure 12 simply correspond to different values of the constants <i>A</i> and <i>B</i>. In fact, as you will see later, by choosing appropriate values for <i>A</i> and <i>B</i>, Equation 3 can be used to represent any straight line that can be drawn on the positiontime graph, except one that runs parallel to the vertical axis. In view of this you will not be surprised to learn that any equation of the general form <i>x</i> = <i>At</i> + <i>B</i> may be referred to as the <b>equation of a straight line</b>, irrespective of the particular constants and variables that it involves.</p><p>Understanding the link between equations (such as Equation 3) and graphs (such as those in Figure 12) is of vital importance throughout physics. Broadly speaking, any equation that relates two variables can be represented as a graph, and any graph showing how one quantity varies with another can be represented by an equation. (There are exceptions to this broad statement, but they tend to be rather unphysical, and won't be considered here.) The skill of looking at a simple equation and visualising it as a graph is one that is well worth developing. Many physicists find that visualisation helps to bring equations to life and makes it possible to look 'into' equations rather than merely looking 'at' them. The equation of a straight line that describes uniform motion is a good place to start developing this skill.</p><p>In the next two subsections you will learn how the constants <i>A</i> and <i>B</i> in the equation of a straight line determine the slope and positioning of the corresponding line, and what those features represent physically in the case of a positiontime graph.</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_005"><h3 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 5:</b> Displacement of the car from the pedestrian at time <i>t</i>, according to Table 4</h3><div class="oucontenttablewrapper"><table><tr><th scope="col"><i>t</i>/s</th><th scope="col">(<i>x</i><sub>car</sub> − <i>x</i><sub>ped</sub>)/m</th></tr><tr><td>0</td><td class="oucontenttablemiddle ">50</td></tr><tr><td>10</td><td class="oucontenttablemiddle ">70</td></tr><tr><td>20</td><td class="oucontenttablemiddle ">90</td></tr><tr><td>30</td><td class="oucontenttablemiddle ">110</td></tr><tr><td>40</td><td class="oucontenttablemiddle ">130</td></tr><tr><td>50</td><td class="oucontenttablemiddle ">150</td></tr><tr><td>60</td><td class="oucontenttablemiddle ">170</td></tr></table></div><div class="oucontentsourcereference"></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_006"><div class="oucontentouterbox"><h3 class="oucontenth3 oucontentheading oucontentnonumber">Question 7</h3><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Use Table 5 to plot a graph showing how the displacement of the car from the pedestrian varies with time. Describe your graph in words and write down the general form of the equation that describes your graph.</p></div>
<div class="oucontentsaqanswer"><h4 class="oucontenth4">Answer</h4><p>The displacement of the car from the pedestrian is plotted in Figure 13. The significant feature is that we get a straight line and consequently the relative motion is uniform, i.e. at constant speed. The straight line does <i>not</i> run through the origin, so this curve is of the general form <i>x</i> = <i>At</i> + <i>B</i>, where <i>x</i> stands for <i>x</i><sub>car</sub> − <i>x</i><sub>ped</sub> in this case.</p><div class="oucontentfigure" style="width:394px;" id="fig004_038"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/d9c7c459/s207_2_039i.jpg" alt="" width="394" height="261" style="maxwidth:394px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 13:</b> The result of plotting the displacements in Table 5</span></div></div></div></div></div></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

3.2 Constant velocity and the gradient of the positiontime graph
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection3.2
Tue, 12 Apr 2016 23:00:00 GMT
<p>Two things you will almost certainly want to know about any particle undergoing uniform motion are 'how fast is it travelling?' and 'in which direction is it moving?' The physical quantity that provides both these items of information is the particle's <b>velocity</b>. This is defined as the rate of change of the particle's position with respect to time, and has a constant value for each case of uniform motion along a line.</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox oucontentsnoheading " id="box001_004"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>velocity = rate of change of position with respect to time.</p></div></div></div><p>In the case of the car whose positiontime graph is shown in Figure 12b, if we choose two different times, say, <i>t</i>﻿<sub>1</sub> = 40 s and <i>t</i>﻿<sub>2</sub> = 50 s, then in the time interval between <i>t</i>﻿<sub>1</sub> and <i>t</i>﻿<sub>2</sub>, the car moves from <i>x</i>﻿<sub>1</sub> = 150 m to <i>x</i>﻿<sub>2</sub> = 180 m. It follows that the rate of change of position of the car is given by the ratio</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_003"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2c0bd19f/s207_1_ie002i.gif" alt=""/></div><p>We therefore say that the velocity of the car is 3.0 metres per second along the <i>x</i>axis, which we write using the abbreviation m s<sup>−1</sup>.</p><p>More generally, if any particle moves uniformly along the <i>x</i>axis, so that its positiontime graph is a straight line, then the constant velocity <i>v</i> <sub><i>x</i></sub> (' <i>v</i> sub <i>x</i>') of that particle is given by</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_004"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/9abd1255/s207_1_ie003i.gif" alt=""/></div><p>where <i>x</i>﻿<sub>1</sub> is the particle's position at time <i>t</i><sub>1</sub>, and <i>x</i>﻿<sub>2</sub> is its position at time <i>t</i>﻿<sub>2</sub>. Note that the velocity may be positive or negative, depending on whether the particle's position coordinate is increasing or decreasing with time. Also note that for uniform motion the velocity is independent of the particular values of <i>t</i>﻿<sub>1</sub> and <i>t</i>﻿<sub>2</sub> that are chosen.</p><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_007"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 8</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Determine the velocity of the car by making measurements on the graph in Figure 12b in the interval from 50 s to 60 s.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Using <span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/9abd1255/s207_1_ie003i.gif" alt="" width="87" height="31" style="maxwidth:87px;" class="oucontentinlinefigureimage"/></span>and Figure 12b we obtain</p><p/><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_002"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/0da8af02/s207_1_ie037i.gif" alt=""/></div><p>A point to note here is that choosing such closely separated values as <i>t</i> = 50 s and <i>t</i> = 60 s makes the evaluation of the velocity trickier, and more prone to error than would have been the case if more widely separated values had been chosen. When evaluating a gradient from a graph, it is always wise to use the widest convenient range of values on the horizontal axis.</p></div></div></div></div><p>In everyday speech the terms <i>velocity</i> and <i>speed</i> are used interchangeably. However, in physics, the term <b>speed</b> is reserved for the <i>magnitude</i> of the velocity, i.e. its value neglecting any overall minus sign. So, if a particle has velocity <i>v</i>﻿﻿<sub>﻿<i>x</i>﻿</sub> = −﻿5 m s<sup>﻿−1</sup>, then its speed is <i>v</i>﻿ = 5 m s<sup>﻿−﻿1</sup>, where</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_005"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/94e758fb/s207_1_ue004i.gif" alt=""/></div><p>Speed is a <i>positive</i> quantity telling us how rapidly the particle is moving but revealing nothing about its direction of motion.</p><p>The subscript <i>x</i> in <i>v</i>﻿﻿<sub>﻿<i>x</i></sub> may seem a bit cumbersome, but it is an essential part of the notation. The subscript reminds us that we are dealing with motion along the <i>x</i>axis. In later work, we will need to use two or sometimes three axes, so by including the subscript at this stage, we will be able to use the same notation throughout. Be careful to include the subscript <i>x</i> in your written work and make sure that it is small enough and low enough to be read as a subscript  don't risk having <i>v</i>﻿﻿<sub>﻿<i>x</i>﻿</sub> misinterpreted as <i>vx</i>, i.e. <i>v</i> times <i>x</i>.</p><p>Another piece of shorthand that you will find useful concerns the upper case Greek letter delta, Δ. If a quantity such as <i>x</i> changes its value from <i>x</i><sub>1</sub> to <i>x</i><sub>2</sub>, then the <i>change</i> in the value of <i>x</i> is conventionally written as Δ<i>x</i>, and read as 'delta ex'. Thus</p><div class="oucontentquote oucontentsbox" id="q014"><blockquote><p>Δ<i>x</i> = <i>x</i><sub>2</sub> − <i>x</i><sub>1</sub>.</p></blockquote></div><p>So, Δ<i>x</i> and Δ<i>t</i> mean the changes in <i>x</i> and <i>t</i>; they do <i>not</i> mean Δ times <i>x</i> or Δ times <i>t</i>. You should always think of the Δ and the symbol that follows it as a single entity; the Δ symbol by itself has no quantitative meaning. Using this notation, the expression for <i>v</i>﻿﻿<sub><i>x</i> </sub> above can be written as</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_007"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/680a03dc/s207_1_ue005i.gif" alt=""/></div><p>In graphical terms, the velocity Δ﻿<i>x</i>﻿/﻿Δ﻿<i>t</i> of a uniformly moving particle is indicated by the <i>slope</i> of its positiontime graph. The steepness of the line represents the speed of the particle, while the orientation of the line  bottom left to top right, or top left to bottom right  indicates the direction of motion. If the graph of <i>x</i> against <i>t</i> is a straight line, then the ratio of the change in <i>x</i> to the corresponding change in <i>t</i> (that is Δ﻿<i>x</i>﻿/﻿Δ﻿<i>t</i>﻿) is called the <b>gradient</b> of the graph. A line that slopes from bottom left to top right indicates that a positive change in <i>t</i> corresponds to a positive change in <i>x</i>, consequently such a line has positive gradient. Similarly, a line sloping from top left to bottom right indicates that a positive change in <i>t</i> corresponds to a negative change in <i>x</i> and consequently to a negative gradient. This gives us another way of describing velocity:</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox oucontentsnoheading " id="box001_005"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>the velocity of a particle = the gradient of its positiontime graph</p></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_008"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 9</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Figure 14 shows the positiontime graphs for four different objects (A, B, C and D) each moving uniformly with a different constant velocity. The position and time scales are the same in each case.</p><p>(a) List the objects in order of increasing speed.</p><p>(b) Which of the objects have positive velocity?</p><p>(c) List the objects in order of increasing velocity.</p><div class="oucontentfigure" style="width:511px;" id="fig001_011"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/ec9d6fe1/s207_2_012i.jpg" alt="" width="511" height="124" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 14:</b> Straight line positiontime graphs for Selfassessment question 9</span></div></div></div></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) The speed is given by the magnitude of Δ<i>x</i>/Δ<i>t</i> obtained from the velocitytime graph. In the order of increasing speed, the objects are B, D, A, C.</p><p>(b) The objects with positive velocity have positive values of Δ<i>x</i>/Δ<i>t</i>, these are A and B.</p><p>(c) Treating large negative values as being <i>less</i> than small negative values, and any negative value as less than any positive value, as is conventional, the list is C, D, B, A.</p></div></div></div></div><p>The gradient of a graph is one of the most important concepts of this course. It is important that you should be able to evaluate gradients, and that you should be able to distinguish graphs with positive gradients from those with negative gradients. It is also important that you should be able to interpret gradients physically as rates of change. In this discussion of positiontime graphs the gradient represents velocity, but in other contexts, where quantities other than <i>x</i> and <i>t</i> are plotted, the gradient may have a very different interpretation. Figure 15, for example, shows how the average temperature <i>T</i> of the atmosphere depends on the height <i>h</i> above sealevel. The gradient of this graph, Δ﻿<i>T</i>﻿/﻿Δ<i>h</i>, describes the rate at which temperature changes with height.</p><div class="oucontentfigure oucontentmediamini" id="fig001_012"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/9fdaf845/s207_2_013i.jpg" alt="" width="237" height="186" style="maxwidth:237px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 15:</b> A graph of average atmospheric temperature <i>T</i> against height <i>h</i> above sealevel</span></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_009"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 10</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Estimate the gradient of the graph in Figure 15. (The use of the word 'estimate' implies that you do not need to take great care over the values you read from the graph, but you should take care over matters such as signs and units of measurement.)</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>The gradient of the temperatureheight graph is</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_003"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/17d67eb5/s207_1_ie038i.gif" alt=""/></div><p>With <i>h</i><sub>1</sub> = 0 and <i>h</i><sub>2</sub> = 2 km, Figure 15 shows that <i>T</i><sub>1</sub> = 20 °C and <i>T</i><sub>2</sub> = 6 °C.</p><p>It follows that the gradient is</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_004"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/4d4fc99a/s207_1_ie039i.gif" alt=""/></div><p>Note that care has to be taken to arrange the values in the correct order when performing the subtractions if the right sign (minus in this case) is to be obtained.</p></div></div></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection3.2
3.2 Constant velocity and the gradient of the positiontime graphS207_2<p>Two things you will almost certainly want to know about any particle undergoing uniform motion are 'how fast is it travelling?' and 'in which direction is it moving?' The physical quantity that provides both these items of information is the particle's <b>velocity</b>. This is defined as the rate of change of the particle's position with respect to time, and has a constant value for each case of uniform motion along a line.</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox
oucontentsnoheading
" id="box001_004"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>velocity = rate of change of position with respect to time.</p></div></div></div><p>In the case of the car whose positiontime graph is shown in Figure 12b, if we choose two different times, say, <i>t</i><sub>1</sub> = 40 s and <i>t</i><sub>2</sub> = 50 s, then in the time interval between <i>t</i><sub>1</sub> and <i>t</i><sub>2</sub>, the car moves from <i>x</i><sub>1</sub> = 150 m to <i>x</i><sub>2</sub> = 180 m. It follows that the rate of change of position of the car is given by the ratio</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_003"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2c0bd19f/s207_1_ie002i.gif" alt=""/></div><p>We therefore say that the velocity of the car is 3.0 metres per second along the <i>x</i>axis, which we write using the abbreviation m s<sup>−1</sup>.</p><p>More generally, if any particle moves uniformly along the <i>x</i>axis, so that its positiontime graph is a straight line, then the constant velocity <i>v</i> <sub><i>x</i></sub> (' <i>v</i> sub <i>x</i>') of that particle is given by</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_004"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/9abd1255/s207_1_ie003i.gif" alt=""/></div><p>where <i>x</i><sub>1</sub> is the particle's position at time <i>t</i><sub>1</sub>, and <i>x</i><sub>2</sub> is its position at time <i>t</i><sub>2</sub>. Note that the velocity may be positive or negative, depending on whether the particle's position coordinate is increasing or decreasing with time. Also note that for uniform motion the velocity is independent of the particular values of <i>t</i><sub>1</sub> and <i>t</i><sub>2</sub> that are chosen.</p><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_007"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 8</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Determine the velocity of the car by making measurements on the graph in Figure 12b in the interval from 50 s to 60 s.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Using <span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/9abd1255/s207_1_ie003i.gif" alt="" width="87" height="31" style="maxwidth:87px;" class="oucontentinlinefigureimage"/></span>and Figure 12b we obtain</p><p/><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_002"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/0da8af02/s207_1_ie037i.gif" alt=""/></div><p>A point to note here is that choosing such closely separated values as <i>t</i> = 50 s and <i>t</i> = 60 s makes the evaluation of the velocity trickier, and more prone to error than would have been the case if more widely separated values had been chosen. When evaluating a gradient from a graph, it is always wise to use the widest convenient range of values on the horizontal axis.</p></div></div></div></div><p>In everyday speech the terms <i>velocity</i> and <i>speed</i> are used interchangeably. However, in physics, the term <b>speed</b> is reserved for the <i>magnitude</i> of the velocity, i.e. its value neglecting any overall minus sign. So, if a particle has velocity <i>v</i><sub><i>x</i></sub> = −5 m s<sup>−1</sup>, then its speed is <i>v</i> = 5 m s<sup>−1</sup>, where</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_005"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/94e758fb/s207_1_ue004i.gif" alt=""/></div><p>Speed is a <i>positive</i> quantity telling us how rapidly the particle is moving but revealing nothing about its direction of motion.</p><p>The subscript <i>x</i> in <i>v</i><sub><i>x</i></sub> may seem a bit cumbersome, but it is an essential part of the notation. The subscript reminds us that we are dealing with motion along the <i>x</i>axis. In later work, we will need to use two or sometimes three axes, so by including the subscript at this stage, we will be able to use the same notation throughout. Be careful to include the subscript <i>x</i> in your written work and make sure that it is small enough and low enough to be read as a subscript  don't risk having <i>v</i><sub><i>x</i></sub> misinterpreted as <i>vx</i>, i.e. <i>v</i> times <i>x</i>.</p><p>Another piece of shorthand that you will find useful concerns the upper case Greek letter delta, Δ. If a quantity such as <i>x</i> changes its value from <i>x</i><sub>1</sub> to <i>x</i><sub>2</sub>, then the <i>change</i> in the value of <i>x</i> is conventionally written as Δ<i>x</i>, and read as 'delta ex'. Thus</p><div class="oucontentquote oucontentsbox" id="q014"><blockquote><p>Δ<i>x</i> = <i>x</i><sub>2</sub> − <i>x</i><sub>1</sub>.</p></blockquote></div><p>So, Δ<i>x</i> and Δ<i>t</i> mean the changes in <i>x</i> and <i>t</i>; they do <i>not</i> mean Δ times <i>x</i> or Δ times <i>t</i>. You should always think of the Δ and the symbol that follows it as a single entity; the Δ symbol by itself has no quantitative meaning. Using this notation, the expression for <i>v</i><sub><i>x</i> </sub> above can be written as</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_007"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/680a03dc/s207_1_ue005i.gif" alt=""/></div><p>In graphical terms, the velocity Δ<i>x</i>/Δ<i>t</i> of a uniformly moving particle is indicated by the <i>slope</i> of its positiontime graph. The steepness of the line represents the speed of the particle, while the orientation of the line  bottom left to top right, or top left to bottom right  indicates the direction of motion. If the graph of <i>x</i> against <i>t</i> is a straight line, then the ratio of the change in <i>x</i> to the corresponding change in <i>t</i> (that is Δ<i>x</i>/Δ<i>t</i>) is called the <b>gradient</b> of the graph. A line that slopes from bottom left to top right indicates that a positive change in <i>t</i> corresponds to a positive change in <i>x</i>, consequently such a line has positive gradient. Similarly, a line sloping from top left to bottom right indicates that a positive change in <i>t</i> corresponds to a negative change in <i>x</i> and consequently to a negative gradient. This gives us another way of describing velocity:</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox
oucontentsnoheading
" id="box001_005"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>the velocity of a particle = the gradient of its positiontime graph</p></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_008"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 9</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Figure 14 shows the positiontime graphs for four different objects (A, B, C and D) each moving uniformly with a different constant velocity. The position and time scales are the same in each case.</p><p>(a) List the objects in order of increasing speed.</p><p>(b) Which of the objects have positive velocity?</p><p>(c) List the objects in order of increasing velocity.</p><div class="oucontentfigure" style="width:511px;" id="fig001_011"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/ec9d6fe1/s207_2_012i.jpg" alt="" width="511" height="124" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 14:</b> Straight line positiontime graphs for Selfassessment question 9</span></div></div></div></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) The speed is given by the magnitude of Δ<i>x</i>/Δ<i>t</i> obtained from the velocitytime graph. In the order of increasing speed, the objects are B, D, A, C.</p><p>(b) The objects with positive velocity have positive values of Δ<i>x</i>/Δ<i>t</i>, these are A and B.</p><p>(c) Treating large negative values as being <i>less</i> than small negative values, and any negative value as less than any positive value, as is conventional, the list is C, D, B, A.</p></div></div></div></div><p>The gradient of a graph is one of the most important concepts of this course. It is important that you should be able to evaluate gradients, and that you should be able to distinguish graphs with positive gradients from those with negative gradients. It is also important that you should be able to interpret gradients physically as rates of change. In this discussion of positiontime graphs the gradient represents velocity, but in other contexts, where quantities other than <i>x</i> and <i>t</i> are plotted, the gradient may have a very different interpretation. Figure 15, for example, shows how the average temperature <i>T</i> of the atmosphere depends on the height <i>h</i> above sealevel. The gradient of this graph, Δ<i>T</i>/Δ<i>h</i>, describes the rate at which temperature changes with height.</p><div class="oucontentfigure oucontentmediamini" id="fig001_012"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/9fdaf845/s207_2_013i.jpg" alt="" width="237" height="186" style="maxwidth:237px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 15:</b> A graph of average atmospheric temperature <i>T</i> against height <i>h</i> above sealevel</span></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_009"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 10</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Estimate the gradient of the graph in Figure 15. (The use of the word 'estimate' implies that you do not need to take great care over the values you read from the graph, but you should take care over matters such as signs and units of measurement.)</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>The gradient of the temperatureheight graph is</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_003"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/17d67eb5/s207_1_ie038i.gif" alt=""/></div><p>With <i>h</i><sub>1</sub> = 0 and <i>h</i><sub>2</sub> = 2 km, Figure 15 shows that <i>T</i><sub>1</sub> = 20 °C and <i>T</i><sub>2</sub> = 6 °C.</p><p>It follows that the gradient is</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_004"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/4d4fc99a/s207_1_ie039i.gif" alt=""/></div><p>Note that care has to be taken to arrange the values in the correct order when performing the subtractions if the right sign (minus in this case) is to be obtained.</p></div></div></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

3.3 Initial position and the intercept of the positiontime graph
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection3.3
Tue, 12 Apr 2016 23:00:00 GMT
<p>The uniform motion of a particle is such a simple form of motion that apart from enquiring about the particle's velocity, the only other kinematic question you can ask is 'where was the particle at some particular time?' The most common way of answering this question is to specify the <b>initial position</b> of the particle, that is, its position at time <i>t</i> = 0 s.</p><p>Although it is common to refer to the position at <i>t</i> = 0 as the 'initial position' it is also possible, and sometimes more convenient, to associate the initial position with some other time.</p><p>The initial position of a uniformly moving particle is easily determined from its positiontime graph. It's just the value of <i>x</i> when <i>t</i> = 0, i.e. the value of <i>x</i> at which the straight line crosses the vertical axis through the origin. In Figure 12a, for example, it is <i>x</i> = −20 m. This value is generally referred to as the <b>intercept</b>.</p><p>One point to bear in mind though; it is sometimes advantageous to draw positiontime graphs that do not include the origin (for instance, you might be asked to draw a graph for the period from <i>t</i> = 100 s to <i>t</i> = 110 s). When dealing with such graphs do not make the mistake of thinking that the value at which the line crosses the vertical axis is the intercept. You can only read the intercept directly from the graph if the vertical axis passes through the zero value on the horizontal axis.</p><p>A little thought should convince you that the gradient and the intercept of a straight line entirely determine that line. In the same way, the velocity and the initial position of a uniformly moving particle entirely determine the motion of that particle. In the next subsection you will learn how these graphical and physical statements can be represented algebraically, in terms of equations.</p>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection3.3
3.3 Initial position and the intercept of the positiontime graphS207_2<p>The uniform motion of a particle is such a simple form of motion that apart from enquiring about the particle's velocity, the only other kinematic question you can ask is 'where was the particle at some particular time?' The most common way of answering this question is to specify the <b>initial position</b> of the particle, that is, its position at time <i>t</i> = 0 s.</p><p>Although it is common to refer to the position at <i>t</i> = 0 as the 'initial position' it is also possible, and sometimes more convenient, to associate the initial position with some other time.</p><p>The initial position of a uniformly moving particle is easily determined from its positiontime graph. It's just the value of <i>x</i> when <i>t</i> = 0, i.e. the value of <i>x</i> at which the straight line crosses the vertical axis through the origin. In Figure 12a, for example, it is <i>x</i> = −20 m. This value is generally referred to as the <b>intercept</b>.</p><p>One point to bear in mind though; it is sometimes advantageous to draw positiontime graphs that do not include the origin (for instance, you might be asked to draw a graph for the period from <i>t</i> = 100 s to <i>t</i> = 110 s). When dealing with such graphs do not make the mistake of thinking that the value at which the line crosses the vertical axis is the intercept. You can only read the intercept directly from the graph if the vertical axis passes through the zero value on the horizontal axis.</p><p>A little thought should convince you that the gradient and the intercept of a straight line entirely determine that line. In the same way, the velocity and the initial position of a uniformly moving particle entirely determine the motion of that particle. In the next subsection you will learn how these graphical and physical statements can be represented algebraically, in terms of equations.</p>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

3.4 The equations of uniform motion
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection3.4
Tue, 12 Apr 2016 23:00:00 GMT
<p>It has already been said that the straightline graph of any uniform motion can be represented by an equation of the general form</p><p/><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_009"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2e3eb78b/s207_1_ue003i.gif" alt=""/></div><p>where <i>A</i> and <i>B</i> are constants. Different cases of uniform motion simply correspond to different values for the constants <i>A</i> and <i>B</i>. Let us now investigate this equation to see how it conveys information about gradients and intercepts, or equivalently, uniform velocities and initial positions.</p><p>To start with, note that according to Equation 3, the position of the particle when <i>t</i> = 0 is just <i>x</i> = <i>B</i>. Thus <i>B</i> represents the initial position of the particle, the intercept of its positiontime graph.</p><p>In a similar way, note that according to Equation 3, at time <i>t</i> = <i>t</i>﻿<sub>1</sub> the position of the particle, let's call it <i>x</i>﻿<sub>1</sub>, is given by <i>x</i>﻿<sub>1</sub> = <i>At</i><sub>1</sub> + <i>B</i>; and at some later time <i>t</i> = <i>t</i>﻿<sub>2</sub> the position of the particle is <i>x</i>﻿<sub>2</sub> = <i>At</i>﻿<sub>2</sub> + <i>B</i>. Now, as we have already seen, the velocity of a uniformly moving particle is defined by</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_010"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/680a03dc/s207_1_ue005i.gif" alt=""/></div><p>so, substituting the expressions for <i>x</i>﻿<sub>1</sub> and <i>x</i>﻿<sub>2</sub> that we have just obtained from Equation 3 we find that in this particular case</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_011"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/bfc7c298/s207_1_ie005i.gif" alt=""/></div><p>Thus, the constant <i>A</i> in Equation 3 represents the particle's velocity, i.e. the gradient of its positiontime graph.</p><p>We can now read the equation of a straight line, interpreting it graphically or physically. It is</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_012"><a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=2298&extra=thumbnailfigure_idp3588880" title="View larger image"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/785ca7fb/s207_1_i001i.jpg" alt=""/></a><div class="oucontentfiguretext"><div class="oucontentthumbnaillink"><a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=2298&extra=thumbnailfigure_idp3588880">View larger image</a></div><div class="oucontentcaption oucontentnonumber oucontentcaptionplaceholder"> </div></div><a id="back_thumbnailfigure_idp3588880"></a></div><p>The physical significance of the constants <i>A</i> and <i>B</i> can be emphasised by using the symbols <i>v</i>﻿<sub><i>x</i>﻿</sub> and <i>x</i><sub>0</sub> in their place. Doing this, we obtain the standard form of the <b>uniform motion equations</b></p><div class="oucontentbox oucontentsheavybox1 oucontentsbox oucontentsnoheading " id="box001_006"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p/><div class="oucontentequation oucontentequationequation oucontentnocaption" id="equ001_013"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7e240e9c/s207_1_ue006i.gif" alt=""/></div></div></div></div><p>So, the uniform motion of the car we have been considering, which has velocity <i>v</i><sub><i>x</i></sub> = 3.0 m s<sup>−1</sup> and initial position <i>x</i><sub>0</sub> = 30 m, can be described by the equation</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_015"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/509d1a7f/s207_1_ie006i.gif" alt=""/></div><p>This contains just as much information as the graph in Figure 12b.</p><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_010"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 11</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Write down the equation that describes the uniform motion of the pedestrian who was the subject of Figure 12a.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>The equation will have the general form of Equation 6a, with velocity <i>v<sub>x</sub> </i> = 1 m s<sup>﻿−﻿1</sup> (found from the gradient) and initial position <i>x</i>﻿<sub>0</sub> = − 20 m (found from the value of x when <i>t</i> = 0). Consequently the required equation may be written:</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_005"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/06da0324/s207_1_ie040i.gif" alt=""/></div></div></div></div></div><p>A pictorial interpretation of Equation 6a is given in Figure 16.</p><div class="oucontentfigure" style="width:511px;" id="fig001_01573"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/1c63648d/s207_2_014i.jpg" alt="" width="511" height="136" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 16:</b> A pictorial interpretation of the equation of uniform motion</span></div></div></div><p>Equations have the great advantage that they can be rearranged to make them fit the problem at hand. For instance, if you were flying at constant velocity from Paris to Rome, you might well be more interested in your displacement from Paris rather than your position in some arbitrary coordinate system. In mathematical terms, you might be more interested in <i>s<sub>x</sub></i> = <i>x</i> − <i>x</i>﻿<sub>0</sub> rather than <i>x</i> itself. Equation 6a can be rearranged to suit this interest by subtracting <i>x</i> <sub>0</sub> from each side:</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_016"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/48714241/s207_1_ie007i.gif" alt=""/></div><p>and this may be rewritten in the compact form</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox oucontentsnoheading " id="box001_007"><div class="oucontentouterbox"><div class="oucontentinnerbox"><div class="oucontentequation oucontentequationequation oucontentnocaption" id="fig001_01676"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2bc501a6/s207_1_ue007i.gif" alt=""/></div></div></div></div><p>Remembering that <i>v</i><sub><i>x</i></sub> is a constant when the motion is uniform, Equation 7 is equivalent to Equation 6a. It relates the displacement from the initial position directly to the time of flight and thus facilitates working out the distance travelled,</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/6c1e4df0/s207_1_ue008i.gif" alt="" width="283" height="16" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>The crucial point to remember when manipulating an equation is that <i>both sides of an equation represent the same value</i>. So, if you add or subtract a term on one side of an equation, you must add or subtract an identical term on the other side. The same principle applies to multiplication and division, and to other operations such as taking magnitudes, squaring or taking square roots. (In the latter case you must remember that a positive quantity generally has two square roots; 2 is a square root of 4, but so is −2.) There are two rules to bear in mind when rearranging equations:</p><ol class="oucontentnumbered"><li><p>When one side of an equation consists of several terms added together, each of those terms must be treated in the same way. (It's no good multiplying the first term on one side of an equation by some quantity if you forget to do the same to all the other terms on that side.)</p></li><li><p>When dividing both sides of an equation by some quantity you must ensure that the quantity is <i>not</i> zero. (For instance, it is only legitimate to divide both sides of Equation 7 by <i>x</i> − <i>x</i><sub>0</sub> <i>if x</i> ≠ <i>x</i><sub>0</sub>, i.e. if <i>x</i> is <i>not equal</i> to <i>x</i>﻿<sub>0</sub>.)</p></li></ol><p>The following question is quite straightforward, and there are many ways of answering it, but each will require you to manipulate the uniform motion equations to some extent. As you work out your answer think carefully about each of the procedures you are performing and write it out in words.</p><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_011"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 12</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>When the time on a certain stopwatch is 100 s, a vehicle is positioned at <i>x</i> = −2 m with respect to a certain onedimensional coordinate system. If the velocity of the vehicle is −12 m s<sup>−1</sup>, find its position when the stopwatch shows 250 s.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Since <i>v<sub>x</sub> </i> = constant in this case, the motion is described by the uniform motion equation <i>x</i> = <i>v<sub>x</sub>t</i> + <i>x</i><sub>0</sub>. However, on this occasion we are not given the value of <i>x</i><sub>0</sub>, though we are given enough information to work it out. Substituting <i>x</i> = −2 m and <i>t</i> = 100 s into the equation we find</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_006"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/1aea0b3d/s207_1_ie041i.gif" alt=""/></div><p>Adding 1200 m to both sides shows that 1198 m = <i>x</i><sub>0</sub>.</p><p>It follows that for this particular motion the equation of uniform motion takes the form <i>x</i> = (−12 m s<sup>−1</sup>)<i>t</i> + (1198 m).</p><p>Consequently, when <i>t</i> = 250 s,</p><div class="oucontentquote oucontentsbox" id="q001"><blockquote><p><i>x</i> = −3000 m + 1198 m = − 1802 m.</p></blockquote></div><p>An alternative method that requires slightly fewer manipulations may be based on the definition of the gradient and the given value of the velocity.</p></div></div></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection3.4
3.4 The equations of uniform motionS207_2<p>It has already been said that the straightline graph of any uniform motion can be represented by an equation of the general form</p><p/><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_009"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2e3eb78b/s207_1_ue003i.gif" alt=""/></div><p>where <i>A</i> and <i>B</i> are constants. Different cases of uniform motion simply correspond to different values for the constants <i>A</i> and <i>B</i>. Let us now investigate this equation to see how it conveys information about gradients and intercepts, or equivalently, uniform velocities and initial positions.</p><p>To start with, note that according to Equation 3, the position of the particle when <i>t</i> = 0 is just <i>x</i> = <i>B</i>. Thus <i>B</i> represents the initial position of the particle, the intercept of its positiontime graph.</p><p>In a similar way, note that according to Equation 3, at time <i>t</i> = <i>t</i><sub>1</sub> the position of the particle, let's call it <i>x</i><sub>1</sub>, is given by <i>x</i><sub>1</sub> = <i>At</i><sub>1</sub> + <i>B</i>; and at some later time <i>t</i> = <i>t</i><sub>2</sub> the position of the particle is <i>x</i><sub>2</sub> = <i>At</i><sub>2</sub> + <i>B</i>. Now, as we have already seen, the velocity of a uniformly moving particle is defined by</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_010"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/680a03dc/s207_1_ue005i.gif" alt=""/></div><p>so, substituting the expressions for <i>x</i><sub>1</sub> and <i>x</i><sub>2</sub> that we have just obtained from Equation 3 we find that in this particular case</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_011"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/bfc7c298/s207_1_ie005i.gif" alt=""/></div><p>Thus, the constant <i>A</i> in Equation 3 represents the particle's velocity, i.e. the gradient of its positiontime graph.</p><p>We can now read the equation of a straight line, interpreting it graphically or physically. It is</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_012"><a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=2298&extra=thumbnailfigure_idp3588880" title="View larger image"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/785ca7fb/s207_1_i001i.jpg" alt=""/></a><div class="oucontentfiguretext"><div class="oucontentthumbnaillink"><a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=2298&extra=thumbnailfigure_idp3588880">View larger image</a></div><div class="oucontentcaption oucontentnonumber oucontentcaptionplaceholder"> </div></div><a id="back_thumbnailfigure_idp3588880"></a></div><p>The physical significance of the constants <i>A</i> and <i>B</i> can be emphasised by using the symbols <i>v</i><sub><i>x</i></sub> and <i>x</i><sub>0</sub> in their place. Doing this, we obtain the standard form of the <b>uniform motion equations</b></p><div class="oucontentbox oucontentsheavybox1 oucontentsbox
oucontentsnoheading
" id="box001_006"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p/><div class="oucontentequation oucontentequationequation oucontentnocaption" id="equ001_013"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7e240e9c/s207_1_ue006i.gif" alt=""/></div></div></div></div><p>So, the uniform motion of the car we have been considering, which has velocity <i>v</i><sub><i>x</i></sub> = 3.0 m s<sup>−1</sup> and initial position <i>x</i><sub>0</sub> = 30 m, can be described by the equation</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_015"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/509d1a7f/s207_1_ie006i.gif" alt=""/></div><p>This contains just as much information as the graph in Figure 12b.</p><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_010"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 11</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Write down the equation that describes the uniform motion of the pedestrian who was the subject of Figure 12a.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>The equation will have the general form of Equation 6a, with velocity <i>v<sub>x</sub> </i> = 1 m s<sup>−1</sup> (found from the gradient) and initial position <i>x</i><sub>0</sub> = − 20 m (found from the value of x when <i>t</i> = 0). Consequently the required equation may be written:</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_005"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/06da0324/s207_1_ie040i.gif" alt=""/></div></div></div></div></div><p>A pictorial interpretation of Equation 6a is given in Figure 16.</p><div class="oucontentfigure" style="width:511px;" id="fig001_01573"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/1c63648d/s207_2_014i.jpg" alt="" width="511" height="136" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 16:</b> A pictorial interpretation of the equation of uniform motion</span></div></div></div><p>Equations have the great advantage that they can be rearranged to make them fit the problem at hand. For instance, if you were flying at constant velocity from Paris to Rome, you might well be more interested in your displacement from Paris rather than your position in some arbitrary coordinate system. In mathematical terms, you might be more interested in <i>s<sub>x</sub></i> = <i>x</i> − <i>x</i><sub>0</sub> rather than <i>x</i> itself. Equation 6a can be rearranged to suit this interest by subtracting <i>x</i> <sub>0</sub> from each side:</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_016"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/48714241/s207_1_ie007i.gif" alt=""/></div><p>and this may be rewritten in the compact form</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox
oucontentsnoheading
" id="box001_007"><div class="oucontentouterbox"><div class="oucontentinnerbox"><div class="oucontentequation oucontentequationequation oucontentnocaption" id="fig001_01676"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2bc501a6/s207_1_ue007i.gif" alt=""/></div></div></div></div><p>Remembering that <i>v</i><sub><i>x</i></sub> is a constant when the motion is uniform, Equation 7 is equivalent to Equation 6a. It relates the displacement from the initial position directly to the time of flight and thus facilitates working out the distance travelled,</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/6c1e4df0/s207_1_ue008i.gif" alt="" width="283" height="16" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>The crucial point to remember when manipulating an equation is that <i>both sides of an equation represent the same value</i>. So, if you add or subtract a term on one side of an equation, you must add or subtract an identical term on the other side. The same principle applies to multiplication and division, and to other operations such as taking magnitudes, squaring or taking square roots. (In the latter case you must remember that a positive quantity generally has two square roots; 2 is a square root of 4, but so is −2.) There are two rules to bear in mind when rearranging equations:</p><ol class="oucontentnumbered"><li><p>When one side of an equation consists of several terms added together, each of those terms must be treated in the same way. (It's no good multiplying the first term on one side of an equation by some quantity if you forget to do the same to all the other terms on that side.)</p></li><li><p>When dividing both sides of an equation by some quantity you must ensure that the quantity is <i>not</i> zero. (For instance, it is only legitimate to divide both sides of Equation 7 by <i>x</i> − <i>x</i><sub>0</sub> <i>if x</i> ≠ <i>x</i><sub>0</sub>, i.e. if <i>x</i> is <i>not equal</i> to <i>x</i><sub>0</sub>.)</p></li></ol><p>The following question is quite straightforward, and there are many ways of answering it, but each will require you to manipulate the uniform motion equations to some extent. As you work out your answer think carefully about each of the procedures you are performing and write it out in words.</p><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_011"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 12</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>When the time on a certain stopwatch is 100 s, a vehicle is positioned at <i>x</i> = −2 m with respect to a certain onedimensional coordinate system. If the velocity of the vehicle is −12 m s<sup>−1</sup>, find its position when the stopwatch shows 250 s.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Since <i>v<sub>x</sub> </i> = constant in this case, the motion is described by the uniform motion equation <i>x</i> = <i>v<sub>x</sub>t</i> + <i>x</i><sub>0</sub>. However, on this occasion we are not given the value of <i>x</i><sub>0</sub>, though we are given enough information to work it out. Substituting <i>x</i> = −2 m and <i>t</i> = 100 s into the equation we find</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_006"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/1aea0b3d/s207_1_ie041i.gif" alt=""/></div><p>Adding 1200 m to both sides shows that 1198 m = <i>x</i><sub>0</sub>.</p><p>It follows that for this particular motion the equation of uniform motion takes the form <i>x</i> = (−12 m s<sup>−1</sup>)<i>t</i> + (1198 m).</p><p>Consequently, when <i>t</i> = 250 s,</p><div class="oucontentquote oucontentsbox" id="q001"><blockquote><p><i>x</i> = −3000 m + 1198 m = − 1802 m.</p></blockquote></div><p>An alternative method that requires slightly fewer manipulations may be based on the definition of the gradient and the given value of the velocity.</p></div></div></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

3.5 Velocitytime and speedtime graphs
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection3.5
Tue, 12 Apr 2016 23:00:00 GMT
<p>Just as we may plot the positiontime graph or the displacementtime graph of a particular motion, so we may plot a <b>velocitytime graph</b> for that motion. By convention, velocity is plotted on the vertical axis (since velocity is the dependent variable) and time (the independent variable) is plotted on the horizontal axis. In the special case of uniform motion, the velocitytime graph takes a particularly simple form  it is just a horizontal line, i.e. the gradient is zero. Examples are given in Figure 17; notice that the velocity can be positive or negative, depending on the direction of motion.</p><div class="oucontentfigure" style="width:342px;" id="fig001_01781"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/dfe5c091/s207_2_015i.jpg" alt="" width="342" height="288" style="maxwidth:342px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 17:</b> Velocitytime graphs for uniform motion</span></div></div></div><p>Rather than plotting the velocity it is sometimes useful to plot the magnitude of the velocity, in other words, the speed. The resulting plots are <b>speedtime graphs</b> and examples are shown in Figure 18. Notice how all of the speeds are positive; in particular, the velocity of −1.5 m s<sup>−1</sup> in Figure 17 corresponds to a speed of 1.5 m s<sup>−1</sup> in Figure 18.</p><p>Clearly, a speedtime graph provides less information than a velocitytime graph, but it may be sufficient. The next time you buy a Ferrari, you may well enquire about its top speed, but you are unlikely to ask 'in which direction?'</p><div class="oucontentfigure" style="width:342px;" id="fig001_01883"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/889a31f5/s207_2_016i.jpg" alt="" width="342" height="198" style="maxwidth:342px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 18:</b> Speedtime graphs for uniform motion, corresponding to the velocitytime graphs shown in Figure 17 </span></div></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection3.5
3.5 Velocitytime and speedtime graphsS207_2<p>Just as we may plot the positiontime graph or the displacementtime graph of a particular motion, so we may plot a <b>velocitytime graph</b> for that motion. By convention, velocity is plotted on the vertical axis (since velocity is the dependent variable) and time (the independent variable) is plotted on the horizontal axis. In the special case of uniform motion, the velocitytime graph takes a particularly simple form  it is just a horizontal line, i.e. the gradient is zero. Examples are given in Figure 17; notice that the velocity can be positive or negative, depending on the direction of motion.</p><div class="oucontentfigure" style="width:342px;" id="fig001_01781"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/dfe5c091/s207_2_015i.jpg" alt="" width="342" height="288" style="maxwidth:342px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 17:</b> Velocitytime graphs for uniform motion</span></div></div></div><p>Rather than plotting the velocity it is sometimes useful to plot the magnitude of the velocity, in other words, the speed. The resulting plots are <b>speedtime graphs</b> and examples are shown in Figure 18. Notice how all of the speeds are positive; in particular, the velocity of −1.5 m s<sup>−1</sup> in Figure 17 corresponds to a speed of 1.5 m s<sup>−1</sup> in Figure 18.</p><p>Clearly, a speedtime graph provides less information than a velocitytime graph, but it may be sufficient. The next time you buy a Ferrari, you may well enquire about its top speed, but you are unlikely to ask 'in which direction?'</p><div class="oucontentfigure" style="width:342px;" id="fig001_01883"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/889a31f5/s207_2_016i.jpg" alt="" width="342" height="198" style="maxwidth:342px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 18:</b> Speedtime graphs for uniform motion, corresponding to the velocitytime graphs shown in Figure 17 </span></div></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

3.6 The signed area under a constant velocitytime graph
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection3.6
Tue, 12 Apr 2016 23:00:00 GMT
<p>There is a simple feature of uniform velocitytime graphs that will be particularly useful to know about when we come to consider nonuniform motion in the next section. It concerns the relationship between the velocitytime graph and the change in position over a given time interval. Consider the following problem. A vehicle travels at a velocity <i>v</i><sub><i>x</i></sub> = 12 m s<sup>−1</sup> for 4 s. By how much does its position change over that interval?</p><p>The answer, from Equation 7, is 48 m. However, for our present purposes it is more instructive to work from the definition of uniform velocity (Equation 5), which may be rearranged by multiplying both sides of the equation by (<i>t</i><sub>2</sub> − <i>t</i><sub>1</sub>) to give</p><div class="oucontentquote oucontentsbox" id="q002"><blockquote><p><i>x</i><sub>2</sub> − <i>x</i><sub>1</sub> = <i>v<sub>x</sub></i> (<i>t</i><sub>2</sub> − <i>t</i><sub>1</sub>).</p></blockquote></div><p>This tells us that the change in position during a given time interval is equal to the velocity multiplied by the time interval. So, a vehicle which travels at a constant velocity <i>v</i>﻿<sub><i>x</i> </sub> = 12 m s<sup>﻿−1</sup> over a time interval Δ<i>t</i> = 4 s will change its position coordinate by Δ<i>x</i> = 48 m.</p><div class="oucontentfigure" style="width:343px;" id="fig001_01987"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/34969113/s207_2_017i.jpg" alt="" width="343" height="277" style="maxwidth:343px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 19:</b> The area under a velocitytime graph</span></div></div></div><p>Now look at the velocitytime graph for this vehicle, which is given in Figure 19. The colourshaded region corresponding to the 4 s interval between <i>t</i> = 2 s and <i>t</i> = 6 s is an example of an <i>area under the graph</i>: that is an area bounded by the plotted line, the horizontal axis and two vertical lines. The important point to note about this region is its area  not its physical area in terms of m<sup>2</sup> of paper, but rather its area in terms of the units used on the axes of the graph. The area of a rectangle is the product of the lengths of two adjacent sides. So in Figure 19 the area under the graph is</p><div class="oucontentquote oucontentsbox" id="q003"><blockquote><p><i>v<sub>x</sub></i> × Δ<i>t</i> = (12 m s<sup>−1</sup>) × (4 s) = 48 m.</p></blockquote></div><p>Clearly, the area under the velocitytime graph, between the specified times, is exactly equal to the change in position coordinate between those times. In case you are worried that this relationship will break down if the velocity is negative, as in Figure 20, it should be added that areas that hang below the horizontal axis are conventionally regarded as negative areas. In this sense the area under a graph is a signed quantity that may be positive or negative; indeed, it is often referred to as the <b>signed area under a graph</b>.</p><p>Recognising the equality between the signed area under a uniform velocitytime graph and a change in position coordinate is not particularly helpful in itself, but it does pave the way for further developments, as does the next question.</p><div class="oucontentfigure" style="width:342px;" id="fig001_020"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/83d99053/s207_2_018i.jpg" alt="" width="342" height="331" style="maxwidth:342px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 20:</b> The area under a velocitytime graph is a signed quantity that may be negative</span></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_012"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 13</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>An athlete walks with a constant velocity <i>v</i>﻿<sub><i>x</i>1</sub> = 1 m s<sup>﻿−﻿1</sup> for 20 s, and then abruptly starts running with a constant velocity <i>v</i>﻿<sub>﻿<i>x</i>﻿2</sub> = 10 m s<sup>−﻿1</sup>, a velocity that is maintained for a further 20 s. What is the area under the corresponding velocitytime graph, and what physical interpretation could you give to that area?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>The relevant velocitytime graph is shown in Figure 21. In this case there are two rectangular areas to evaluate. Their total area is</p><div class="oucontentquote oucontentsbox" id="q004"><blockquote><p>(20 s) × (1 m s<sup>−1</sup>) + (20 s) × (10 m s<sup>−1</sup>) = 220 m.</p></blockquote></div><p>This is equal to the change in position over the full 40 s duration of the motion.</p><div class="oucontentfigure" style="width:392px;" id="fig001_02194"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/4a09e825/s207_2_040i.jpg" alt="" width="392" height="298" style="maxwidth:392px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 21:</b> The graph for Question 13. Note that the time values are not known, only the durations, which have been indicated by Δ<i>t</i></span></div></div></div></div></div></div></div><p>Here we are using a double subscript on the symbol <i>v</i>. The subscript <i>x</i> denotes the <i>x</i>direction as usual, and the 1 and 2 refer to the different stages of the motion.</p>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection3.6
3.6 The signed area under a constant velocitytime graphS207_2<p>There is a simple feature of uniform velocitytime graphs that will be particularly useful to know about when we come to consider nonuniform motion in the next section. It concerns the relationship between the velocitytime graph and the change in position over a given time interval. Consider the following problem. A vehicle travels at a velocity <i>v</i><sub><i>x</i></sub> = 12 m s<sup>−1</sup> for 4 s. By how much does its position change over that interval?</p><p>The answer, from Equation 7, is 48 m. However, for our present purposes it is more instructive to work from the definition of uniform velocity (Equation 5), which may be rearranged by multiplying both sides of the equation by (<i>t</i><sub>2</sub> − <i>t</i><sub>1</sub>) to give</p><div class="oucontentquote oucontentsbox" id="q002"><blockquote><p><i>x</i><sub>2</sub> − <i>x</i><sub>1</sub> = <i>v<sub>x</sub></i> (<i>t</i><sub>2</sub> − <i>t</i><sub>1</sub>).</p></blockquote></div><p>This tells us that the change in position during a given time interval is equal to the velocity multiplied by the time interval. So, a vehicle which travels at a constant velocity <i>v</i><sub><i>x</i> </sub> = 12 m s<sup>−1</sup> over a time interval Δ<i>t</i> = 4 s will change its position coordinate by Δ<i>x</i> = 48 m.</p><div class="oucontentfigure" style="width:343px;" id="fig001_01987"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/34969113/s207_2_017i.jpg" alt="" width="343" height="277" style="maxwidth:343px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 19:</b> The area under a velocitytime graph</span></div></div></div><p>Now look at the velocitytime graph for this vehicle, which is given in Figure 19. The colourshaded region corresponding to the 4 s interval between <i>t</i> = 2 s and <i>t</i> = 6 s is an example of an <i>area under the graph</i>: that is an area bounded by the plotted line, the horizontal axis and two vertical lines. The important point to note about this region is its area  not its physical area in terms of m<sup>2</sup> of paper, but rather its area in terms of the units used on the axes of the graph. The area of a rectangle is the product of the lengths of two adjacent sides. So in Figure 19 the area under the graph is</p><div class="oucontentquote oucontentsbox" id="q003"><blockquote><p><i>v<sub>x</sub></i> × Δ<i>t</i> = (12 m s<sup>−1</sup>) × (4 s) = 48 m.</p></blockquote></div><p>Clearly, the area under the velocitytime graph, between the specified times, is exactly equal to the change in position coordinate between those times. In case you are worried that this relationship will break down if the velocity is negative, as in Figure 20, it should be added that areas that hang below the horizontal axis are conventionally regarded as negative areas. In this sense the area under a graph is a signed quantity that may be positive or negative; indeed, it is often referred to as the <b>signed area under a graph</b>.</p><p>Recognising the equality between the signed area under a uniform velocitytime graph and a change in position coordinate is not particularly helpful in itself, but it does pave the way for further developments, as does the next question.</p><div class="oucontentfigure" style="width:342px;" id="fig001_020"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/83d99053/s207_2_018i.jpg" alt="" width="342" height="331" style="maxwidth:342px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 20:</b> The area under a velocitytime graph is a signed quantity that may be negative</span></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_012"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 13</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>An athlete walks with a constant velocity <i>v</i><sub><i>x</i>1</sub> = 1 m s<sup>−1</sup> for 20 s, and then abruptly starts running with a constant velocity <i>v</i><sub><i>x</i>2</sub> = 10 m s<sup>−1</sup>, a velocity that is maintained for a further 20 s. What is the area under the corresponding velocitytime graph, and what physical interpretation could you give to that area?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>The relevant velocitytime graph is shown in Figure 21. In this case there are two rectangular areas to evaluate. Their total area is</p><div class="oucontentquote oucontentsbox" id="q004"><blockquote><p>(20 s) × (1 m s<sup>−1</sup>) + (20 s) × (10 m s<sup>−1</sup>) = 220 m.</p></blockquote></div><p>This is equal to the change in position over the full 40 s duration of the motion.</p><div class="oucontentfigure" style="width:392px;" id="fig001_02194"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/4a09e825/s207_2_040i.jpg" alt="" width="392" height="298" style="maxwidth:392px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 21:</b> The graph for Question 13. Note that the time values are not known, only the durations, which have been indicated by Δ<i>t</i></span></div></div></div></div></div></div></div><p>Here we are using a double subscript on the symbol <i>v</i>. The subscript <i>x</i> denotes the <i>x</i>direction as usual, and the 1 and 2 refer to the different stages of the motion.</p>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

3.7 A note on straightline graphs and their gradients
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection3.7
Tue, 12 Apr 2016 23:00:00 GMT
<p>We end this section by reviewing some of the important features of straightline graphs, though we do so in terms of two general variables <i>z</i> and <i>y</i>, rather than <i>x</i> and <i>t</i>, in order to emphasise their generality. If the graph of <i>z</i> against <i>y</i> is a straight line of the kind shown in Figure 22, then <i>z</i> and <i>y</i> are related by an equation of the form</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_021"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/9f23a6d2/s207_1_ue009i.gif" alt=""/></div><p>where <i>m</i> and <i>c</i> are constants. Here <i>c</i> represents the intercept of the graph and is equal to the value of <i>z</i> at which the plotted line crosses the <i>z</i>axis (provided the <i>z</i>axis passes through <i>y</i> = 0). The constant <i>m</i> represents the gradient of the graph and is obtained by dividing the change in <i>z</i> by the corresponding change in <i>y</i>, taking full account of the sign of each change.</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_022"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/8e9f8c95/s207_1_ue010i.gif" alt=""/></div><div class="oucontentfigure oucontentmediamini" id="fig001_02299"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/a6a80b89/s207_2_019i.jpg" alt="" width="231" height="209" style="maxwidth:231px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 22:</b> A straightline graph of <i>z</i> against <i>y</i></span></div></div></div><p>You may find it useful to remember that the gradient of a graph is given by its 'rise' over its 'run'.</p><p>There are several points to notice about this definition.</p><ol class="oucontentnumbered"><li><p>It applies only to straightline graphs.</p></li><li><p>The units of the gradient are the units of <i>z</i> divided by the units of <i>y</i>.</p></li><li><p>The gradient of a straightline graph is a constant (a number, multiplied by an appropriate unit). The same constant is obtained, no matter which two points (see P<sub>1</sub> and P<sub>2</sub> in Figure 22) are used to determine it.</p></li><li><p>The gradient of a straightline graph can be positive, negative or zero. Equation 10 assigns a positive gradient to a graph sloping from bottom left to top right, as in Figure 22 for example, and a negative gradient to a graph sloping from top left to bottom right, as in Figure 15. A horizontal line has zero gradient.</p></li></ol><p>Although the appearance of a graph can be changed by plotting the points on different scales (compare Figures 23a and b for example) the gradient, defined by Equation 10, is independent of the shape or size of the graph paper or display screen.</p><div class="oucontentfigure" style="width:511px;" id="fig001_023101"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/8900f66d/s207_2_020i.jpg" alt="" width="511" height="183" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 23:</b> Two graphs constructed from the same data but plotted with different scales. The gradients are the same, 4 ×10<sup>3</sup> kg m<sup>−3</sup></span></div></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection3.7
3.7 A note on straightline graphs and their gradientsS207_2<p>We end this section by reviewing some of the important features of straightline graphs, though we do so in terms of two general variables <i>z</i> and <i>y</i>, rather than <i>x</i> and <i>t</i>, in order to emphasise their generality. If the graph of <i>z</i> against <i>y</i> is a straight line of the kind shown in Figure 22, then <i>z</i> and <i>y</i> are related by an equation of the form</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_021"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/9f23a6d2/s207_1_ue009i.gif" alt=""/></div><p>where <i>m</i> and <i>c</i> are constants. Here <i>c</i> represents the intercept of the graph and is equal to the value of <i>z</i> at which the plotted line crosses the <i>z</i>axis (provided the <i>z</i>axis passes through <i>y</i> = 0). The constant <i>m</i> represents the gradient of the graph and is obtained by dividing the change in <i>z</i> by the corresponding change in <i>y</i>, taking full account of the sign of each change.</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_022"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/8e9f8c95/s207_1_ue010i.gif" alt=""/></div><div class="oucontentfigure oucontentmediamini" id="fig001_02299"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/a6a80b89/s207_2_019i.jpg" alt="" width="231" height="209" style="maxwidth:231px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 22:</b> A straightline graph of <i>z</i> against <i>y</i></span></div></div></div><p>You may find it useful to remember that the gradient of a graph is given by its 'rise' over its 'run'.</p><p>There are several points to notice about this definition.</p><ol class="oucontentnumbered"><li><p>It applies only to straightline graphs.</p></li><li><p>The units of the gradient are the units of <i>z</i> divided by the units of <i>y</i>.</p></li><li><p>The gradient of a straightline graph is a constant (a number, multiplied by an appropriate unit). The same constant is obtained, no matter which two points (see P<sub>1</sub> and P<sub>2</sub> in Figure 22) are used to determine it.</p></li><li><p>The gradient of a straightline graph can be positive, negative or zero. Equation 10 assigns a positive gradient to a graph sloping from bottom left to top right, as in Figure 22 for example, and a negative gradient to a graph sloping from top left to bottom right, as in Figure 15. A horizontal line has zero gradient.</p></li></ol><p>Although the appearance of a graph can be changed by plotting the points on different scales (compare Figures 23a and b for example) the gradient, defined by Equation 10, is independent of the shape or size of the graph paper or display screen.</p><div class="oucontentfigure" style="width:511px;" id="fig001_023101"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/8900f66d/s207_2_020i.jpg" alt="" width="511" height="183" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 23:</b> Two graphs constructed from the same data but plotted with different scales. The gradients are the same, 4 ×10<sup>3</sup> kg m<sup>−3</sup></span></div></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

4.1 Instantaneous velocity
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.1
Tue, 12 Apr 2016 23:00:00 GMT
<p>Uniform motion is simple to describe, but is rarely achieved in practice. Most objects do not move at a precisely constant velocity. If you drop an apple it will fall downwards, but it will pick up speed as it does so (Figure 24), and if you drive along a straight road you are likely to encounter some traffic that will force you to vary your speed from time to time. For the most part, real motions are <b>nonuniform motions</b>.</p><div class="oucontentfigure oucontentmediamini" id="fig001_024"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7c421650/s207_2_021i.jpg" alt="" width="252" height="502" style="maxwidth:252px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 24:</b> A falling apple provides an example of nonuniform motion. A sequence of pictures taken at equal intervals of time reveals the increasing speed of the apple as it falls</span></div></div></div><p>Figure 25 shows the positiontime graph of an object that has an increasing velocity over the period <i>t</i> = 0 to <i>t</i> = 20 s; a car accelerating from rest. As you can see, the positiontime graph is curved. There is relatively little change in position during the first few seconds of the motion but as the velocity increases the car is able to change its position by increasingly large amounts over a given interval of time. This is shown by the increasing steepness of the graph. In everyday language we would say that the graph has an increasing gradient, but you saw in the last section that the term gradient has a precise technical meaning in the context of straightline graphs. Is it legitimate to extend this terminology to cover curved graphs, and if so, how exactly should it be done?</p><div class="oucontentfigure" style="width:342px;" id="fig001_025106"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/f5b8d7e7/s207_2_022i.jpg" alt="" width="342" height="266" style="maxwidth:342px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 25:</b> The positiontime graph for a car accelerating from rest</span></div></div></div><p>Extending the concept of gradient to the case of curved graphs is actually quite straightforward. The crucial point to recognise is that if you look closely enough at a small part of a smooth curve, then it generally becomes indistinguishable from a straight line. (In a similar way, the surface of the Earth is clearly curved when viewed from space, but each region is approximately flat when seen closeup.) So, if we choose a point on a curve we can usually draw a straight line passing through that point which has the same slope as the curve at the point of contact. This straight line is said to be the <b>tangent</b> to the curve at the point in question. Now, we already know how to determine the gradient of a straight line, so we can define the gradient at any point on a curve to be the gradient of the tangent to the curve at that point, provided the curve is sufficiently smooth that a tangent exists.</p><p>Figure 26 repeats the positiontime graph of the accelerating car, but this time tangents have been added at <i>t</i> = 5 s and <i>t</i> = 10 s. The gradient of any such tangent represents a velocity and is referred to as the <b>instantaneous velocity</b> of the car at the relevant time. At least, that's what it should be called; in practice the word 'instantaneous' is often omitted, so references to 'velocity' should generally be taken to mean 'instantaneous velocity'. Allowing ourselves this informality, we can say:</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox oucontentsnoheading " id="box001_008"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>velocity at time <i>t</i> = gradient of positiontime graph at time <i>t</i>.</p></div></div></div><p>Although we have had to extend the meaning of gradient, we can still regard it as a measure of the rate of change of one variable with respect to another, so we can also say:</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox oucontentsnoheading " id="box001_009"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>velocity at time <i>t</i> = rate of change of position with respect to time at time <i>t</i>.</p></div></div></div><div class="oucontentfigure" style="width:511px;" id="fig001_026110"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/a8e8d041/s207_2_023i.jpg" alt="" width="511" height="381" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 26:</b> The instantaneous velocity at <i>t</i> = 5 s and at <i>t</i> = 10 s is determined by the gradient of the tangent to the positiontime graph at each of those times</span></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_013"><div class="oucontentouterbox"><h3 class="oucontenth3 oucontentheading oucontentnonumber">Question 14</h3><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Estimate the (instantaneous) velocity of the car at <i>t</i> = 5 s and at <i>t</i> = 10 s (from Figure 26), and write down your answers taking care to distinguish one velocity from the other.</p></div>
<div class="oucontentsaqanswer"><h4 class="oucontenth4">Answer</h4><p>The relevant tangents are shown in Figure 26. Their gradients give the following estimates for the instantaneous velocities</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_009"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/c923f3ff/s207_1_ie044i.gif" alt=""/></div></div></div></div></div><p>How did you distinguish the velocity at <i>t</i> = 5 s from the velocity at <i>t</i> = 10 s? The conventional method is to use a common symbol for velocity, <i>v</i>﻿<sub><i>x</i></sub>, but to follow it by the relevant value of time enclosed in parentheses, as in <i>v</i>﻿<sub><i>x</i></sub>﻿(﻿5 s﻿) and <i>v</i>﻿﻿<sub><i>x</i></sub>﻿(﻿10 s﻿). This notation can also be used to indicate the velocity at any time <i>t</i>, by writing <i>v</i>﻿<sub>﻿<i>x</i></sub>(<i>t</i>), even if the value of <i>t</i> is unspecified.</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox oucontentsnoheading " id="box0001_0010"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>It is important to remember that <i>v</i>﻿<sub><i>x</i>﻿</sub>(﻿<i>t</i>﻿) represents the (instantaneous) velocity at time <i>t</i>. It does not mean <i>v</i>﻿﻿<sub><i>x</i> </sub> multiplied by <i>t</i>.</p></div></div></div><p>You will not be surprised to learn that the positive quantity ﻿<i>v</i>﻿<sub><i>x</i>﻿</sub>(﻿<i>t</i>﻿)﻿ representing the magnitude of the instantaneous velocity at time <i>t</i> is called the <b>instantaneous speed</b> at time <i>t</i>. If we denote this by <i>v</i>﻿(﻿<i>t</i>﻿), we can write</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/e116bdf8/s207_1_ue011i.gif" alt="" width="283" height="16" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>Speed and velocity are measured in the same units, m s<sup>−1</sup>. Some typical values of physically interesting speeds are listed in Figure 27.</p><div class="oucontentfigure" style="width:511px;" id="fig001_027115"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/816a94db/s207_2_024i.jpg" alt="" width="511" height="680" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 27:</b> Some physically interesting speeds</span></div></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.1
4.1 Instantaneous velocityS207_2<p>Uniform motion is simple to describe, but is rarely achieved in practice. Most objects do not move at a precisely constant velocity. If you drop an apple it will fall downwards, but it will pick up speed as it does so (Figure 24), and if you drive along a straight road you are likely to encounter some traffic that will force you to vary your speed from time to time. For the most part, real motions are <b>nonuniform motions</b>.</p><div class="oucontentfigure oucontentmediamini" id="fig001_024"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7c421650/s207_2_021i.jpg" alt="" width="252" height="502" style="maxwidth:252px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 24:</b> A falling apple provides an example of nonuniform motion. A sequence of pictures taken at equal intervals of time reveals the increasing speed of the apple as it falls</span></div></div></div><p>Figure 25 shows the positiontime graph of an object that has an increasing velocity over the period <i>t</i> = 0 to <i>t</i> = 20 s; a car accelerating from rest. As you can see, the positiontime graph is curved. There is relatively little change in position during the first few seconds of the motion but as the velocity increases the car is able to change its position by increasingly large amounts over a given interval of time. This is shown by the increasing steepness of the graph. In everyday language we would say that the graph has an increasing gradient, but you saw in the last section that the term gradient has a precise technical meaning in the context of straightline graphs. Is it legitimate to extend this terminology to cover curved graphs, and if so, how exactly should it be done?</p><div class="oucontentfigure" style="width:342px;" id="fig001_025106"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/f5b8d7e7/s207_2_022i.jpg" alt="" width="342" height="266" style="maxwidth:342px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 25:</b> The positiontime graph for a car accelerating from rest</span></div></div></div><p>Extending the concept of gradient to the case of curved graphs is actually quite straightforward. The crucial point to recognise is that if you look closely enough at a small part of a smooth curve, then it generally becomes indistinguishable from a straight line. (In a similar way, the surface of the Earth is clearly curved when viewed from space, but each region is approximately flat when seen closeup.) So, if we choose a point on a curve we can usually draw a straight line passing through that point which has the same slope as the curve at the point of contact. This straight line is said to be the <b>tangent</b> to the curve at the point in question. Now, we already know how to determine the gradient of a straight line, so we can define the gradient at any point on a curve to be the gradient of the tangent to the curve at that point, provided the curve is sufficiently smooth that a tangent exists.</p><p>Figure 26 repeats the positiontime graph of the accelerating car, but this time tangents have been added at <i>t</i> = 5 s and <i>t</i> = 10 s. The gradient of any such tangent represents a velocity and is referred to as the <b>instantaneous velocity</b> of the car at the relevant time. At least, that's what it should be called; in practice the word 'instantaneous' is often omitted, so references to 'velocity' should generally be taken to mean 'instantaneous velocity'. Allowing ourselves this informality, we can say:</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox
oucontentsnoheading
" id="box001_008"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>velocity at time <i>t</i> = gradient of positiontime graph at time <i>t</i>.</p></div></div></div><p>Although we have had to extend the meaning of gradient, we can still regard it as a measure of the rate of change of one variable with respect to another, so we can also say:</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox
oucontentsnoheading
" id="box001_009"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>velocity at time <i>t</i> = rate of change of position with respect to time at time <i>t</i>.</p></div></div></div><div class="oucontentfigure" style="width:511px;" id="fig001_026110"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/a8e8d041/s207_2_023i.jpg" alt="" width="511" height="381" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 26:</b> The instantaneous velocity at <i>t</i> = 5 s and at <i>t</i> = 10 s is determined by the gradient of the tangent to the positiontime graph at each of those times</span></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_013"><div class="oucontentouterbox"><h3 class="oucontenth3 oucontentheading oucontentnonumber">Question 14</h3><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Estimate the (instantaneous) velocity of the car at <i>t</i> = 5 s and at <i>t</i> = 10 s (from Figure 26), and write down your answers taking care to distinguish one velocity from the other.</p></div>
<div class="oucontentsaqanswer"><h4 class="oucontenth4">Answer</h4><p>The relevant tangents are shown in Figure 26. Their gradients give the following estimates for the instantaneous velocities</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_009"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/c923f3ff/s207_1_ie044i.gif" alt=""/></div></div></div></div></div><p>How did you distinguish the velocity at <i>t</i> = 5 s from the velocity at <i>t</i> = 10 s? The conventional method is to use a common symbol for velocity, <i>v</i><sub><i>x</i></sub>, but to follow it by the relevant value of time enclosed in parentheses, as in <i>v</i><sub><i>x</i></sub>(5 s) and <i>v</i><sub><i>x</i></sub>(10 s). This notation can also be used to indicate the velocity at any time <i>t</i>, by writing <i>v</i><sub><i>x</i></sub>(<i>t</i>), even if the value of <i>t</i> is unspecified.</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox
oucontentsnoheading
" id="box0001_0010"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>It is important to remember that <i>v</i><sub><i>x</i></sub>(<i>t</i>) represents the (instantaneous) velocity at time <i>t</i>. It does not mean <i>v</i><sub><i>x</i> </sub> multiplied by <i>t</i>.</p></div></div></div><p>You will not be surprised to learn that the positive quantity <i>v</i><sub><i>x</i></sub>(<i>t</i>) representing the magnitude of the instantaneous velocity at time <i>t</i> is called the <b>instantaneous speed</b> at time <i>t</i>. If we denote this by <i>v</i>(<i>t</i>), we can write</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/e116bdf8/s207_1_ue011i.gif" alt="" width="283" height="16" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>Speed and velocity are measured in the same units, m s<sup>−1</sup>. Some typical values of physically interesting speeds are listed in Figure 27.</p><div class="oucontentfigure" style="width:511px;" id="fig001_027115"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/816a94db/s207_2_024i.jpg" alt="" width="511" height="680" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 27:</b> Some physically interesting speeds</span></div></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

4.2 Instantaneous acceleration
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.2
Tue, 12 Apr 2016 23:00:00 GMT
<p>The procedure of Question 15 for determining the instantaneous velocity of the car can be carried out for a whole set of different times and the resulting values of <i>v</i>﻿﻿<sub><i>x</i> </sub> can be plotted against <i>t</i> to form a graph. This has been done in Figure 28, which shows how the velocity varies with time. At time <i>t</i> = 0 s, the car has zero velocity because it starts from rest. At later times, the velocity is positive because the car moves in the direction of increasing <i>x</i>. The velocity increases rapidly at first, as the car picks up speed. Subsequently, the velocity increases more slowly, and eventually the car settles down to a steady velocity of just over 3 m s<sup>−1</sup>. We have already come across graphs of this general type in Section 3; they are known as velocitytime graphs. The crucial new feature here is that the velocity now depends on time.</p><div class="oucontentfigure" style="width:343px;" id="fig001_028118"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/df42f99b/s207_2_025i.jpg" alt="" width="343" height="206" style="maxwidth:343px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 28:</b> A velocitytime graph for the moving car</span></div></div></div><p>The time dependence of velocity can have dramatic consequences. If you are on board a train, moving at a constant velocity, you might not even be aware of your motion and you will have no difficulty in, say, drinking a cup of coffee. However, drinking coffee can become distinctly hazardous if the driver suddenly changes the velocity of the train by putting on the brakes. In such situations the important physical quantity is the <i>rate of change of velocity with respect to time</i>, as measured by the gradient of the velocitytime graph. This is the quantity that we usually call acceleration, though once again it should more properly be called <b>instantaneous acceleration</b>. Thus</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox oucontentsnoheading " id="box001_011"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>acceleration at time <i>t</i> = rate of change of velocity with respect to time at time <i>t</i>.</p></div></div></div><p>or, if you prefer</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox oucontentsnoheading " id="box001_012"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>acceleration at time <i>t</i> = gradient of velocitytime graph at time <i>t</i>.</p></div></div></div><p>Acceleration is a key idea in physics. It was Newton's recognition of the crucial role that acceleration played in determining the link between motion and force that formed the centrepiece of the Newtonian revolution. The detailed study of that revolution would take us too far from our present theme. For the moment let's concentrate on some basic questions about acceleration itself. In particular, in what units should acceleration be measured, and what are typical values of acceleration in various physical contexts? The first of these you can answer for yourself.</p><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="ique001_002"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 15</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>What are suitable SI units for the measurement of acceleration?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Since acceleration is the rate of change of velocity with respect to time, the units of acceleration are the units of velocity (m s﻿<sup>−﻿1</sup>﻿) divided by the units of time (﻿s﻿), so acceleration is measured in metres per second per second, which is abbreviated to m s<sup>﻿−﻿2</sup>.</p></div></div></div></div><p>As for typical values of acceleration, some physically interesting values are shown in Figure 29.</p><div class="oucontentfigure" style="width:511px;" id="fig001_029122"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7669ce7f/s207_2_026i.jpg" alt="" width="511" height="683" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"><b>Figure 29:</b> Some physically interesting values of acceleration</span></div></div></div><p>When dealing with nonuniform motion along the <i>x</i>axis, the symbol <i>a﻿<sub>x</sub>﻿</i>(﻿<i>t</i>﻿) is normally used to denote instantaneous acceleration. As usual, <i>a﻿<sub>x</sub>﻿</i>(﻿<i>t</i>﻿) will be positive if the velocity is increasing with time, though, as you will see below, this statement needs careful interpretation. In contrast to the relationship between velocity and speed, there is no special name for the magnitude of an acceleration, though we shall use the symbol <i>a</i>﻿(﻿<i>t</i>﻿) for this quantity, so we may write</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_028"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/5b327d30/s207_1_ue012i.gif" alt=""/></div><p>In physics, the concept of acceleration is precise and quantitative. It is important to realise that this precise definition differs, in some respects, from everyday usage. In ordinary speech, 'accelerating' is a synonym for 'speeding up'. This is not true in physics. In physics, a particle accelerates if it changes its velocity in <i>any</i> way. A particle travelling along a straight line may accelerate by speeding up <i>or</i> by slowing down. It is tempting to suppose that a positive acceleration corresponds to speeding up and a negative acceleration corresponds to slowing down, but this is not always true either, as the following exercise shows.</p><div class="oucontentfigure oucontentmediamini" id="fig001_030125"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/261785dc/s207_2_027i.jpg" alt="" width="235" height="181" style="maxwidth:235px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 30:</b> The velocitytime graph used in Question 16</span></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_014"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 16</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>The velocitytime graph of Figure 30 is divided into four regions, marked AD.</p><p>(a) In which regions does the particle move in the direction of increasing <i>x</i>? In which regions is it moving in the direction of decreasing <i>x</i>?</p><p>(b) In which regions does the particle speed up? In which regions does it slow down?</p><p>(c) In which regions does the particle have positive acceleration? In which regions does the particle have negative acceleration?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) The particle moves in the direction of increasing <i>x</i> when its velocity is positive. This occurs in regions A and B. Conversely, the particle moves in the direction of decreasing <i>x</i> when its velocity is negative. This occurs in regions C and D.</p><p>(b) The particle is speeding up when the <i>magnitude</i> of its velocity is increasing. This occurs in regions A and C. Conversely, the particle is slowing down in regions B and D.</p><p>(c) The particle will have a positive acceleration when the gradient of the velocitytime graph is positive. This occurs in regions A and D. Conversely the acceleration is negative in regions B and C.</p></div></div></div></div><p>So, for a particle with negative velocity, an increase in velocity (a positive acceleration) may result in a decrease in speed. You may think that it is a nuisance for physics to use words in such a nonstandard way. However, the definitions of velocity and acceleration given in this course are essential if we are to develop a study of kinematics that is both simple and comprehensive. It does mean, however, that you must be careful in using words like velocity, speed and acceleration. In everyday speech, the word deceleration is used to mean 'slowing down', but this term is seldom encountered in physics since it is already covered by the scientific definition of acceleration.</p>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.2
4.2 Instantaneous accelerationS207_2<p>The procedure of Question 15 for determining the instantaneous velocity of the car can be carried out for a whole set of different times and the resulting values of <i>v</i><sub><i>x</i> </sub> can be plotted against <i>t</i> to form a graph. This has been done in Figure 28, which shows how the velocity varies with time. At time <i>t</i> = 0 s, the car has zero velocity because it starts from rest. At later times, the velocity is positive because the car moves in the direction of increasing <i>x</i>. The velocity increases rapidly at first, as the car picks up speed. Subsequently, the velocity increases more slowly, and eventually the car settles down to a steady velocity of just over 3 m s<sup>−1</sup>. We have already come across graphs of this general type in Section 3; they are known as velocitytime graphs. The crucial new feature here is that the velocity now depends on time.</p><div class="oucontentfigure" style="width:343px;" id="fig001_028118"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/df42f99b/s207_2_025i.jpg" alt="" width="343" height="206" style="maxwidth:343px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 28:</b> A velocitytime graph for the moving car</span></div></div></div><p>The time dependence of velocity can have dramatic consequences. If you are on board a train, moving at a constant velocity, you might not even be aware of your motion and you will have no difficulty in, say, drinking a cup of coffee. However, drinking coffee can become distinctly hazardous if the driver suddenly changes the velocity of the train by putting on the brakes. In such situations the important physical quantity is the <i>rate of change of velocity with respect to time</i>, as measured by the gradient of the velocitytime graph. This is the quantity that we usually call acceleration, though once again it should more properly be called <b>instantaneous acceleration</b>. Thus</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox
oucontentsnoheading
" id="box001_011"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>acceleration at time <i>t</i> = rate of change of velocity with respect to time at time <i>t</i>.</p></div></div></div><p>or, if you prefer</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox
oucontentsnoheading
" id="box001_012"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>acceleration at time <i>t</i> = gradient of velocitytime graph at time <i>t</i>.</p></div></div></div><p>Acceleration is a key idea in physics. It was Newton's recognition of the crucial role that acceleration played in determining the link between motion and force that formed the centrepiece of the Newtonian revolution. The detailed study of that revolution would take us too far from our present theme. For the moment let's concentrate on some basic questions about acceleration itself. In particular, in what units should acceleration be measured, and what are typical values of acceleration in various physical contexts? The first of these you can answer for yourself.</p><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="ique001_002"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 15</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>What are suitable SI units for the measurement of acceleration?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Since acceleration is the rate of change of velocity with respect to time, the units of acceleration are the units of velocity (m s<sup>−1</sup>) divided by the units of time (s), so acceleration is measured in metres per second per second, which is abbreviated to m s<sup>−2</sup>.</p></div></div></div></div><p>As for typical values of acceleration, some physically interesting values are shown in Figure 29.</p><div class="oucontentfigure" style="width:511px;" id="fig001_029122"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7669ce7f/s207_2_026i.jpg" alt="" width="511" height="683" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"><b>Figure 29:</b> Some physically interesting values of acceleration</span></div></div></div><p>When dealing with nonuniform motion along the <i>x</i>axis, the symbol <i>a<sub>x</sub></i>(<i>t</i>) is normally used to denote instantaneous acceleration. As usual, <i>a<sub>x</sub></i>(<i>t</i>) will be positive if the velocity is increasing with time, though, as you will see below, this statement needs careful interpretation. In contrast to the relationship between velocity and speed, there is no special name for the magnitude of an acceleration, though we shall use the symbol <i>a</i>(<i>t</i>) for this quantity, so we may write</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_028"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/5b327d30/s207_1_ue012i.gif" alt=""/></div><p>In physics, the concept of acceleration is precise and quantitative. It is important to realise that this precise definition differs, in some respects, from everyday usage. In ordinary speech, 'accelerating' is a synonym for 'speeding up'. This is not true in physics. In physics, a particle accelerates if it changes its velocity in <i>any</i> way. A particle travelling along a straight line may accelerate by speeding up <i>or</i> by slowing down. It is tempting to suppose that a positive acceleration corresponds to speeding up and a negative acceleration corresponds to slowing down, but this is not always true either, as the following exercise shows.</p><div class="oucontentfigure oucontentmediamini" id="fig001_030125"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/261785dc/s207_2_027i.jpg" alt="" width="235" height="181" style="maxwidth:235px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 30:</b> The velocitytime graph used in Question 16</span></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_014"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 16</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>The velocitytime graph of Figure 30 is divided into four regions, marked AD.</p><p>(a) In which regions does the particle move in the direction of increasing <i>x</i>? In which regions is it moving in the direction of decreasing <i>x</i>?</p><p>(b) In which regions does the particle speed up? In which regions does it slow down?</p><p>(c) In which regions does the particle have positive acceleration? In which regions does the particle have negative acceleration?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) The particle moves in the direction of increasing <i>x</i> when its velocity is positive. This occurs in regions A and B. Conversely, the particle moves in the direction of decreasing <i>x</i> when its velocity is negative. This occurs in regions C and D.</p><p>(b) The particle is speeding up when the <i>magnitude</i> of its velocity is increasing. This occurs in regions A and C. Conversely, the particle is slowing down in regions B and D.</p><p>(c) The particle will have a positive acceleration when the gradient of the velocitytime graph is positive. This occurs in regions A and D. Conversely the acceleration is negative in regions B and C.</p></div></div></div></div><p>So, for a particle with negative velocity, an increase in velocity (a positive acceleration) may result in a decrease in speed. You may think that it is a nuisance for physics to use words in such a nonstandard way. However, the definitions of velocity and acceleration given in this course are essential if we are to develop a study of kinematics that is both simple and comprehensive. It does mean, however, that you must be careful in using words like velocity, speed and acceleration. In everyday speech, the word deceleration is used to mean 'slowing down', but this term is seldom encountered in physics since it is already covered by the scientific definition of acceleration.</p>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

4.3 A note on functions and derivatives
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3
Tue, 12 Apr 2016 23:00:00 GMT
<p>This subsection introduces two crucially important mathematical ideas, <i>functions</i> and <i>derivatives</i>, both of which are used throughout physics.</p>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3
4.3 A note on functions and derivativesS207_2<p>This subsection introduces two crucially important mathematical ideas, <i>functions</i> and <i>derivatives</i>, both of which are used throughout physics.</p>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

Functions and the function notation
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.1
Tue, 12 Apr 2016 23:00:00 GMT
<p>In Figure 25, the position <i>x</i> of the car depends on the time <i>t</i>. The graph associates a particular value of <i>x</i> with each value of <i>t</i> over the plotted range. In other circumstances we might know an equation that associates a value of <i>x</i> with each value of <i>t</i>, as in the case of the equation <i>x</i> = <i>At</i> + <i>B</i> that we discussed in Section 3. You can invent countless other ways in which <i>x</i> depends on <i>t</i>: for instance <i>x</i> = <i>At</i><sup>2</sup> + <i>B</i> or <i>x</i> = <i>At</i> + <i>Bt</i>﻿<sup>2</sup>. In all such cases we describe the dependence of <i>x</i> on <i>t</i> by saying that <i>x</i> is a <b>function</b> of <i>t</i>. This terminology is widely used and is certainly not restricted to <i>x</i> and <i>t</i>. In the example of nonuniform motion we have just been discussing, the instantaneous velocity <i>v</i>﻿<sub><i>x</i> </sub> is a function of time and so is the instantaneous acceleration <i>a﻿<sub>x</sub> </i>.</p><p>Generally, if the value of a quantity <i>f</i> is determined by the value of another quantity <i>y</i>, then we say that <i>f</i> is a function of <i>y</i> and we use the special notation <i>f</i>(<i>y</i>) to emphasise this relationship. Although we have only just defined what we mean by a function we have already been using this notation, as in <i>v</i>﻿<sub><i>x</i>﻿</sub>(﻿<i>t</i>﻿) and <i>a﻿<sub>x</sub>﻿</i>(﻿<i>t</i>﻿), for some time.</p><p>This function notation has two great merits:</p><ol class="oucontentnumbered"><li><p>Writing <i>f</i>﻿(﻿<i>y</i>﻿), provides a clear visual reminder that <i>f</i> depends on <i>y</i> in a welldefined way. If we happen to know the equation that relates <i>f</i> to <i>y</i>, say <i>f</i> = <i>y</i>﻿<sup>2</sup>, then we can show this explicitly by writing <i>f</i>﻿(﻿<i>y</i>﻿) = <i>y</i>﻿<sup>2</sup>.</p></li><li><p>If we want to indicate the value of <i>f</i> that corresponds to a particular value of <i>y</i>, it is easy to do so. For example, the value of <i>f</i>﻿(﻿<i>y</i>﻿) at <i>y</i> = 2 can be written <i>f</i>﻿(﻿2﻿). We call <i>f</i>﻿(﻿2﻿) the <i>function value</i> at <i>y</i> = 2. Of course, in order to be able to write <i>f</i>﻿(﻿2﻿) as a number we would have to know the explicit form of <i>f</i>﻿(﻿<i>y</i>﻿). For example, if <i>f</i>﻿(﻿<i>y</i>﻿) = <i>y</i>﻿<sup>2</sup>, then we can say <i>f</i>﻿(﻿2﻿) = 2﻿<sup>2</sup> = 2 × 2 = 4.</p></li></ol><p>The only serious disadvantage of the function notation is that you may confuse <i>f</i>(<i>y</i>) with <i>f</i> × <i>y</i>. Be careful! If <i>f</i> is a function, then <i>f</i>﻿(﻿<i>y</i>﻿) means <i>f</i> is a function of <i>y</i>; it does <i>not</i> mean <i>f</i> × <i>y</i>.</p><p>Some functions arise repeatedly and are given special names so that they can be easily identified. You will be familiar with some of these names, even if you are not yet fully familiar with the functions they describe. For instance, if you look at a scientific calculator (see Figure 31 for example) you will sometimes find that there are keys labelled sin, cos, log and e<sup><i>x</i></sup>; each of these is the name of an important function that you will meet later. The corresponding calculator keys are actually called 'function keys'. Electronic calculators are constructed in such a way that when you key in a number and press a function key, the calculator works out the corresponding function value and displays it. If the value you keyed in is not within the acceptable range of input values for the function you selected (usually called the <i>domain</i> of the function), then the calculator will probably display an error message such as 'err'.</p><div class="oucontentfigure oucontentmediamini" id="fig001_031130"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2dfd801e/s207_2_028i.jpg" alt="" width="338" height="341" style="maxwidth:338px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"><b>Figure 31:</b> A calculator has inbuilt routines for evaluating basic functions such as <i>x</i>﻿<sup>2</sup>, sin﻿(﻿<i>x</i>﻿), cos﻿(﻿<i>x</i>﻿), log﻿(﻿<i>x</i>﻿), etc. These are activated by pressing the function keys</span></div></div></div><p>One particularly simple class of functions consists of functions of the form</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/26c79d5d/s207_1_ue013i.gif" alt="" width="283" height="15" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>where <i>A</i> is a constant and <i>n</i> is a positive whole number, such as 0, 1, 2, 3, …. Functions of this kind include <b>squares</b> <i>f</i>﻿(﻿<i>y</i>﻿) = <i>y</i>﻿<sup>2</sup>, and <b>cubes</b> <i>f</i>﻿(﻿<i>y</i>﻿) = <i>y</i>﻿<sup>3</sup>, which correspond to <i>A</i> = 1 with <i>n</i> = 2 and <i>n</i> = 3, respectively. The function that arises when <i>n</i> = 0 is especially noteworthy since, by convention, <i>y</i><sup>0</sup> = 1, so <i>f</i>﻿(﻿<i>y</i>﻿) = <i>A</i> in this case, a constant. Thus, even a simple constant is just a special kind of function. In the case of a constant function each value of <i>y</i> is associated with the same value of <i>f</i>﻿(﻿<i>y</i>﻿), as in the case of the velocitytime graph of a uniformly moving object.</p><p>Next in complexity are sums of squares, cubes, etc. These functions are called <b>polynomial functions</b> and include the following special cases (where <i>A</i>, <i>B</i>, <i>C</i> and <i>D</i> are constants):</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_i006"><div class="oucontenttablewrapper"><table><tr><td><b>linear functions</b> of the form</td><td><i>f</i>﻿(﻿<i>y</i>﻿) = <i>Ay</i> + <i>B</i></td></tr><tr><td><b>quadratic functions</b> of the form</td><td><i>f</i>﻿(﻿<i>y</i>﻿) = <i>Ay</i>﻿<sup>2</sup> + <i>By</i> + <i>C</i></td></tr><tr><td><b>cubic functions</b> of the form</td><td><i>f</i>﻿(﻿<i>y</i>﻿) = <i>Ay</i>﻿<sup>3</sup> + <i>By</i>﻿<sup>2</sup> + <i>Cy</i> + <i>D</i>.</td></tr></table></div><div class="oucontentsourcereference"></div></div><p>In the case of uniform motion (described in Section 3), the positiontime graph is a straight line described by an equation of the form <i>x</i> = <i>At</i> + <i>B</i>. It should now be clear that another way of describing this relationship is to say that the position is a <i>linear function</i> of the time, <i>x</i>﻿(﻿<i>t</i>﻿) = <i>At</i> + <i>B</i>. In a similar way, the nonuniform motion of a test vehicle in the NASA dropshaft described at the beginning of this course can be described by a <i>quadratic function x</i>(<i>t</i>) = <i>At</i>﻿<sup>2</sup> + <i>Bt</i> + <i>C</i>, and many other forms of onedimensional motion can be described, at least approximately, by suitably chosen polynomial functions.</p>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.1
Functions and the function notationS207_2<p>In Figure 25, the position <i>x</i> of the car depends on the time <i>t</i>. The graph associates a particular value of <i>x</i> with each value of <i>t</i> over the plotted range. In other circumstances we might know an equation that associates a value of <i>x</i> with each value of <i>t</i>, as in the case of the equation <i>x</i> = <i>At</i> + <i>B</i> that we discussed in Section 3. You can invent countless other ways in which <i>x</i> depends on <i>t</i>: for instance <i>x</i> = <i>At</i><sup>2</sup> + <i>B</i> or <i>x</i> = <i>At</i> + <i>Bt</i><sup>2</sup>. In all such cases we describe the dependence of <i>x</i> on <i>t</i> by saying that <i>x</i> is a <b>function</b> of <i>t</i>. This terminology is widely used and is certainly not restricted to <i>x</i> and <i>t</i>. In the example of nonuniform motion we have just been discussing, the instantaneous velocity <i>v</i><sub><i>x</i> </sub> is a function of time and so is the instantaneous acceleration <i>a<sub>x</sub> </i>.</p><p>Generally, if the value of a quantity <i>f</i> is determined by the value of another quantity <i>y</i>, then we say that <i>f</i> is a function of <i>y</i> and we use the special notation <i>f</i>(<i>y</i>) to emphasise this relationship. Although we have only just defined what we mean by a function we have already been using this notation, as in <i>v</i><sub><i>x</i></sub>(<i>t</i>) and <i>a<sub>x</sub></i>(<i>t</i>), for some time.</p><p>This function notation has two great merits:</p><ol class="oucontentnumbered"><li><p>Writing <i>f</i>(<i>y</i>), provides a clear visual reminder that <i>f</i> depends on <i>y</i> in a welldefined way. If we happen to know the equation that relates <i>f</i> to <i>y</i>, say <i>f</i> = <i>y</i><sup>2</sup>, then we can show this explicitly by writing <i>f</i>(<i>y</i>) = <i>y</i><sup>2</sup>.</p></li><li><p>If we want to indicate the value of <i>f</i> that corresponds to a particular value of <i>y</i>, it is easy to do so. For example, the value of <i>f</i>(<i>y</i>) at <i>y</i> = 2 can be written <i>f</i>(2). We call <i>f</i>(2) the <i>function value</i> at <i>y</i> = 2. Of course, in order to be able to write <i>f</i>(2) as a number we would have to know the explicit form of <i>f</i>(<i>y</i>). For example, if <i>f</i>(<i>y</i>) = <i>y</i><sup>2</sup>, then we can say <i>f</i>(2) = 2<sup>2</sup> = 2 × 2 = 4.</p></li></ol><p>The only serious disadvantage of the function notation is that you may confuse <i>f</i>(<i>y</i>) with <i>f</i> × <i>y</i>. Be careful! If <i>f</i> is a function, then <i>f</i>(<i>y</i>) means <i>f</i> is a function of <i>y</i>; it does <i>not</i> mean <i>f</i> × <i>y</i>.</p><p>Some functions arise repeatedly and are given special names so that they can be easily identified. You will be familiar with some of these names, even if you are not yet fully familiar with the functions they describe. For instance, if you look at a scientific calculator (see Figure 31 for example) you will sometimes find that there are keys labelled sin, cos, log and e<sup><i>x</i></sup>; each of these is the name of an important function that you will meet later. The corresponding calculator keys are actually called 'function keys'. Electronic calculators are constructed in such a way that when you key in a number and press a function key, the calculator works out the corresponding function value and displays it. If the value you keyed in is not within the acceptable range of input values for the function you selected (usually called the <i>domain</i> of the function), then the calculator will probably display an error message such as 'err'.</p><div class="oucontentfigure oucontentmediamini" id="fig001_031130"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2dfd801e/s207_2_028i.jpg" alt="" width="338" height="341" style="maxwidth:338px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"><b>Figure 31:</b> A calculator has inbuilt routines for evaluating basic functions such as <i>x</i><sup>2</sup>, sin(<i>x</i>), cos(<i>x</i>), log(<i>x</i>), etc. These are activated by pressing the function keys</span></div></div></div><p>One particularly simple class of functions consists of functions of the form</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/26c79d5d/s207_1_ue013i.gif" alt="" width="283" height="15" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>where <i>A</i> is a constant and <i>n</i> is a positive whole number, such as 0, 1, 2, 3, …. Functions of this kind include <b>squares</b> <i>f</i>(<i>y</i>) = <i>y</i><sup>2</sup>, and <b>cubes</b> <i>f</i>(<i>y</i>) = <i>y</i><sup>3</sup>, which correspond to <i>A</i> = 1 with <i>n</i> = 2 and <i>n</i> = 3, respectively. The function that arises when <i>n</i> = 0 is especially noteworthy since, by convention, <i>y</i><sup>0</sup> = 1, so <i>f</i>(<i>y</i>) = <i>A</i> in this case, a constant. Thus, even a simple constant is just a special kind of function. In the case of a constant function each value of <i>y</i> is associated with the same value of <i>f</i>(<i>y</i>), as in the case of the velocitytime graph of a uniformly moving object.</p><p>Next in complexity are sums of squares, cubes, etc. These functions are called <b>polynomial functions</b> and include the following special cases (where <i>A</i>, <i>B</i>, <i>C</i> and <i>D</i> are constants):</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_i006"><div class="oucontenttablewrapper"><table><tr><td><b>linear functions</b> of the form</td><td><i>f</i>(<i>y</i>) = <i>Ay</i> + <i>B</i></td></tr><tr><td><b>quadratic functions</b> of the form</td><td><i>f</i>(<i>y</i>) = <i>Ay</i><sup>2</sup> + <i>By</i> + <i>C</i></td></tr><tr><td><b>cubic functions</b> of the form</td><td><i>f</i>(<i>y</i>) = <i>Ay</i><sup>3</sup> + <i>By</i><sup>2</sup> + <i>Cy</i> + <i>D</i>.</td></tr></table></div><div class="oucontentsourcereference"></div></div><p>In the case of uniform motion (described in Section 3), the positiontime graph is a straight line described by an equation of the form <i>x</i> = <i>At</i> + <i>B</i>. It should now be clear that another way of describing this relationship is to say that the position is a <i>linear function</i> of the time, <i>x</i>(<i>t</i>) = <i>At</i> + <i>B</i>. In a similar way, the nonuniform motion of a test vehicle in the NASA dropshaft described at the beginning of this course can be described by a <i>quadratic function x</i>(<i>t</i>) = <i>At</i><sup>2</sup> + <i>Bt</i> + <i>C</i>, and many other forms of onedimensional motion can be described, at least approximately, by suitably chosen polynomial functions.</p>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

Derived functions and derivative notation
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.2
Tue, 12 Apr 2016 23:00:00 GMT
<p>Given the function <i>x</i>(<i>t</i>) that describes some particular motion, you could plot the corresponding positiontime graph, measure its gradient at a variety of times to find the instantaneous velocity at those times and then plot the velocitytime graph. If you had some time left, you might go on to measure the gradient of the velocitytime graph at various times, and then plot the accelerationtime graph for the motion. This would effectively complete the description of the motion, but it would be enormously time consuming and, given the difficulty of reading graphs, not particularly accurate.</p><p>Fortunately this graphical procedure can usually be entirely avoided. Starting again from the function <i>x</i>(<i>t</i>), there exists a mathematical procedure, called <b>differentiation</b>, that makes it possible to determine the velocity <i>v</i><sub><i>x</i></sub>(<i>t</i>) directly, by algebra alone. We shall not try to describe the principles that underpin differentiation, but we will introduce the notation of the subject and list some of the basic results. To make this introduction as general as possible we shall initially consider a general function <i>f</i>(<i>y</i>) rather than the position function <i>x</i>(<i>t</i>).</p><p>The central idea is this:</p><p>Remember, the gradient of a graph at a given point is defined by the gradient of its tangent at that point.</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox oucontentsnoheading " id="box001_015"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>Given a function <i>f</i>(<i>y</i>) it is often possible to determine a related function of <i>y</i>, called the <b>derived function</b>, with the property that, at each value of <i>y</i>, the derived function is equal to the gradient of the graph of <i>f</i> against <i>y</i> at that same value of <i>y</i>.</p></div></div></div><p>The derived function is usually referred to as the <b>derivative</b> of <i>f</i> with respect to <i>y</i> (often abbreviated to <i>derivative</i>) and may be represented by the symbol <span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/685f295d/s207_1_ie028i.gif" alt="" width="18" height="32" style="maxwidth:18px;" class="oucontentinlinefigureimage"/></span> or, more formally <span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/dd027258/s207_1_ie029i.gif" alt="" width="35" height="32" style="maxwidth:35px;" class="oucontentinlinefigureimage"/></span>. The <span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/685f295d/s207_1_ie028i.gif" alt="" width="18" height="32" style="maxwidth:18px;" class="oucontentinlinefigureimage"/></span> notation is reminiscent of the <span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/3c5539d2/s207_1_ie030i.gif" alt="" width="18" height="29" style="maxwidth:18px;" class="oucontentinlinefigureimage"/></span> notation that was used when discussing the gradient of a straight line and thus provides a clear reminder of the link between the derived function and the gradient of the <i>f</i> against <i>y</i> graph. However, it is important to remember that <span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/685f295d/s207_1_ie028i.gif" alt="" width="18" height="32" style="maxwidth:18px;" class="oucontentinlinefigureimage"/></span> is a single symbol representing the derived function, it is <i>not</i> the ratio of two quantities d<i>f</i> and d<i>y</i>.</p><p>Although there are systematic ways of finding derived functions from first principles, you will not be required to use them in this course. Indeed, physicists are rarely required to do this because tables of derivatives already exist for all the wellknown functions, and derivatives of more complicated functions can usually be expressed as combinations of those basic derivatives. Table 6 lists a few of the basic derivatives along with the simplest of the rules for combining them  it also gives some explicit examples of functions and their derivatives. Computer packages are now available that implement the rules of differentiation, these are often used to determine the derivatives of more complicated functions (Figure 32).</p><div class="oucontentfigure" style="width:511px;" id="fig001_032135"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/0d0ff481/s207_2_029i.jpg" alt="" width="511" height="294" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 32:</b> A function and its derivative, as displayed by an algebraic computing package</span></div></div></div><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_006"><h2 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 6:</b> Some simple derivatives. The functions <i>f</i>, <i>g</i> and <i>h</i> depend on the variable <i>y</i>. The quantities <i>A</i> and <i>n</i> are constants, which may be positive, negative or zero. Note that <i>n</i> is not necessarily an integer</h2><div class="oucontenttablewrapper"><table><tr><th scope="col"><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/0c1fd155/tablenudge.gif" alt="" width="2" height="37" style="maxwidth:2px;" class="oucontentinlinefigureimage"/></span>Function <i>f</i>(<i>y</i>)<span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/0c1fd155/tablenudge.gif" alt="" width="2" height="37" style="maxwidth:2px;" class="oucontentinlinefigureimage"/></span></th><th scope="col">Derivative <span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/f06d6331/s207_1_ie028iwhite.gif" alt="" width="18" height="32" style="maxwidth:18px;" class="oucontentinlinefigureimage"/></span></th><th scope="col"><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/0c1fd155/tablenudge.gif" alt="" width="2" height="37" style="maxwidth:2px;" class="oucontentinlinefigureimage"/></span>Example<span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/0c1fd155/tablenudge.gif" alt="" width="2" height="37" style="maxwidth:2px;" class="oucontentinlinefigureimage"/></span></th></tr><tr><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/23096b4d/s207_1_ie031ai.gif" alt="" width="80" height="33" style="maxwidth:80px;" class="oucontentinlinefigureimage"/></span></td><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/54488502/s207_1_ie031bi.gif" alt="" width="42" height="32" style="maxwidth:42px;" class="oucontentinlinefigureimage"/></span></td><td><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/68777d1b/s207_1_ie031ci.gif" alt="" width="50" height="52" style="maxwidth:50px;" class="oucontentinlinefigureimage"/></span></td></tr><tr><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/aa481c66/s207_1_ie032ai.gif" alt="" width="95" height="35" style="maxwidth:95px;" class="oucontentinlinefigureimage"/></span></td><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/cfeea7e2/s207_1_ie032bi.gif" alt="" width="74" height="32" style="maxwidth:74px;" class="oucontentinlinefigureimage"/></span></td><td><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/d50b190c/s207_1_ie032ci.gif" alt="" width="66" height="55" style="maxwidth:66px;" class="oucontentinlinefigureimage"/></span></td></tr><tr><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/c88eb806/s207_1_ie033ai.gif" alt="" width="189" height="35" style="maxwidth:189px;" class="oucontentinlinefigureimage"/></span></td><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/fc43a929/s207_1_ie033bi.gif" alt="" width="84" height="32" style="maxwidth:84px;" class="oucontentinlinefigureimage"/></span></td><td><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7f65f91d/s207_1_ie033ci.gif" alt="" width="66" height="55" style="maxwidth:66px;" class="oucontentinlinefigureimage"/></span></td></tr><tr><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/5504dca8/s207_1_ie034ai.gif" alt="" width="135" height="33" style="maxwidth:135px;" class="oucontentinlinefigureimage"/></span></td><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/5485d140/s207_1_ie034bi.gif" alt="" width="139" height="53" style="maxwidth:139px;" class="oucontentinlinefigureimage"/></span></td><td><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/f94af1ce/s207_1_ie034ci.gif" alt="" width="100" height="55" style="maxwidth:100px;" class="oucontentinlinefigureimage"/></span></td></tr><tr><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2e8b33f0/s207_1_ie035ai.gif" alt="" width="172" height="33" style="maxwidth:172px;" class="oucontentinlinefigureimage"/></span></td><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/3e951870/s207_1_ie035bi.gif" alt="" width="176" height="53" style="maxwidth:176px;" class="oucontentinlinefigureimage"/></span></td><td><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/8fc1b38a/s207_1_ie035ci.gif" alt="" width="147" height="55" style="maxwidth:147px;" class="oucontentinlinefigureimage"/></span></td></tr></table></div><div class="oucontentsourcereference"></div></div><p>The idea of a derivative may be new to you and, if so, may seem rather strange. However, if you know the explicit form of a function, then there are several crucial advantages in using derivatives to determine gradients, rather than making measurements on a graph. In particular, looking up the derivative of a function in a table should be completely accurate, whereas measuring the gradient of the tangent to a graph is always approximate. For example, if <i>f</i>﻿(<i>y</i>﻿) = <i>y</i>﻿<sup>2</sup> then the derivative of <i>f</i>﻿(﻿<i>y</i>﻿) is d﻿<i>f</i>﻿/﻿d﻿<i>y</i> = 2<i>y</i> and evaluating the derivative at <i>y</i> = 3 to find the gradient at that particular value of <i>y</i> gives 6. This is an <i>exact</i> result that could not have been obtained with such precision from measurements on a graph. Moreover, if we want to know the gradient at many different values of <i>y</i>, all we need to do is to substitute each of those values into the general expression for the derivative, d<i>f</i>/d<i>y</i> = 2<i>y</i>. This is much simpler than drawing many different tangents and measuring their individual gradients.</p>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.2
Derived functions and derivative notationS207_2<p>Given the function <i>x</i>(<i>t</i>) that describes some particular motion, you could plot the corresponding positiontime graph, measure its gradient at a variety of times to find the instantaneous velocity at those times and then plot the velocitytime graph. If you had some time left, you might go on to measure the gradient of the velocitytime graph at various times, and then plot the accelerationtime graph for the motion. This would effectively complete the description of the motion, but it would be enormously time consuming and, given the difficulty of reading graphs, not particularly accurate.</p><p>Fortunately this graphical procedure can usually be entirely avoided. Starting again from the function <i>x</i>(<i>t</i>), there exists a mathematical procedure, called <b>differentiation</b>, that makes it possible to determine the velocity <i>v</i><sub><i>x</i></sub>(<i>t</i>) directly, by algebra alone. We shall not try to describe the principles that underpin differentiation, but we will introduce the notation of the subject and list some of the basic results. To make this introduction as general as possible we shall initially consider a general function <i>f</i>(<i>y</i>) rather than the position function <i>x</i>(<i>t</i>).</p><p>The central idea is this:</p><p>Remember, the gradient of a graph at a given point is defined by the gradient of its tangent at that point.</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox
oucontentsnoheading
" id="box001_015"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>Given a function <i>f</i>(<i>y</i>) it is often possible to determine a related function of <i>y</i>, called the <b>derived function</b>, with the property that, at each value of <i>y</i>, the derived function is equal to the gradient of the graph of <i>f</i> against <i>y</i> at that same value of <i>y</i>.</p></div></div></div><p>The derived function is usually referred to as the <b>derivative</b> of <i>f</i> with respect to <i>y</i> (often abbreviated to <i>derivative</i>) and may be represented by the symbol <span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/685f295d/s207_1_ie028i.gif" alt="" width="18" height="32" style="maxwidth:18px;" class="oucontentinlinefigureimage"/></span> or, more formally <span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/dd027258/s207_1_ie029i.gif" alt="" width="35" height="32" style="maxwidth:35px;" class="oucontentinlinefigureimage"/></span>. The <span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/685f295d/s207_1_ie028i.gif" alt="" width="18" height="32" style="maxwidth:18px;" class="oucontentinlinefigureimage"/></span> notation is reminiscent of the <span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/3c5539d2/s207_1_ie030i.gif" alt="" width="18" height="29" style="maxwidth:18px;" class="oucontentinlinefigureimage"/></span> notation that was used when discussing the gradient of a straight line and thus provides a clear reminder of the link between the derived function and the gradient of the <i>f</i> against <i>y</i> graph. However, it is important to remember that <span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/685f295d/s207_1_ie028i.gif" alt="" width="18" height="32" style="maxwidth:18px;" class="oucontentinlinefigureimage"/></span> is a single symbol representing the derived function, it is <i>not</i> the ratio of two quantities d<i>f</i> and d<i>y</i>.</p><p>Although there are systematic ways of finding derived functions from first principles, you will not be required to use them in this course. Indeed, physicists are rarely required to do this because tables of derivatives already exist for all the wellknown functions, and derivatives of more complicated functions can usually be expressed as combinations of those basic derivatives. Table 6 lists a few of the basic derivatives along with the simplest of the rules for combining them  it also gives some explicit examples of functions and their derivatives. Computer packages are now available that implement the rules of differentiation, these are often used to determine the derivatives of more complicated functions (Figure 32).</p><div class="oucontentfigure" style="width:511px;" id="fig001_032135"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/0d0ff481/s207_2_029i.jpg" alt="" width="511" height="294" style="maxwidth:511px;" class="oucontentfigureimage oucontentmediawide"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 32:</b> A function and its derivative, as displayed by an algebraic computing package</span></div></div></div><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_006"><h2 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 6:</b> Some simple derivatives. The functions <i>f</i>, <i>g</i> and <i>h</i> depend on the variable <i>y</i>. The quantities <i>A</i> and <i>n</i> are constants, which may be positive, negative or zero. Note that <i>n</i> is not necessarily an integer</h2><div class="oucontenttablewrapper"><table><tr><th scope="col"><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/0c1fd155/tablenudge.gif" alt="" width="2" height="37" style="maxwidth:2px;" class="oucontentinlinefigureimage"/></span>Function <i>f</i>(<i>y</i>)<span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/0c1fd155/tablenudge.gif" alt="" width="2" height="37" style="maxwidth:2px;" class="oucontentinlinefigureimage"/></span></th><th scope="col">Derivative <span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/f06d6331/s207_1_ie028iwhite.gif" alt="" width="18" height="32" style="maxwidth:18px;" class="oucontentinlinefigureimage"/></span></th><th scope="col"><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/0c1fd155/tablenudge.gif" alt="" width="2" height="37" style="maxwidth:2px;" class="oucontentinlinefigureimage"/></span>Example<span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/0c1fd155/tablenudge.gif" alt="" width="2" height="37" style="maxwidth:2px;" class="oucontentinlinefigureimage"/></span></th></tr><tr><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/23096b4d/s207_1_ie031ai.gif" alt="" width="80" height="33" style="maxwidth:80px;" class="oucontentinlinefigureimage"/></span></td><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/54488502/s207_1_ie031bi.gif" alt="" width="42" height="32" style="maxwidth:42px;" class="oucontentinlinefigureimage"/></span></td><td><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/68777d1b/s207_1_ie031ci.gif" alt="" width="50" height="52" style="maxwidth:50px;" class="oucontentinlinefigureimage"/></span></td></tr><tr><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/aa481c66/s207_1_ie032ai.gif" alt="" width="95" height="35" style="maxwidth:95px;" class="oucontentinlinefigureimage"/></span></td><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/cfeea7e2/s207_1_ie032bi.gif" alt="" width="74" height="32" style="maxwidth:74px;" class="oucontentinlinefigureimage"/></span></td><td><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/d50b190c/s207_1_ie032ci.gif" alt="" width="66" height="55" style="maxwidth:66px;" class="oucontentinlinefigureimage"/></span></td></tr><tr><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/c88eb806/s207_1_ie033ai.gif" alt="" width="189" height="35" style="maxwidth:189px;" class="oucontentinlinefigureimage"/></span></td><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/fc43a929/s207_1_ie033bi.gif" alt="" width="84" height="32" style="maxwidth:84px;" class="oucontentinlinefigureimage"/></span></td><td><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7f65f91d/s207_1_ie033ci.gif" alt="" width="66" height="55" style="maxwidth:66px;" class="oucontentinlinefigureimage"/></span></td></tr><tr><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/5504dca8/s207_1_ie034ai.gif" alt="" width="135" height="33" style="maxwidth:135px;" class="oucontentinlinefigureimage"/></span></td><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/5485d140/s207_1_ie034bi.gif" alt="" width="139" height="53" style="maxwidth:139px;" class="oucontentinlinefigureimage"/></span></td><td><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/f94af1ce/s207_1_ie034ci.gif" alt="" width="100" height="55" style="maxwidth:100px;" class="oucontentinlinefigureimage"/></span></td></tr><tr><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2e8b33f0/s207_1_ie035ai.gif" alt="" width="172" height="33" style="maxwidth:172px;" class="oucontentinlinefigureimage"/></span></td><td class="oucontenttablemiddle "><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/3e951870/s207_1_ie035bi.gif" alt="" width="176" height="53" style="maxwidth:176px;" class="oucontentinlinefigureimage"/></span></td><td><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/8fc1b38a/s207_1_ie035ci.gif" alt="" width="147" height="55" style="maxwidth:147px;" class="oucontentinlinefigureimage"/></span></td></tr></table></div><div class="oucontentsourcereference"></div></div><p>The idea of a derivative may be new to you and, if so, may seem rather strange. However, if you know the explicit form of a function, then there are several crucial advantages in using derivatives to determine gradients, rather than making measurements on a graph. In particular, looking up the derivative of a function in a table should be completely accurate, whereas measuring the gradient of the tangent to a graph is always approximate. For example, if <i>f</i>(<i>y</i>) = <i>y</i><sup>2</sup> then the derivative of <i>f</i>(<i>y</i>) is d<i>f</i>/d<i>y</i> = 2<i>y</i> and evaluating the derivative at <i>y</i> = 3 to find the gradient at that particular value of <i>y</i> gives 6. This is an <i>exact</i> result that could not have been obtained with such precision from measurements on a graph. Moreover, if we want to know the gradient at many different values of <i>y</i>, all we need to do is to substitute each of those values into the general expression for the derivative, d<i>f</i>/d<i>y</i> = 2<i>y</i>. This is much simpler than drawing many different tangents and measuring their individual gradients.</p>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

4.4 Velocity and acceleration as derivatives
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.4
Tue, 12 Apr 2016 23:00:00 GMT
<p>Recalling that the instantaneous velocity of a particle at time <i>t</i> is given by the gradient of its positiontime graph at that time, we can now use the terminology of functions and derivatives to say that the velocity of the particle is given by the derivative of its position function. In terms of symbols:</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox oucontentsnoheading " id="box001_016"><div class="oucontentouterbox"><div class="oucontentinnerbox"><div class="oucontentequation oucontentequationequation oucontentnocaption" id="equ001_033"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/6a0d5dc2/s207_1_ue014i.gif" alt=""/></div></div></div></div><p>Similarly, we can say that the instantaneous acceleration of a particle is given by the derivative of the velocity function:</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox oucontentsnoheading " id="box001_017"><div class="oucontentouterbox"><div class="oucontentinnerbox"><div class="oucontentequation oucontentequationequation oucontentnocaption" id="equ001_034"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/a831e95a/s207_1_ue015i.gif" alt=""/></div></div></div></div><p>What's more we can use derivatives to simplify problems, as Example 1 shows.</p><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="exe001_001"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Example 1</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>The position <i>x</i> of a particle at time <i>t</i> is given by the function <i>x</i>﻿(﻿<i>t</i>﻿) = <i>kt</i>﻿<sup>2</sup> where <i>k</i> = −5 m s<sup> −2</sup>. Find (a) the velocity as a function of time; (b) the velocity at time <i>t</i> = 3 s.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) The velocity <i>v</i><sub><i>x</i></sub> is the derivative with respect to time of the position function <i>x</i>﻿(<i>t</i>) which is of the form <i>At﻿<sup>n</sup> </i>, with <i>A</i> = <i>k</i> and <i>n</i> = 2. It therefore follows from the third of the standard results in Table 6 that</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="eqn001_016"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/b77acd46/s207_1_ue016i.gif" alt=""/></div><p>This is the required answer. No measuring of the gradients of tangents to curves is involved!</p><p>(b) Remembering that <i>k</i> is given as −5 m s<sup>−2</sup>, the velocity at time <i>t</i> = 3 s is now easily obtained from Equation 16, as follows</p><div class="oucontentquote oucontentsbox" id="q005"><blockquote><p><i>v<sub>x</sub></i>(3 s) = 2<i>kt</i> = 2 × (−5 m s<sup>−2</sup>) × (3 s) = −30 m s<sup>−1</sup></p></blockquote></div><p>Note that we have multiplied m s﻿<sup>−2</sup> by s to give m s<sup>﻿−﻿1</sup>, i.e. the units have been treated just like algebraic quantities.</p></div></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_015"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 17</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>If the velocity <i>v</i>﻿<sub><i>x</i>﻿</sub>(﻿<i>t</i>﻿) of a particle is given by <i>v</i>﻿<sub><i>x</i>﻿</sub>(﻿<i>t</i>﻿) = <i>kt</i>﻿<sup>2</sup>, where <i>k</i> = 4 m s<sup>﻿−3</sup>, find a general expression for the acceleration <i>a<sub>x</sub></i>(<i>t</i>). What is the value of <i>a﻿<sub>x</sub>﻿</i>(﻿3 s﻿)?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>The expression for the acceleration is given by</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_011"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/84d20669/s207_1_ie045i.gif" alt=""/></div><p>using Table 6. Therefore <i>a<sub>x</sub></i>(3 s) = (2 × 4 × 3) m s<sup>−2</sup> = 24 m s<sup>−2</sup>.</p></div></div></div></div><p>Although it would be quite wrong to think of d<i>x</i>/d<i>t</i> as a ratio of the quantities d<i>x</i> and d<i>t</i>, it is useful to regard d<i>x</i>/d<i>t</i> as consisting of an entity d/d<i>t</i> that acts on the function <i>x</i>(<i>t</i>). The entity d/d<i>t</i> is a mathematical instruction to differentiate the function that follows, <i>x</i>﻿(﻿<i>t</i>﻿) in this case. It is an example of what a mathematician would call an <b>operator</b>. Adopting this view, we can say that if, for example, <i>x</i>﻿(﻿<i>t</i>﻿) = <i>kt</i>﻿<sup>2</sup> + <i>ct</i>, then</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_034"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/688e0591/s207_1_ie011i.gif" alt=""/></div><p>Using the rule for differentiating the sum of two functions, from Table 6, we may write this as</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_035"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7385a3be/s207_1_ie012i.gif" alt=""/></div><p>Using the third result in <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.2#tbl001006">Table 6</a> we may work out both the derivatives on the righthand side to obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_036"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/e5411077/s207_1_ie013i.gif" alt=""/></div><p>Regarding d/d<i>t</i> as an operator also suggests another way of writing the acceleration. We already know that</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_037"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/04782207/s207_1_ie014i.gif" alt=""/></div><p>so it seems sensible to write</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/66ee07c5/s207_1_ue017i.gif" alt="" width="283" height="36" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>This emphasises that the acceleration of a particle is the rate of change of the rate of change of the position, or if you prefer, the derivative of the derivative of the position function. Either of these formulations is a bit of a mouthful, so it is more conventional to refer to the acceleration as the <b>second derivative</b> of <i>x</i>(<i>t</i>) and to represent it symbolically by</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox oucontentsnoheading " id="box001_018"><div class="oucontentouterbox"><div class="oucontentinnerbox"><div class="oucontentequation oucontentequationequation oucontentnocaption" id="fig001_035154"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/5a95df5a/s207_1_ue018i.gif" alt=""/></div></div></div></div><p>Once again, it would be quite wrong to think of this as some kind of ratio of d﻿<sup>2</sup>﻿<i>x</i> and d﻿<i>t</i>﻿<sup>2</sup>; it simply indicates that a particle's position function, <i>x</i>﻿(﻿<i>t</i>﻿), must be differentiated twice in order to find the particle's acceleration.</p><p><i>Note:</i> If you are not already familiar with derivatives you should pay particular attention to the positioning of the superscripts in the second derivative symbol. Newcomers to differentiation often make the mistake of writing d<i>x</i>﻿<sup>2</sup>﻿/﻿d﻿<i>t</i>﻿<sup>2</sup> when they mean d﻿<sup>2</sup>﻿<i>x</i>﻿/﻿d﻿<i>t</i>﻿<sup>2</sup>. Remembering that it is the operator d/dt that is to be squared rather than the function <i>x</i>﻿(﻿<i>t</i>﻿) may help you to avoid this error.</p><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="exe001_002"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Example 2</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Suppose, as in Example 1, the position of a particle is given by <i>x</i>﻿(﻿<i>t</i>﻿) = <i>kt</i>﻿<sup>2</sup> where <i>k</i> is a constant. Find <i>a﻿<sub>x</sub>﻿</i>(﻿<i>t</i>﻿).</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>We use Equation 14 and the third result in <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.2#tbl001006">Table 6</a> with <i>A</i> = <i>k</i>, <i>y</i> = <i>t</i> and <i>n</i> = 2, to obtain</p><p/><div class="oucontentequation oucontentequationequation oucontentnocaption" id="fig001_036157"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/b77acd46/s207_1_ue016i.gif" alt=""/></div><p>It then follows from Equation 18, that</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_041"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/4d3401ca/s207_1_ie076i.gif" alt=""/></div><p>where we have again used the third result in <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.2#tbl001006">Table 6</a> but with <i>A</i> = 2<i>k</i> and <i>n</i> = 1.</p></div></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_016"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 18</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Suppose that the vertical position <i>x</i> of a test vehicle falling down a dropshaft is given by the quadratic function</p><div class="oucontentquote oucontentsbox" id="q006"><blockquote><p><i>x</i>(<i>t</i>) = <i>k</i><sub>0</sub> + <i>k</i><sub>1</sub><i>t</i> + <i>k</i><sub>2</sub><i>t</i><sup>2</sup></p></blockquote></div><p>where <i>k</i><sub>0</sub>, <i>k</i><sub>1</sub> and <i>k</i><sub>2</sub> are constants. Work out an expression for the vehicle's acceleration <i>a﻿<sub>x</sub>﻿</i>(﻿<i>t</i>﻿) in terms of <i>k</i>﻿<sub>0</sub>, <i>k</i>﻿<sub>1</sub> and <i>k</i>﻿<sub>2</sub>, taking care to indicate each step in your working.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Using Equations 14 and 15 and <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.2#tbl001006">Table 6</a>, we find</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_012"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/929f7841/s207_1_ie046i.gif" alt=""/></div></div></div></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.4
4.4 Velocity and acceleration as derivativesS207_2<p>Recalling that the instantaneous velocity of a particle at time <i>t</i> is given by the gradient of its positiontime graph at that time, we can now use the terminology of functions and derivatives to say that the velocity of the particle is given by the derivative of its position function. In terms of symbols:</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox
oucontentsnoheading
" id="box001_016"><div class="oucontentouterbox"><div class="oucontentinnerbox"><div class="oucontentequation oucontentequationequation oucontentnocaption" id="equ001_033"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/6a0d5dc2/s207_1_ue014i.gif" alt=""/></div></div></div></div><p>Similarly, we can say that the instantaneous acceleration of a particle is given by the derivative of the velocity function:</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox
oucontentsnoheading
" id="box001_017"><div class="oucontentouterbox"><div class="oucontentinnerbox"><div class="oucontentequation oucontentequationequation oucontentnocaption" id="equ001_034"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/a831e95a/s207_1_ue015i.gif" alt=""/></div></div></div></div><p>What's more we can use derivatives to simplify problems, as Example 1 shows.</p><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="exe001_001"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Example 1</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>The position <i>x</i> of a particle at time <i>t</i> is given by the function <i>x</i>(<i>t</i>) = <i>kt</i><sup>2</sup> where <i>k</i> = −5 m s<sup> −2</sup>. Find (a) the velocity as a function of time; (b) the velocity at time <i>t</i> = 3 s.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) The velocity <i>v</i><sub><i>x</i></sub> is the derivative with respect to time of the position function <i>x</i>(<i>t</i>) which is of the form <i>At<sup>n</sup> </i>, with <i>A</i> = <i>k</i> and <i>n</i> = 2. It therefore follows from the third of the standard results in Table 6 that</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="eqn001_016"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/b77acd46/s207_1_ue016i.gif" alt=""/></div><p>This is the required answer. No measuring of the gradients of tangents to curves is involved!</p><p>(b) Remembering that <i>k</i> is given as −5 m s<sup>−2</sup>, the velocity at time <i>t</i> = 3 s is now easily obtained from Equation 16, as follows</p><div class="oucontentquote oucontentsbox" id="q005"><blockquote><p><i>v<sub>x</sub></i>(3 s) = 2<i>kt</i> = 2 × (−5 m s<sup>−2</sup>) × (3 s) = −30 m s<sup>−1</sup></p></blockquote></div><p>Note that we have multiplied m s<sup>−2</sup> by s to give m s<sup>−1</sup>, i.e. the units have been treated just like algebraic quantities.</p></div></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_015"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 17</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>If the velocity <i>v</i><sub><i>x</i></sub>(<i>t</i>) of a particle is given by <i>v</i><sub><i>x</i></sub>(<i>t</i>) = <i>kt</i><sup>2</sup>, where <i>k</i> = 4 m s<sup>−3</sup>, find a general expression for the acceleration <i>a<sub>x</sub></i>(<i>t</i>). What is the value of <i>a<sub>x</sub></i>(3 s)?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>The expression for the acceleration is given by</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_011"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/84d20669/s207_1_ie045i.gif" alt=""/></div><p>using Table 6. Therefore <i>a<sub>x</sub></i>(3 s) = (2 × 4 × 3) m s<sup>−2</sup> = 24 m s<sup>−2</sup>.</p></div></div></div></div><p>Although it would be quite wrong to think of d<i>x</i>/d<i>t</i> as a ratio of the quantities d<i>x</i> and d<i>t</i>, it is useful to regard d<i>x</i>/d<i>t</i> as consisting of an entity d/d<i>t</i> that acts on the function <i>x</i>(<i>t</i>). The entity d/d<i>t</i> is a mathematical instruction to differentiate the function that follows, <i>x</i>(<i>t</i>) in this case. It is an example of what a mathematician would call an <b>operator</b>. Adopting this view, we can say that if, for example, <i>x</i>(<i>t</i>) = <i>kt</i><sup>2</sup> + <i>ct</i>, then</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_034"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/688e0591/s207_1_ie011i.gif" alt=""/></div><p>Using the rule for differentiating the sum of two functions, from Table 6, we may write this as</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_035"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7385a3be/s207_1_ie012i.gif" alt=""/></div><p>Using the third result in <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.2#tbl001006">Table 6</a> we may work out both the derivatives on the righthand side to obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_036"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/e5411077/s207_1_ie013i.gif" alt=""/></div><p>Regarding d/d<i>t</i> as an operator also suggests another way of writing the acceleration. We already know that</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_037"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/04782207/s207_1_ie014i.gif" alt=""/></div><p>so it seems sensible to write</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/66ee07c5/s207_1_ue017i.gif" alt="" width="283" height="36" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>This emphasises that the acceleration of a particle is the rate of change of the rate of change of the position, or if you prefer, the derivative of the derivative of the position function. Either of these formulations is a bit of a mouthful, so it is more conventional to refer to the acceleration as the <b>second derivative</b> of <i>x</i>(<i>t</i>) and to represent it symbolically by</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox
oucontentsnoheading
" id="box001_018"><div class="oucontentouterbox"><div class="oucontentinnerbox"><div class="oucontentequation oucontentequationequation oucontentnocaption" id="fig001_035154"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/5a95df5a/s207_1_ue018i.gif" alt=""/></div></div></div></div><p>Once again, it would be quite wrong to think of this as some kind of ratio of d<sup>2</sup><i>x</i> and d<i>t</i><sup>2</sup>; it simply indicates that a particle's position function, <i>x</i>(<i>t</i>), must be differentiated twice in order to find the particle's acceleration.</p><p><i>Note:</i> If you are not already familiar with derivatives you should pay particular attention to the positioning of the superscripts in the second derivative symbol. Newcomers to differentiation often make the mistake of writing d<i>x</i><sup>2</sup>/d<i>t</i><sup>2</sup> when they mean d<sup>2</sup><i>x</i>/d<i>t</i><sup>2</sup>. Remembering that it is the operator d/dt that is to be squared rather than the function <i>x</i>(<i>t</i>) may help you to avoid this error.</p><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="exe001_002"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Example 2</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Suppose, as in Example 1, the position of a particle is given by <i>x</i>(<i>t</i>) = <i>kt</i><sup>2</sup> where <i>k</i> is a constant. Find <i>a<sub>x</sub></i>(<i>t</i>).</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>We use Equation 14 and the third result in <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.2#tbl001006">Table 6</a> with <i>A</i> = <i>k</i>, <i>y</i> = <i>t</i> and <i>n</i> = 2, to obtain</p><p/><div class="oucontentequation oucontentequationequation oucontentnocaption" id="fig001_036157"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/b77acd46/s207_1_ue016i.gif" alt=""/></div><p>It then follows from Equation 18, that</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_041"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/4d3401ca/s207_1_ie076i.gif" alt=""/></div><p>where we have again used the third result in <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.2#tbl001006">Table 6</a> but with <i>A</i> = 2<i>k</i> and <i>n</i> = 1.</p></div></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_016"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 18</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Suppose that the vertical position <i>x</i> of a test vehicle falling down a dropshaft is given by the quadratic function</p><div class="oucontentquote oucontentsbox" id="q006"><blockquote><p><i>x</i>(<i>t</i>) = <i>k</i><sub>0</sub> + <i>k</i><sub>1</sub><i>t</i> + <i>k</i><sub>2</sub><i>t</i><sup>2</sup></p></blockquote></div><p>where <i>k</i><sub>0</sub>, <i>k</i><sub>1</sub> and <i>k</i><sub>2</sub> are constants. Work out an expression for the vehicle's acceleration <i>a<sub>x</sub></i>(<i>t</i>) in terms of <i>k</i><sub>0</sub>, <i>k</i><sub>1</sub> and <i>k</i><sub>2</sub>, taking care to indicate each step in your working.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Using Equations 14 and 15 and <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.2#tbl001006">Table 6</a>, we find</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_012"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/929f7841/s207_1_ie046i.gif" alt=""/></div></div></div></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

4.5 The signed area under a general velocitytime graph
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.5
Tue, 12 Apr 2016 23:00:00 GMT
<p>We have already seen (in <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection3.6">Section 3.6</a>) that in the context of uniform motion, the signed area under a particle's velocitytime graph, between two given times, represents the change in the particle's position during that time interval, with a positive area corresponding to displacement in the positive direction. In the case of uniform motion, the velocitytime graph was a horizontal line and the area under the graph was rectangular.</p><div class="oucontentfigure oucontentmediamini" id="fig001_037a"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7a798082/s207_2_030i.jpg" alt="" width="341" height="294" style="maxwidth:341px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 33:</b> The area under the velocitytime graph between <i>t</i>﻿<sub>1</sub> and <i>t</i>﻿<sub>2</sub> for an accelerating particle</span></div></div></div><p>Now, in the context of nonuniform motion, it seems natural to ask if the same interpretation can be given to the area under a general velocitytime graph, such as that between <i>t</i><sub>1</sub> and <i>t</i><sub>2</sub> in Figure 33. The answer to this is a definite yes, though a rigorous proof is beyond the scope of this course, so what follows is simply a plausibility argument.</p><p>In Figure 33, the colourshaded area under the graph does not take the shape of a rectangle. However, it may be approximately represented by a sum of rectangular areas, as indicated in Figure 34. To produce Figure 34 we have broken the time between <i>t</i><sub>1</sub> and <i>t</i><sub>2</sub> into a number of small intervals, each of identical duration Δ<i>t</i> and within each small interval, the velocity has been approximated by a constant, which can be taken to be the average velocity during that interval. As a result, the area of each rectangular strip in Figure 34 represents the approximate change of position over a short time Δ<i>t</i> and the sum of those areas represents the approximate change of position over the interval <i>t</i><sub>1</sub> to <i>t</i><sub>2</sub>. Now, if we were to repeat this process while using a smaller value for Δ<i>t</i>, as in Figure 35, then we would have more strips between <i>t</i><sub>1</sub> and <i>t</i><sub>2</sub>; their total area, representing the approximate change in position between <i>t</i><sub>1</sub> and <i>t</i><sub>2</sub>, would be an even closer approximation to the true area shown in Figure 33. Given these results, it seems reasonable to suppose that if we allowed Δ<i>t</i> to become smaller and smaller, while the number of rectangular strips between <i>t</i><sub>1</sub> and <i>t</i><sub>2</sub> became correspondingly larger and larger, then we would eventually find that the area under the graph in Figure 33 was <i>exactly</i> equal to the change in position between <i>t</i><sub>1</sub> and <i>t</i><sub>2</sub>.</p><div class="oucontentfigure" style="width:343px;" id="fig001_038166"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/e547926a/s207_2_031i.jpg" alt="" width="343" height="297" style="maxwidth:343px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 34:</b> The area under the velocitytime graph of Figure 33, broken up into thin rectangular strips</span></div></div></div><div class="oucontentfigure" style="width:345px;" id="fig001_039168"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/bb09dc08/s207_2_032i.jpg" alt="" width="345" height="296" style="maxwidth:345px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 35:</b> The area under the velocitytime graph of Figure 33, broken up into even more rectangular strips by reducing the value of Δ<i>t</i></span></div></div></div><p>This conclusion is in fact correct and can be proved in a rigorous way by considering what mathematicians call a <i>limit</i>, in this case 'the limit as Δ<i>t</i> tends to zero'. We shall not pursue that here, but we should note that it wasn't until the early nineteenth century that the mathematics of limits was properly formulated, although it was in use long before then. It is also worth pointing out that it was the development of the idea of a limit that finally laid Zeno's paradox to rest. Just as the increasing number of diminishing strips can have a finite total area, so the increasing number of smaller steps that Achilles must take to reach the tortoise can have a finite sum and be completed in a finite time. Rigorous mathematical reasoning agrees with our everyday experience in telling us that motion can exist and that athletes outrun tortoises!</p><div class="oucontentfigure oucontentmediamini" id="fig001_041"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/da3b6e00/s207_2_033i.jpg" alt="" width="338" height="299" style="maxwidth:338px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 36:</b> The velocitytime graph for Selfassessment question 19</span></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_017"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 19</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Figure 36 shows the velocitytime graph for a particle with constant acceleration.</p><p>(a) What is the displacement of the particle from its initial position after 6 s?</p><p>(b) What is the distance travelled by the particle between <i>t</i> = 2 s and <i>t</i> = 6 s﻿? (You may find it useful to know that the formula for the area of a triangle is: area = half the base × height, and that for the area of a trapezium is: area = base × (half the sum of the lengths of the parallel sides).)</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) The displacement over the first 6 s is equal to the total signed area between the graph and the taxis and between <i>t</i> = 0 s and <i>t</i> = 6 s in Figure 37. Recall that regions below the axis are regarded as negative areas. In this case the total area is composed of two triangles, one above the axis, one below. The total displacement is therefore given by</p><p><i>s<sub>x</sub></i>(6 s) = (1/2)×(1 s)×(1.2 m s<sup>−1</sup>) −(1/2)×(5 s)×(6 m s<sup>−1</sup>)</p><p>i.e. <i>s﻿<sub>x</sub>﻿</i>(﻿6 s﻿) = (﻿0.60 m﻿) −(15 m) = −14.4 m.</p><p>(b) The distance travelled between <i>t</i> = 2 s and <i>t</i> = 6 s will be the magnitude of the displacement over that time. Note that the displacement will be negative since it is represented by the area of the colourshaded trapezium, which is entirely below the axis. However, the corresponding distance will be positive (since it is a magnitude) and will have the value</p><div class="oucontentquote oucontentsbox" id="q013"><blockquote><p><i>s</i> = [(1/2) × (6 s − 2 s) × (6 m s<sup>−1</sup> − 1.2 m s<sup>−1</sup>)] + [(6 s − 2 s) × 1.2 m s<sup>−1</sup>] = 14.4 m</p></blockquote></div><div class="oucontentfigure" style="width:391px;" id="fig001_040"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/5dd2e090/s207_2_041i.jpg" alt="" width="391" height="310" style="maxwidth:391px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"><b>Figure 37:</b> The velocitytime graph for Selfassessment question 19</span></div></div></div></div></div></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.5
4.5 The signed area under a general velocitytime graphS207_2<p>We have already seen (in <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection3.6">Section 3.6</a>) that in the context of uniform motion, the signed area under a particle's velocitytime graph, between two given times, represents the change in the particle's position during that time interval, with a positive area corresponding to displacement in the positive direction. In the case of uniform motion, the velocitytime graph was a horizontal line and the area under the graph was rectangular.</p><div class="oucontentfigure oucontentmediamini" id="fig001_037a"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7a798082/s207_2_030i.jpg" alt="" width="341" height="294" style="maxwidth:341px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 33:</b> The area under the velocitytime graph between <i>t</i><sub>1</sub> and <i>t</i><sub>2</sub> for an accelerating particle</span></div></div></div><p>Now, in the context of nonuniform motion, it seems natural to ask if the same interpretation can be given to the area under a general velocitytime graph, such as that between <i>t</i><sub>1</sub> and <i>t</i><sub>2</sub> in Figure 33. The answer to this is a definite yes, though a rigorous proof is beyond the scope of this course, so what follows is simply a plausibility argument.</p><p>In Figure 33, the colourshaded area under the graph does not take the shape of a rectangle. However, it may be approximately represented by a sum of rectangular areas, as indicated in Figure 34. To produce Figure 34 we have broken the time between <i>t</i><sub>1</sub> and <i>t</i><sub>2</sub> into a number of small intervals, each of identical duration Δ<i>t</i> and within each small interval, the velocity has been approximated by a constant, which can be taken to be the average velocity during that interval. As a result, the area of each rectangular strip in Figure 34 represents the approximate change of position over a short time Δ<i>t</i> and the sum of those areas represents the approximate change of position over the interval <i>t</i><sub>1</sub> to <i>t</i><sub>2</sub>. Now, if we were to repeat this process while using a smaller value for Δ<i>t</i>, as in Figure 35, then we would have more strips between <i>t</i><sub>1</sub> and <i>t</i><sub>2</sub>; their total area, representing the approximate change in position between <i>t</i><sub>1</sub> and <i>t</i><sub>2</sub>, would be an even closer approximation to the true area shown in Figure 33. Given these results, it seems reasonable to suppose that if we allowed Δ<i>t</i> to become smaller and smaller, while the number of rectangular strips between <i>t</i><sub>1</sub> and <i>t</i><sub>2</sub> became correspondingly larger and larger, then we would eventually find that the area under the graph in Figure 33 was <i>exactly</i> equal to the change in position between <i>t</i><sub>1</sub> and <i>t</i><sub>2</sub>.</p><div class="oucontentfigure" style="width:343px;" id="fig001_038166"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/e547926a/s207_2_031i.jpg" alt="" width="343" height="297" style="maxwidth:343px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 34:</b> The area under the velocitytime graph of Figure 33, broken up into thin rectangular strips</span></div></div></div><div class="oucontentfigure" style="width:345px;" id="fig001_039168"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/bb09dc08/s207_2_032i.jpg" alt="" width="345" height="296" style="maxwidth:345px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 35:</b> The area under the velocitytime graph of Figure 33, broken up into even more rectangular strips by reducing the value of Δ<i>t</i></span></div></div></div><p>This conclusion is in fact correct and can be proved in a rigorous way by considering what mathematicians call a <i>limit</i>, in this case 'the limit as Δ<i>t</i> tends to zero'. We shall not pursue that here, but we should note that it wasn't until the early nineteenth century that the mathematics of limits was properly formulated, although it was in use long before then. It is also worth pointing out that it was the development of the idea of a limit that finally laid Zeno's paradox to rest. Just as the increasing number of diminishing strips can have a finite total area, so the increasing number of smaller steps that Achilles must take to reach the tortoise can have a finite sum and be completed in a finite time. Rigorous mathematical reasoning agrees with our everyday experience in telling us that motion can exist and that athletes outrun tortoises!</p><div class="oucontentfigure oucontentmediamini" id="fig001_041"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/da3b6e00/s207_2_033i.jpg" alt="" width="338" height="299" style="maxwidth:338px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 36:</b> The velocitytime graph for Selfassessment question 19</span></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_017"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 19</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Figure 36 shows the velocitytime graph for a particle with constant acceleration.</p><p>(a) What is the displacement of the particle from its initial position after 6 s?</p><p>(b) What is the distance travelled by the particle between <i>t</i> = 2 s and <i>t</i> = 6 s? (You may find it useful to know that the formula for the area of a triangle is: area = half the base × height, and that for the area of a trapezium is: area = base × (half the sum of the lengths of the parallel sides).)</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) The displacement over the first 6 s is equal to the total signed area between the graph and the taxis and between <i>t</i> = 0 s and <i>t</i> = 6 s in Figure 37. Recall that regions below the axis are regarded as negative areas. In this case the total area is composed of two triangles, one above the axis, one below. The total displacement is therefore given by</p><p><i>s<sub>x</sub></i>(6 s) = (1/2)×(1 s)×(1.2 m s<sup>−1</sup>) −(1/2)×(5 s)×(6 m s<sup>−1</sup>)</p><p>i.e. <i>s<sub>x</sub></i>(6 s) = (0.60 m) −(15 m) = −14.4 m.</p><p>(b) The distance travelled between <i>t</i> = 2 s and <i>t</i> = 6 s will be the magnitude of the displacement over that time. Note that the displacement will be negative since it is represented by the area of the colourshaded trapezium, which is entirely below the axis. However, the corresponding distance will be positive (since it is a magnitude) and will have the value</p><div class="oucontentquote oucontentsbox" id="q013"><blockquote><p><i>s</i> = [(1/2) × (6 s − 2 s) × (6 m s<sup>−1</sup> − 1.2 m s<sup>−1</sup>)] + [(6 s − 2 s) × 1.2 m s<sup>−1</sup>] = 14.4 m</p></blockquote></div><div class="oucontentfigure" style="width:391px;" id="fig001_040"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/5dd2e090/s207_2_041i.jpg" alt="" width="391" height="310" style="maxwidth:391px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"><b>Figure 37:</b> The velocitytime graph for Selfassessment question 19</span></div></div></div></div></div></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

5.1 Describing uniformly accelerated motion
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection5.1
Tue, 12 Apr 2016 23:00:00 GMT
<p>An important special case of nonuniform motion along a line is that which arises when an object is subjected to constant acceleration. This kind of motion is called <b>uniformly accelerated motion</b>. An object falling under gravity near to the surface of the Earth, such as the apple of <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.1#fig001024">Figure 24</a>, provides an approximate realisation of such motion. (Air resistance, which increases with speed, prevents the acceleration from being truly constant in such cases.) A more precise realisation of uniformly accelerated motion is provided by an object falling under gravity close to the surface of an airless body such as the Moon (see Figure 38) or by a falling object in an evacuated (i.e. airless) droptower or dropshaft of the kind discussed in Section 1 of this course.</p><div class="oucontentfigure oucontentmediamini" id="fig001_034"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/20ce2ed7/s207_2_034i.jpg" alt="" width="232" height="223" style="maxwidth:232px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 38:</b> The accelerationtime graph for an object with constant (positive) acceleration</span></div></div></div><p>The accelerationtime graph for a uniformly accelerated body is simple; it's just a horizontal line of the kind shown in <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection5.1#fig001034">Figure 38</a>. The value of this constant acceleration (which may be positive or negative) represents the <i>gradient</i> of the velocitytime graph at any moment. It follows that the velocitytime graph must have the same gradient at all points and must therefore be a straight line of the kind shown in <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection5.1#fig001033">Figure 39</a>. Note that the sign of the acceleration determines whether the velocitytime graph slopes up or down, and the value of the acceleration determines the precise value of the gradient. However, the acceleration does not determine the initial value of <i>v</i>﻿<sub><i>x</i> </sub> so we have arbitrarily chosen a point on the <i>v</i>﻿<sub><i>x</i> </sub>axis to represent this value and labelled it <i>u﻿<sub>x</sub> </i>. From Figure 38, we deduce that if</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/a5437918/s207_1_ue019i.gif" alt="" width="283" height="14" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>then</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/1c2b3b5d/s207_1_ue020i.gif" alt="" width="283" height="14" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>where <i>a﻿<sub>x</sub></i> represents the <i>constant</i> value of <i>a﻿<sub>x</sub></i>(<i>t</i>).</p><div class="oucontentfigure oucontentmediamini" id="fig001_033"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/0e87e0d4/s207_2_035i.jpg" alt="" width="231" height="224" style="maxwidth:231px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 39:</b> A velocitytime graph that is consistent with the accelerationtime graph of Figure 38. Note that the intercept has been chosen arbitrarily; only the gradient is determined by the acceleration</span></div></div></div><p>It's rather more difficult to deduce the positiontime graph that corresponds to uniformly accelerated motion, but the steady change of velocity with time certainly implies that the <i>gradient</i> of the positiontime graph must also change steadily with time. In fact, given a velocitytime graph like that in Figure 39, the corresponding positiontime graph will be of the general form shown in Figure 40.</p><div class="oucontentfigure oucontentmediamini" id="fig001_044179"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/b0e45fa5/s207_2_036i.jpg" alt="" width="232" height="230" style="maxwidth:232px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 40:</b> A positiontime graph that is consistent with the accelerationtime graph of Figure 38 and the velocitytime graph of Figure 39. Note that the intercept has been chosen arbitrarily; only the gradient is determined by velocity</span></div></div></div><p>Once again, the intercept with the vertical axis (representing the initial position at <i>t</i> = 0) is not determined by anything we have said so far; it has therefore been chosen arbitrarily and labelled <i>x</i><sub>0</sub>. The curve however is not arbitrary since the displacement from <i>x</i><sub>0</sub> at any particular time <i>t</i> = <i>T</i> is determined by the area under the velocitytime graph between <i>t</i> = 0 and <i>t</i> = <i>T</i>. That area will be the sum of two parts, a rectangle of height <i>u﻿<sub>x</sub></i> and base length <i>T</i>, and a triangle of height <i>a﻿<sub>x</sub>﻿T</i> and base length <i>T</i>. Since the area of a triangle is half the product of its height and its base length, it follows that the displacement from <i>x</i>﻿<sub>0</sub> at time <i>T</i> will be</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_045"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/d385451e/s207_1_ie016i.gif" alt=""/></div><p>and the position of the uniformly accelerating particle, at any time <i>t</i>, will be <i>x</i>﻿(﻿<i>t</i>﻿) = <i>x</i>﻿<sub>0</sub> + <i>s﻿<sub>x</sub>﻿</i>(﻿<i>t</i>﻿), that is</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_046"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/abfa7e8b/s207_1_ue021i.gif" alt=""/></div><p>Equations 19, 20 and 21 provide an essentially complete description of uniformly accelerated motion in one dimension and have many applications. They are not the most common form of the equations of uniform acceleration. We will discuss those in the next subsection, but before doing so let's use the method of differentiation to confirm the consistency of the equations we have deduced.</p><p>Starting from Equation 21, the righthand side of which is a quadratic function of <i>t</i>, we expect to find that</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_047"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/fc9a397d/s207_1_ie017i.gif" alt=""/></div><p>Using the rules and results of <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.2#tbl001006">Table 6</a> to carry out the differentiation (which is just like that in Question 18) we find that</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_048"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/996f37b8/s207_1_ie018i.gif" alt=""/></div><p>in complete agreement with Equation 20. Similarly, differentiating the linear function that appears on the righthand side of Equation 20, we expect to find</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_049"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/91b8d422/s207_1_ie019i.gif" alt=""/></div><p>Again, performing the differentiation, using <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.2#tbl001006">Table 6</a>, confirms our expectations:</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_050"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/697844f6/s207_1_ie020i.gif" alt=""/></div><p>We see that in this case, <i>a<sub>x</sub> </i>(<i>t</i>) is just the constant acceleration <i>a<sub>x</sub> </i> from which we started.</p><p>This short exercise in checking consistency gives just a hint of the immense power of differentiation to simplify a wide range of tasks and investigations.</p><p>Returning now to the description of uniformly accelerated motion, let's gather together our results so far:</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/b66268f6/s207_1_ue022i.gif" alt="" width="283" height="27" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/cd9b5ff1/s207_1_ue023i.gif" alt="" width="283" height="13" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/03f45dd1/s207_1_ue024i.gif" alt="" width="283" height="13" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>These are the equations we shall use, rearrange and extend in the next subsection.</p>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection5.1
5.1 Describing uniformly accelerated motionS207_2<p>An important special case of nonuniform motion along a line is that which arises when an object is subjected to constant acceleration. This kind of motion is called <b>uniformly accelerated motion</b>. An object falling under gravity near to the surface of the Earth, such as the apple of <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.1#fig001024">Figure 24</a>, provides an approximate realisation of such motion. (Air resistance, which increases with speed, prevents the acceleration from being truly constant in such cases.) A more precise realisation of uniformly accelerated motion is provided by an object falling under gravity close to the surface of an airless body such as the Moon (see Figure 38) or by a falling object in an evacuated (i.e. airless) droptower or dropshaft of the kind discussed in Section 1 of this course.</p><div class="oucontentfigure oucontentmediamini" id="fig001_034"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/20ce2ed7/s207_2_034i.jpg" alt="" width="232" height="223" style="maxwidth:232px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 38:</b> The accelerationtime graph for an object with constant (positive) acceleration</span></div></div></div><p>The accelerationtime graph for a uniformly accelerated body is simple; it's just a horizontal line of the kind shown in <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection5.1#fig001034">Figure 38</a>. The value of this constant acceleration (which may be positive or negative) represents the <i>gradient</i> of the velocitytime graph at any moment. It follows that the velocitytime graph must have the same gradient at all points and must therefore be a straight line of the kind shown in <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection5.1#fig001033">Figure 39</a>. Note that the sign of the acceleration determines whether the velocitytime graph slopes up or down, and the value of the acceleration determines the precise value of the gradient. However, the acceleration does not determine the initial value of <i>v</i><sub><i>x</i> </sub> so we have arbitrarily chosen a point on the <i>v</i><sub><i>x</i> </sub>axis to represent this value and labelled it <i>u<sub>x</sub> </i>. From Figure 38, we deduce that if</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/a5437918/s207_1_ue019i.gif" alt="" width="283" height="14" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>then</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/1c2b3b5d/s207_1_ue020i.gif" alt="" width="283" height="14" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>where <i>a<sub>x</sub></i> represents the <i>constant</i> value of <i>a<sub>x</sub></i>(<i>t</i>).</p><div class="oucontentfigure oucontentmediamini" id="fig001_033"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/0e87e0d4/s207_2_035i.jpg" alt="" width="231" height="224" style="maxwidth:231px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 39:</b> A velocitytime graph that is consistent with the accelerationtime graph of Figure 38. Note that the intercept has been chosen arbitrarily; only the gradient is determined by the acceleration</span></div></div></div><p>It's rather more difficult to deduce the positiontime graph that corresponds to uniformly accelerated motion, but the steady change of velocity with time certainly implies that the <i>gradient</i> of the positiontime graph must also change steadily with time. In fact, given a velocitytime graph like that in Figure 39, the corresponding positiontime graph will be of the general form shown in Figure 40.</p><div class="oucontentfigure oucontentmediamini" id="fig001_044179"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/b0e45fa5/s207_2_036i.jpg" alt="" width="232" height="230" style="maxwidth:232px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 40:</b> A positiontime graph that is consistent with the accelerationtime graph of Figure 38 and the velocitytime graph of Figure 39. Note that the intercept has been chosen arbitrarily; only the gradient is determined by velocity</span></div></div></div><p>Once again, the intercept with the vertical axis (representing the initial position at <i>t</i> = 0) is not determined by anything we have said so far; it has therefore been chosen arbitrarily and labelled <i>x</i><sub>0</sub>. The curve however is not arbitrary since the displacement from <i>x</i><sub>0</sub> at any particular time <i>t</i> = <i>T</i> is determined by the area under the velocitytime graph between <i>t</i> = 0 and <i>t</i> = <i>T</i>. That area will be the sum of two parts, a rectangle of height <i>u<sub>x</sub></i> and base length <i>T</i>, and a triangle of height <i>a<sub>x</sub>T</i> and base length <i>T</i>. Since the area of a triangle is half the product of its height and its base length, it follows that the displacement from <i>x</i><sub>0</sub> at time <i>T</i> will be</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_045"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/d385451e/s207_1_ie016i.gif" alt=""/></div><p>and the position of the uniformly accelerating particle, at any time <i>t</i>, will be <i>x</i>(<i>t</i>) = <i>x</i><sub>0</sub> + <i>s<sub>x</sub></i>(<i>t</i>), that is</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_046"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/abfa7e8b/s207_1_ue021i.gif" alt=""/></div><p>Equations 19, 20 and 21 provide an essentially complete description of uniformly accelerated motion in one dimension and have many applications. They are not the most common form of the equations of uniform acceleration. We will discuss those in the next subsection, but before doing so let's use the method of differentiation to confirm the consistency of the equations we have deduced.</p><p>Starting from Equation 21, the righthand side of which is a quadratic function of <i>t</i>, we expect to find that</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_047"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/fc9a397d/s207_1_ie017i.gif" alt=""/></div><p>Using the rules and results of <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.2#tbl001006">Table 6</a> to carry out the differentiation (which is just like that in Question 18) we find that</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_048"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/996f37b8/s207_1_ie018i.gif" alt=""/></div><p>in complete agreement with Equation 20. Similarly, differentiating the linear function that appears on the righthand side of Equation 20, we expect to find</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_049"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/91b8d422/s207_1_ie019i.gif" alt=""/></div><p>Again, performing the differentiation, using <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.2#tbl001006">Table 6</a>, confirms our expectations:</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_050"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/697844f6/s207_1_ie020i.gif" alt=""/></div><p>We see that in this case, <i>a<sub>x</sub> </i>(<i>t</i>) is just the constant acceleration <i>a<sub>x</sub> </i> from which we started.</p><p>This short exercise in checking consistency gives just a hint of the immense power of differentiation to simplify a wide range of tasks and investigations.</p><p>Returning now to the description of uniformly accelerated motion, let's gather together our results so far:</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/b66268f6/s207_1_ue022i.gif" alt="" width="283" height="27" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/cd9b5ff1/s207_1_ue023i.gif" alt="" width="283" height="13" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/03f45dd1/s207_1_ue024i.gif" alt="" width="283" height="13" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>These are the equations we shall use, rearrange and extend in the next subsection.</p>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

5.2 The equations of uniformly accelerated motion
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection5.2
Tue, 12 Apr 2016 23:00:00 GMT
<p>Equations 22, 23 and 24 provide a complete description of uniformly accelerated motion. By combining them appropriately, it is possible to solve a wide class of problems concerning the kinematics of uniformly accelerated motion. Nonetheless, those particular equations are not always the best starting point for the most common problems. For example, it is often the case that we want to know the displacement from the initial position after some specified period of constant acceleration, rather than the final position <i>x</i>. In such circumstances it is useful to subtract <i>x</i>﻿<sub>0</sub> from both sides of <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection5.1#ueqn001050">Equation 22</a> and use the definition <i>s﻿<sub>x</sub>﻿</i> = <i>x</i> − <i>x</i>﻿<sub>0</sub> to write the resulting equation as</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/8561b729/s207_1_ue025i.gif" alt="" width="283" height="27" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>More significantly, it is often the case that we need to find the final velocity <i>v</i>﻿<sub><i>x</i> </sub> when all we are given is the (constant) acceleration <i>a<sub>x</sub> </i>, the initial velocity <i>u<sub>x</sub> </i> and the displacement <i>s<sub>x</sub> </i>. The problem can be solved using Equations 23 and 25, but doing so involves finding the duration of the motion <i>t</i>, which is not required as part of the answer. It would be more convenient to use an equation that related <i>v</i>﻿<sub><i>x</i> </sub> to <i>a﻿<sub>x</sub> </i>, <i>u﻿<sub>x</sub> </i> and <i>s﻿<sub>x</sub> </i> directly, thus avoiding the need to work out <i>t</i> altogether. Fortunately, it is possible to find such an equation by using a standard mathematical procedure called <b>elimination</b>.</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/cd9b5ff1/s207_1_ue023i.gif" alt="" width="283" height="13" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>The first step in the process is to identify a set of equations that contain the variables we want to relate, along with at least one variable we can eliminate. In this case we want to eliminate <i>t</i> from Equations 23 and 25. The second step usually involves rearranging one of the equations so that the unwanted variable is isolated on the lefthand side, thus becoming the <b>subject</b> of that equation. In this case, we can subtract <i>u﻿<sub>x</sub> </i> from both sides of Equation 23, divide both sides by <i>a﻿<sub>x</sub> </i> and then interchange the two sides to give</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/d6114d9f/s207_1_ue026i.gif" alt="" width="283" height="34" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>Having obtained this relation from one of the equations, the third step is to use it to eliminate the unwanted variable from all the other equations. In our case this means replacing <i>t</i> by (﻿<i>v</i>﻿﻿﻿<sub><i>x</i></sub> − <i>u﻿<sub>x</sub>﻿</i>)﻿/﻿<i>a﻿<sub>x</sub> </i> throughout Equation 25, so we obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_056"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/c6552cb7/s207_1_ie021i.gif" alt=""/></div><p>so after a little algebra, we obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_057"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/e3f887b6/s207_1_ie022i.gif" alt=""/></div><p>which can be rearranged to give the result</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/4d4eff20/s207_1_ue027i.gif" alt="" width="283" height="18" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>Equations 23, 25 and 27 are the most frequently used equations of uniformly accelerated motion and are usually referred to collectively as the constant acceleration equations (or the uniform acceleration equations).</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox oucontentsnoheading " id="box001_019"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>Constant (or uniform) acceleration equations</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_059"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7ab379df/s207_1_ue028i.gif" alt=""/></div><p/><p/></div></div></div><p>It follows from these equations that</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox oucontentsnoheading " id="box0019a"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p/><div class="oucontentequation oucontentequationequation oucontentnocaption" id="fig001_045"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/1ae6d41d/s207_1_ue028di.gif" alt=""/></div></div></div></div><p>Remember, these are not universal equations that describe every form of motion. They apply <i>only</i> to situations in which the acceleration <i>a﻿<sub>x</sub> </i> is <i>constant</i>.</p><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_018"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 20</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Starting from the constant acceleration equations, use the elimination procedure to derive Equation 28d.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>In this case we wish to eliminate <i>a<sub>x</sub> </i> from Equations 28a and 28b. One way is to rearrange Equation 28b (subtracting <i>u<sub>x</sub> </i> from both sides and dividing both sides by <i>t</i>) to obtain:</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_015"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/81b03e93/s207_1_ie048i.gif" alt=""/></div><p>Substituting this into Equation 28a gives</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_016"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2fead9e2/s207_1_ie049i.gif" alt=""/></div></div></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_019"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 21</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Give a graphical interpretation of Equation 28d in terms of the area under a velocitytime graph for the case of uniformly accelerated motion.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Graphically, <i>s<sub>x</sub> </i> is the signed area under the velocitytime graph between the given times. In this case those times are <i>t</i> = 0 when the initial velocity is <i>u<sub>x</sub> </i> and some later time <i>t</i> when the velocity is <i>v<sub>x</sub> </i>. Since the velocitytime graph for uniformly accelerated motion is a straight line of fixed gradient, the area required will always be either a trapezium (above or below the axis) or a pair of triangles. (<a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection5.2#fig001040a">Figure 37</a> shows a particular case.)</p><div class="oucontentfigure" style="width:391px;" id="fig001_040a"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/5dd2e090/s207_2_041i.jpg" alt="" width="391" height="310" style="maxwidth:391px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"><b>Figure 37:</b> The velocitytime graph for Selfassessment question 19</span></div></div></div></div></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_020"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 22 </h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>An object has a final velocity of 30.0 m s<sup>﻿−﻿1</sup>, after accelerating uniformly at 2.00 m s<sup>﻿−﻿2</sup> over a displacement of 20.0 m. (a) What was the initial speed of the object? (b) For how long was the object accelerated?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) In this case we know <i>s<sub>x</sub> </i> = 20 m, <i>v<sub>x</sub> </i> = 30 m s<sup>−1</sup> and <i>a<sub>x</sub> </i> = 2 m s<sup>−2</sup>, and we need to find <i>u<sub>x</sub></i>. As a first step we should make <i>u<sub>x</sub> </i> the subject of an equation that involves the known quantities. Rearranging Equation 28c gives</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_017"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/b508c3d3/s207_1_ie050i.gif" alt=""/></div><p>Taking the square root of each side</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_018"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/010130ea/s207_1_ie051i.gif" alt=""/></div><p>Substituting the given values,</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_019"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/665d0dc7/s207_1_ie052i.gif" alt=""/></div><p>(b) The duration of the acceleration is given by Equation 26 as <i>t</i> = (<i>v<sub>x</sub> </i> − <i>u<sub>x</sub></i>)<i>/a<sub>x</sub></i>, which was itself obtained by rearranging constant acceleration equations. Substituting the relevant values gives</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_020"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/04d7165d/s207_1_ie053i.gif" alt=""/></div></div></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_021"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 23</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>A train accelerates uniformly along a straight track at 2.00 m s<sup>﻿−﻿2</sup> from an initial velocity of 4.00 m s<sup>﻿−﻿1</sup> to a final velocity of 16.00 m s﻿<sup>−﻿1</sup>. What is the train's displacement from its initial position at the end of this interval?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Using Equation 28c</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_021"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/86963263/s207_1_ie054i.gif" alt=""/></div><p>with <i>u<sub>x</sub> </i> = 4.00 m s<sup>−1</sup>, <i>v<sub>x</sub> </i> = 16.0 m s<sup>−1</sup> and <i>a<sub>x</sub> </i> = 2.00 m s<sup>−2</sup>, we obtain, after rearranging</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_022"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/0364f2ce/s207_1_ie055i.gif" alt=""/></div></div></div></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection5.2
5.2 The equations of uniformly accelerated motionS207_2<p>Equations 22, 23 and 24 provide a complete description of uniformly accelerated motion. By combining them appropriately, it is possible to solve a wide class of problems concerning the kinematics of uniformly accelerated motion. Nonetheless, those particular equations are not always the best starting point for the most common problems. For example, it is often the case that we want to know the displacement from the initial position after some specified period of constant acceleration, rather than the final position <i>x</i>. In such circumstances it is useful to subtract <i>x</i><sub>0</sub> from both sides of <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection5.1#ueqn001050">Equation 22</a> and use the definition <i>s<sub>x</sub></i> = <i>x</i> − <i>x</i><sub>0</sub> to write the resulting equation as</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/8561b729/s207_1_ue025i.gif" alt="" width="283" height="27" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>More significantly, it is often the case that we need to find the final velocity <i>v</i><sub><i>x</i> </sub> when all we are given is the (constant) acceleration <i>a<sub>x</sub> </i>, the initial velocity <i>u<sub>x</sub> </i> and the displacement <i>s<sub>x</sub> </i>. The problem can be solved using Equations 23 and 25, but doing so involves finding the duration of the motion <i>t</i>, which is not required as part of the answer. It would be more convenient to use an equation that related <i>v</i><sub><i>x</i> </sub> to <i>a<sub>x</sub> </i>, <i>u<sub>x</sub> </i> and <i>s<sub>x</sub> </i> directly, thus avoiding the need to work out <i>t</i> altogether. Fortunately, it is possible to find such an equation by using a standard mathematical procedure called <b>elimination</b>.</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/cd9b5ff1/s207_1_ue023i.gif" alt="" width="283" height="13" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>The first step in the process is to identify a set of equations that contain the variables we want to relate, along with at least one variable we can eliminate. In this case we want to eliminate <i>t</i> from Equations 23 and 25. The second step usually involves rearranging one of the equations so that the unwanted variable is isolated on the lefthand side, thus becoming the <b>subject</b> of that equation. In this case, we can subtract <i>u<sub>x</sub> </i> from both sides of Equation 23, divide both sides by <i>a<sub>x</sub> </i> and then interchange the two sides to give</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/d6114d9f/s207_1_ue026i.gif" alt="" width="283" height="34" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>Having obtained this relation from one of the equations, the third step is to use it to eliminate the unwanted variable from all the other equations. In our case this means replacing <i>t</i> by (<i>v</i><sub><i>x</i></sub> − <i>u<sub>x</sub></i>)/<i>a<sub>x</sub> </i> throughout Equation 25, so we obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_056"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/c6552cb7/s207_1_ie021i.gif" alt=""/></div><p>so after a little algebra, we obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_057"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/e3f887b6/s207_1_ie022i.gif" alt=""/></div><p>which can be rearranged to give the result</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/4d4eff20/s207_1_ue027i.gif" alt="" width="283" height="18" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>Equations 23, 25 and 27 are the most frequently used equations of uniformly accelerated motion and are usually referred to collectively as the constant acceleration equations (or the uniform acceleration equations).</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox
oucontentsnoheading
" id="box001_019"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>Constant (or uniform) acceleration equations</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_059"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7ab379df/s207_1_ue028i.gif" alt=""/></div><p/><p/></div></div></div><p>It follows from these equations that</p><div class="oucontentbox oucontentsheavybox1 oucontentsbox
oucontentsnoheading
" id="box0019a"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p/><div class="oucontentequation oucontentequationequation oucontentnocaption" id="fig001_045"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/1ae6d41d/s207_1_ue028di.gif" alt=""/></div></div></div></div><p>Remember, these are not universal equations that describe every form of motion. They apply <i>only</i> to situations in which the acceleration <i>a<sub>x</sub> </i> is <i>constant</i>.</p><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_018"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 20</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Starting from the constant acceleration equations, use the elimination procedure to derive Equation 28d.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>In this case we wish to eliminate <i>a<sub>x</sub> </i> from Equations 28a and 28b. One way is to rearrange Equation 28b (subtracting <i>u<sub>x</sub> </i> from both sides and dividing both sides by <i>t</i>) to obtain:</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_015"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/81b03e93/s207_1_ie048i.gif" alt=""/></div><p>Substituting this into Equation 28a gives</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_016"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2fead9e2/s207_1_ie049i.gif" alt=""/></div></div></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_019"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 21</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Give a graphical interpretation of Equation 28d in terms of the area under a velocitytime graph for the case of uniformly accelerated motion.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Graphically, <i>s<sub>x</sub> </i> is the signed area under the velocitytime graph between the given times. In this case those times are <i>t</i> = 0 when the initial velocity is <i>u<sub>x</sub> </i> and some later time <i>t</i> when the velocity is <i>v<sub>x</sub> </i>. Since the velocitytime graph for uniformly accelerated motion is a straight line of fixed gradient, the area required will always be either a trapezium (above or below the axis) or a pair of triangles. (<a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection5.2#fig001040a">Figure 37</a> shows a particular case.)</p><div class="oucontentfigure" style="width:391px;" id="fig001_040a"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/5dd2e090/s207_2_041i.jpg" alt="" width="391" height="310" style="maxwidth:391px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"><b>Figure 37:</b> The velocitytime graph for Selfassessment question 19</span></div></div></div></div></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_020"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 22 </h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>An object has a final velocity of 30.0 m s<sup>−1</sup>, after accelerating uniformly at 2.00 m s<sup>−2</sup> over a displacement of 20.0 m. (a) What was the initial speed of the object? (b) For how long was the object accelerated?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) In this case we know <i>s<sub>x</sub> </i> = 20 m, <i>v<sub>x</sub> </i> = 30 m s<sup>−1</sup> and <i>a<sub>x</sub> </i> = 2 m s<sup>−2</sup>, and we need to find <i>u<sub>x</sub></i>. As a first step we should make <i>u<sub>x</sub> </i> the subject of an equation that involves the known quantities. Rearranging Equation 28c gives</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_017"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/b508c3d3/s207_1_ie050i.gif" alt=""/></div><p>Taking the square root of each side</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_018"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/010130ea/s207_1_ie051i.gif" alt=""/></div><p>Substituting the given values,</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_019"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/665d0dc7/s207_1_ie052i.gif" alt=""/></div><p>(b) The duration of the acceleration is given by Equation 26 as <i>t</i> = (<i>v<sub>x</sub> </i> − <i>u<sub>x</sub></i>)<i>/a<sub>x</sub></i>, which was itself obtained by rearranging constant acceleration equations. Substituting the relevant values gives</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_020"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/04d7165d/s207_1_ie053i.gif" alt=""/></div></div></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_021"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 23</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>A train accelerates uniformly along a straight track at 2.00 m s<sup>−2</sup> from an initial velocity of 4.00 m s<sup>−1</sup> to a final velocity of 16.00 m s<sup>−1</sup>. What is the train's displacement from its initial position at the end of this interval?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Using Equation 28c</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_021"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/86963263/s207_1_ie054i.gif" alt=""/></div><p>with <i>u<sub>x</sub> </i> = 4.00 m s<sup>−1</sup>, <i>v<sub>x</sub> </i> = 16.0 m s<sup>−1</sup> and <i>a<sub>x</sub> </i> = 2.00 m s<sup>−2</sup>, we obtain, after rearranging</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_022"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/0364f2ce/s207_1_ie055i.gif" alt=""/></div></div></div></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

5.3 The acceleration due to gravity
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection5.3
Tue, 12 Apr 2016 23:00:00 GMT
<p>In the absence of air resistance, an object falling freely under the influence of the Earth's gravity, close to the surface of the Earth, experiences an acceleration of about 9.81 m s﻿<sup>−﻿2</sup> in the downward direction. The precise value of the magnitude is indicated by the symbol <i>g</i> and varies slightly from place to place due to variations in surface altitude, the effect of the Earth's rotation and variations in the internal composition of the Earth. Some typical values for the <b>magnitude of the acceleration due to gravity</b>, <i>g</i>, at various points on the Earth's surface, are given in Table 7.</p><p>Incidentally, one of the reasons that the value of <i>g</i> is well known across much of the Earth's surface is that extensive surveys have been carried out in which <i>g</i> has been accurately measured by timing the swing of very carefully constructed pendulums.</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_007"><h2 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 7:</b> The magnitude of the acceleration due to gravity at various points on the Earth's surface</h2><div class="oucontenttablewrapper"><table><tr><th scope="col">Location</th><th scope="col"><i>g</i>/m s<sup>−2</sup></th></tr><tr><td>North Pole</td><td>9.83</td></tr><tr><td>London</td><td>9.81</td></tr><tr><td>New York</td><td>9.80</td></tr><tr><td>Equator</td><td>9.78</td></tr><tr><td>Sydney</td><td>9.80</td></tr></table></div><div class="oucontentsourcereference"></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection5.3
5.3 The acceleration due to gravityS207_2<p>In the absence of air resistance, an object falling freely under the influence of the Earth's gravity, close to the surface of the Earth, experiences an acceleration of about 9.81 m s<sup>−2</sup> in the downward direction. The precise value of the magnitude is indicated by the symbol <i>g</i> and varies slightly from place to place due to variations in surface altitude, the effect of the Earth's rotation and variations in the internal composition of the Earth. Some typical values for the <b>magnitude of the acceleration due to gravity</b>, <i>g</i>, at various points on the Earth's surface, are given in Table 7.</p><p>Incidentally, one of the reasons that the value of <i>g</i> is well known across much of the Earth's surface is that extensive surveys have been carried out in which <i>g</i> has been accurately measured by timing the swing of very carefully constructed pendulums.</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_007"><h2 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 7:</b> The magnitude of the acceleration due to gravity at various points on the Earth's surface</h2><div class="oucontenttablewrapper"><table><tr><th scope="col">Location</th><th scope="col"><i>g</i>/m s<sup>−2</sup></th></tr><tr><td>North Pole</td><td>9.83</td></tr><tr><td>London</td><td>9.81</td></tr><tr><td>New York</td><td>9.80</td></tr><tr><td>Equator</td><td>9.78</td></tr><tr><td>Sydney</td><td>9.80</td></tr></table></div><div class="oucontentsourcereference"></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

5.4 Droptowers revisited
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection5.4
Tue, 12 Apr 2016 23:00:00 GMT
<p>In Section 1 we described how research into near weightless conditions can be carried out on Earth by using a droptower or a dropshaft (Figure 41). We are now in a position to examine dropshafts in more detail (Example 3).</p><div class="oucontentfigure oucontentmediamini" id="fig001_046"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/a0ee1d98/s207_2_037i.jpg" alt="" width="238" height="422" style="maxwidth:238px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 41:</b> The test vehicle at the Bremen droptower</span></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="exe001_003"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Example 3</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Consider a vertical shaft of total length 700 m, with free fall taking place for the first 500 m and constant deceleration for the final 200 m. Work out:</p><ol class="oucontentnumbered"><li><p>The time to fall the first 500 m. (This would be the time during which near weightless experiments could be done.)</p></li><li><p>The velocity at the end of the first 500 m.</p></li><li><p>The acceleration needed to reduce the velocity to zero at the bottom of the shaft.</p></li></ol></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>It is convenient to split the calculations into three corresponding parts.</p><p><b>Part 1: Finding the time to fall the first 500 m</b></p><p>When carrying out calculations it is always wise to choose a coordinate system that fits the problem. In this case it seems natural to choose an <i>x</i>﻿﻿axis that points down the shaft, so that downward displacements and velocities will be positive and the acceleration due to gravity will be <i>a﻿<sub>x</sub></i> = + <i>g</i>. It is also sensible to use the equations that lead to the least amount of work. In this case, we know only the initial velocity (assuming the object is initially at rest), the acceleration (<i>g</i>) and the final displacement (500 m). So, in order to work out the time to fall, the best approach is to use Equation 28a</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/8561b729/s207_1_ue025i.gif" alt="" width="283" height="27" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>with <i>s﻿<sub>x</sub>﻿</i> = 500 m, <i>u﻿<sub>x</sub></i> = 0 m s<sup>﻿−﻿1</sup> and <i>a﻿<sub>x</sub></i> = <i>g</i> = 9.81 m s<sup>﻿−﻿2</sup>. Setting <i>u﻿<sub>x</sub></i> = 0 m s<sup>﻿−1</sup> and rearranging Equation 28a we obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_064"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/a41c28c7/s207_1_ie023i.gif" alt=""/></div><p><b>Part 2: Finding the velocity at the end of the first 500 m</b></p><p>To calculate the velocity after a free fall of 500 m, there are two convenient methods. We could either use</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/cd9b5ff1/s207_1_ue023i.gif" alt="" width="283" height="13" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>with <i>u<sub>x</sub> </i> = 0 m s<sup>−1</sup>, <i>a<sub>x</sub> </i> = <i>g</i> = 9.81 m s<sup>−2</sup> and <i>t</i> = 10.1 s, or</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/4d4eff20/s207_1_ue027i.gif" alt="" width="283" height="18" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>with <i>u<sub>x</sub> </i> = 0 m s<sup>−1</sup>, <i>a<sub>x</sub> </i> = <i>g</i> = 9.81 m s<sup>−2</sup> and <i>s<sub>x</sub> </i> = 500 m.</p><p>We choose the second method since it uses only values that we have been given rather than ones we have calculated. (It is always possible that we made some slip in our calculation.) So using Equation 28c with <i>u﻿<sub>x</sub></i> = 0 m s<sup>﻿−﻿1</sup>, we obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_067"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/bb8c2645/s207_1_ie024i.gif" alt=""/></div><p><b>Part 3: Finding the acceleration over the final 200 m</b></p><p>To find the uniform acceleration necessary to bring the object to rest at the bottom of the shaft we can again make use of Equation 28c. This time, the quantity <i>a<sub>x</sub> </i> is what we want to calculate and the known parameters are <i>v</i>﻿<sub><i>x</i></sub> = 0 m s<sup>﻿−﻿1</sup>, <i>u﻿<sub>x</sub></i> = 99.0 m s<sup>−1</sup> and <i>s﻿<sub>x</sub></i> = 200 m. Substituting <i>v</i>﻿<sub>﻿<i>x</i></sub> = 0 m s<sup>﻿−﻿1</sup> and rearranging Equation 28c gives</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_068"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/46b1af95/s207_1_ie025i.gif" alt=""/></div></div></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_022"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 24</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Consider the following proposal to roughly double the period of weightlessness in the 140 m Bremen droptower. The basic idea is that the dropvehicle should be launched from the bottom of the tower with just the right velocity to enable it to reach the top of the tower with zero velocity. If the vehicle is uniformly accelerated over the first 10 m to give it the necessary upward velocity it can then rise freely for the remaining 130 m, pass through its highest point, and fall freely for another 130 m before being brought to rest over the final 10 m of its descent. Using a coordinate system in which the upward direction is positive, answer the following questions.</p><p>(a) What must be the launch velocity of the vehicle if it is to freely rise 130 m?</p><p>(b) What constant total acceleration must be applied over the first 10 m if the vehicle is to attain the required launch velocity? How long will this last?</p><p>(c) How long will the vehicle actually spend in free motion?</p><p>(d) What will be the acceleration of the vehicle when at its highest point?</p><p>(e) What will be the displacement of the vehicle from its starting point when it returns to that point at the end of the trip, and through what distance will it have travelled during the round trip?</p><p>(f) Sketch a rough accelerationtime graph for the entire motion, and briefly comment on the feasibility of the whole proposal.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) Using a coordinate system in which up is the positive direction means that the acceleration due to gravity will be negative, that is <i>a<sub>x</sub> </i> = −<i>g</i>. With <i>v<sub>x</sub> </i> = 0 m s<sup>−1</sup> and <i>s<sub>x</sub> </i> = 130 m, the initial velocity will be given (as in Question 22) by</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_023"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/529f1360/s207_1_ie056i.gif" alt=""/></div><p>(b) In this case we know <i>u<sub>x</sub> </i> = 0 m s<sup>−1</sup>, <i>v<sub>x</sub> </i> = 50.5 m s<sup>−1</sup> and <i>s<sub>x</sub> </i> = 10.0 m. (Note that what was an initial velocity in the last part of the question has become a final velocity in this part.) Rearranging Equation 28c and substituting the given values</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_024"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/41ddea28/s207_1_ie057i.gif" alt=""/></div><p>To find the duration of the acceleration rearrange Equation 28b</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_025"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/b6989800/s207_1_ie058i.gif" alt=""/></div><p>(c) The time in free motion is given by Equation 26 with <i>u<sub>x</sub> </i> = 50.5 m s<sup>−1</sup>, <i>v<sub>x</sub> </i> = −50.5 m s<sup>−1</sup> and <i>a<sub>x</sub> </i> = −9.81 m s<sup>−2</sup></p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_051"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/92ba6ec1/s207_1_ie059i.gif" alt=""/></div><p>(Pay attention to signs here, remember that up is positive so the initial velocity will be positive, but the final velocity will be negative.)</p><p>(d) When at its highest point the acceleration of the vehicle will be <i>a<sub>x</sub> </i> = −<i>g</i>. (The fact that the vehicle's velocity is momentarily zero as it passes through the highest point does not affect the (constant) acceleration.)</p><p>(e) When the particle returns to its starting point its displacement from that point will be zero. The distance travelled, however, will be twice the height of the tower, 280 m.</p><p>(f) The accelerationtime graph is given in Figure 42. Note that the initial and final accelerations are both positive because both increase the vehicle's velocity, even though the final acceleration reduces the vehicle's speed. (This subtlety concerning acceleration was discussed in Section 4.2.) The plan is feasible. In fact a similar system is in use at various droptowers and dropshafts.</p><div class="oucontentfigure" style="width:391px;" id="fig001_047"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2898ca2b/s207_2_042i.jpg" alt="" width="391" height="437" style="maxwidth:391px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 42:</b> The accelerationtime graph for Question 24(f)</span></div></div></div></div></div></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection5.4
5.4 Droptowers revisitedS207_2<p>In Section 1 we described how research into near weightless conditions can be carried out on Earth by using a droptower or a dropshaft (Figure 41). We are now in a position to examine dropshafts in more detail (Example 3).</p><div class="oucontentfigure oucontentmediamini" id="fig001_046"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/a0ee1d98/s207_2_037i.jpg" alt="" width="238" height="422" style="maxwidth:238px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 41:</b> The test vehicle at the Bremen droptower</span></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="exe001_003"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Example 3</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Consider a vertical shaft of total length 700 m, with free fall taking place for the first 500 m and constant deceleration for the final 200 m. Work out:</p><ol class="oucontentnumbered"><li><p>The time to fall the first 500 m. (This would be the time during which near weightless experiments could be done.)</p></li><li><p>The velocity at the end of the first 500 m.</p></li><li><p>The acceleration needed to reduce the velocity to zero at the bottom of the shaft.</p></li></ol></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>It is convenient to split the calculations into three corresponding parts.</p><p><b>Part 1: Finding the time to fall the first 500 m</b></p><p>When carrying out calculations it is always wise to choose a coordinate system that fits the problem. In this case it seems natural to choose an <i>x</i>axis that points down the shaft, so that downward displacements and velocities will be positive and the acceleration due to gravity will be <i>a<sub>x</sub></i> = + <i>g</i>. It is also sensible to use the equations that lead to the least amount of work. In this case, we know only the initial velocity (assuming the object is initially at rest), the acceleration (<i>g</i>) and the final displacement (500 m). So, in order to work out the time to fall, the best approach is to use Equation 28a</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/8561b729/s207_1_ue025i.gif" alt="" width="283" height="27" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>with <i>s<sub>x</sub></i> = 500 m, <i>u<sub>x</sub></i> = 0 m s<sup>−1</sup> and <i>a<sub>x</sub></i> = <i>g</i> = 9.81 m s<sup>−2</sup>. Setting <i>u<sub>x</sub></i> = 0 m s<sup>−1</sup> and rearranging Equation 28a we obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_064"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/a41c28c7/s207_1_ie023i.gif" alt=""/></div><p><b>Part 2: Finding the velocity at the end of the first 500 m</b></p><p>To calculate the velocity after a free fall of 500 m, there are two convenient methods. We could either use</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/cd9b5ff1/s207_1_ue023i.gif" alt="" width="283" height="13" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>with <i>u<sub>x</sub> </i> = 0 m s<sup>−1</sup>, <i>a<sub>x</sub> </i> = <i>g</i> = 9.81 m s<sup>−2</sup> and <i>t</i> = 10.1 s, or</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/4d4eff20/s207_1_ue027i.gif" alt="" width="283" height="18" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span></p><p>with <i>u<sub>x</sub> </i> = 0 m s<sup>−1</sup>, <i>a<sub>x</sub> </i> = <i>g</i> = 9.81 m s<sup>−2</sup> and <i>s<sub>x</sub> </i> = 500 m.</p><p>We choose the second method since it uses only values that we have been given rather than ones we have calculated. (It is always possible that we made some slip in our calculation.) So using Equation 28c with <i>u<sub>x</sub></i> = 0 m s<sup>−1</sup>, we obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_067"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/bb8c2645/s207_1_ie024i.gif" alt=""/></div><p><b>Part 3: Finding the acceleration over the final 200 m</b></p><p>To find the uniform acceleration necessary to bring the object to rest at the bottom of the shaft we can again make use of Equation 28c. This time, the quantity <i>a<sub>x</sub> </i> is what we want to calculate and the known parameters are <i>v</i><sub><i>x</i></sub> = 0 m s<sup>−1</sup>, <i>u<sub>x</sub></i> = 99.0 m s<sup>−1</sup> and <i>s<sub>x</sub></i> = 200 m. Substituting <i>v</i><sub><i>x</i></sub> = 0 m s<sup>−1</sup> and rearranging Equation 28c gives</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_068"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/46b1af95/s207_1_ie025i.gif" alt=""/></div></div></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_022"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 24</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Consider the following proposal to roughly double the period of weightlessness in the 140 m Bremen droptower. The basic idea is that the dropvehicle should be launched from the bottom of the tower with just the right velocity to enable it to reach the top of the tower with zero velocity. If the vehicle is uniformly accelerated over the first 10 m to give it the necessary upward velocity it can then rise freely for the remaining 130 m, pass through its highest point, and fall freely for another 130 m before being brought to rest over the final 10 m of its descent. Using a coordinate system in which the upward direction is positive, answer the following questions.</p><p>(a) What must be the launch velocity of the vehicle if it is to freely rise 130 m?</p><p>(b) What constant total acceleration must be applied over the first 10 m if the vehicle is to attain the required launch velocity? How long will this last?</p><p>(c) How long will the vehicle actually spend in free motion?</p><p>(d) What will be the acceleration of the vehicle when at its highest point?</p><p>(e) What will be the displacement of the vehicle from its starting point when it returns to that point at the end of the trip, and through what distance will it have travelled during the round trip?</p><p>(f) Sketch a rough accelerationtime graph for the entire motion, and briefly comment on the feasibility of the whole proposal.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) Using a coordinate system in which up is the positive direction means that the acceleration due to gravity will be negative, that is <i>a<sub>x</sub> </i> = −<i>g</i>. With <i>v<sub>x</sub> </i> = 0 m s<sup>−1</sup> and <i>s<sub>x</sub> </i> = 130 m, the initial velocity will be given (as in Question 22) by</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_023"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/529f1360/s207_1_ie056i.gif" alt=""/></div><p>(b) In this case we know <i>u<sub>x</sub> </i> = 0 m s<sup>−1</sup>, <i>v<sub>x</sub> </i> = 50.5 m s<sup>−1</sup> and <i>s<sub>x</sub> </i> = 10.0 m. (Note that what was an initial velocity in the last part of the question has become a final velocity in this part.) Rearranging Equation 28c and substituting the given values</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_024"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/41ddea28/s207_1_ie057i.gif" alt=""/></div><p>To find the duration of the acceleration rearrange Equation 28b</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_025"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/b6989800/s207_1_ie058i.gif" alt=""/></div><p>(c) The time in free motion is given by Equation 26 with <i>u<sub>x</sub> </i> = 50.5 m s<sup>−1</sup>, <i>v<sub>x</sub> </i> = −50.5 m s<sup>−1</sup> and <i>a<sub>x</sub> </i> = −9.81 m s<sup>−2</sup></p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_051"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/92ba6ec1/s207_1_ie059i.gif" alt=""/></div><p>(Pay attention to signs here, remember that up is positive so the initial velocity will be positive, but the final velocity will be negative.)</p><p>(d) When at its highest point the acceleration of the vehicle will be <i>a<sub>x</sub> </i> = −<i>g</i>. (The fact that the vehicle's velocity is momentarily zero as it passes through the highest point does not affect the (constant) acceleration.)</p><p>(e) When the particle returns to its starting point its displacement from that point will be zero. The distance travelled, however, will be twice the height of the tower, 280 m.</p><p>(f) The accelerationtime graph is given in Figure 42. Note that the initial and final accelerations are both positive because both increase the vehicle's velocity, even though the final acceleration reduces the vehicle's speed. (This subtlety concerning acceleration was discussed in Section 4.2.) The plan is feasible. In fact a similar system is in use at various droptowers and dropshafts.</p><div class="oucontentfigure" style="width:391px;" id="fig001_047"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2898ca2b/s207_2_042i.jpg" alt="" width="391" height="437" style="maxwidth:391px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 42:</b> The accelerationtime graph for Question 24(f)</span></div></div></div></div></div></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

6.1 course summary
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection6.1
Tue, 12 Apr 2016 23:00:00 GMT
<p>1. A coordinate system provides a systematic means of specifying the position of a particle. A system in one dimension involves choosing an origin and a positive direction in which values of the position coordinate increase. Values of the position coordinate are positive or negative numbers multiplied by an appropriate unit of length, usually the SI unit of length, the metre (m).</p><p>2. The movement of a particle along a line can be described graphically by plotting values of the particle's position <i>x</i>, against the corresponding times <i>t</i>, to produce a positiontime graph. Alternatively, by choosing an appropriate reference position <i>x</i>﻿<sub>ref</sub> and defining the displacement from that point by <i>s﻿<sub>x</sub></i> = <i>x</i> − <i>x</i> <sub>ref</sub>, the motion may be described by means of a displacementtime graph.</p><p>3. Uniform motion along a line is characterised by a straightline positiontime graph that may be described by the equation</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_006a"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/6bc4d638/s207_1_ue006ai.gif" alt=""/></div><p>where <i>v</i><sub><i>x</i></sub> and <i>x</i><sub>0</sub> are constants. Physically, <i>v</i><sub><i>x</i></sub> represents the particle's velocity, the rate of change of its position with respect to time, and is determined by the gradient of the positiontime graph</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="eqn001_005a"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/680a03dc/s207_1_ue005i.gif" alt=""/></div><p><i>x</i>﻿<sub>0</sub> represents the particle's initial position, its position at <i>t</i> = 0, and is determined by the intercept of the positiontime graph, the value of <i>x</i> at which the plotted line crosses the <i>x</i>axis, provided that axis has been drawn through <i>t</i> = 0.</p><p>4. Nonuniform motion along a line is characterised by a positiontime graph that is not a straight line. In such circumstances the rate of change of position with respect to time may vary from moment to moment and defines the instantaneous velocity. Its value at any particular time is determined by the gradient of the tangent to the positiontime graph at that time.</p><p>5. More generally, if the position of a particle varies with time in the way described by a function <i>x</i>﻿(﻿<i>t</i>﻿), then the way in which the (instantaneous) velocity varies with time will be described by the associated derived function or derivative</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="eqn001_014"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/6a0d5dc2/s207_1_ue014i.gif" alt=""/></div><p>6. The instantaneous acceleration is the rate of change of the instantaneous velocity with respect to time. Its value at any time is determined by the gradient of the tangent to the velocitytime graph at that time. More generally, the way in which the (instantaneous) acceleration varies with time will be described by the derivative of the function that describes the instantaneous velocity, or, equivalently, the second derivative of the function that describes the position:</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="eqn001_0182"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/a4361a81/s207_1_ue018_2i.gif" alt=""/></div><p>7. Results and rules relating to differentiation and the determination of derivatives are contained in <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.2#tbl001006">Table 6</a>. The derivative of a constant is zero, the derivative of <i>f</i>﻿(﻿<i>y</i>﻿) = <i>﻿Ay﻿<sup>n</sup>﻿</i> is d﻿<i>f</i>﻿/﻿d﻿<i>y</i> = <i>nAy<sup>n</sup> </i> <sup> −1</sup>.</p><p>8. The signed area under a velocitytime graph, between specified values of time, represents the change in position of the particle during that interval.</p><p>9. Uniformly accelerated motion is a special case of nonuniform motion characterised by a constant value of the acceleration, <i>a<sub>x</sub> </i> = constant. In such circumstances the velocity is a linear function of time (<i>v</i>﻿<sub><i>x</i></sub>﻿(﻿<i>t</i>﻿) = <i>u﻿<sub>x</sub></i> + <i>a﻿<sub>x</sub>﻿t</i>), and the position is a quadratic function of time (﻿<span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2b4bb3e6/s207_1_ie077i.gif" alt="" width="156" height="27" style="maxwidth:156px;" class="oucontentinlinefigureimage"/></span>).</p><p>10. The most widely used equations describing uniformly accelerated motion are</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_028a"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7ab379df/s207_1_ue028i.gif" alt=""/></div><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_028b"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/1ae6d41d/s207_1_ue028di.gif" alt=""/></div><p>11. Position <i>x</i>, displacement <i>s<sub>x</sub> </i>, velocity <i>v</i> <sub> <i>x</i> </sub>, and acceleration, <i>a<sub>x</sub> </i>, are all signed quantities that may be positive or negative, depending on the associated direction. The magnitude of each of these quantities is a positive quantity that is devoid of directional information. The magnitude of the displacement of one point from another, <i>s</i> = ﻿<i>s﻿<sub>x</sub>﻿</i>, represents the distance between those two points, while the magnitude of a particle's velocity, <i>v</i> = ﻿<i>v</i>﻿<sub><i>x</i></sub>﻿, represents the speed of the particle. The magnitude of the acceleration due to gravity is represented by the symbol <i>g</i>, and has the approximate value 9.81 m s<sup>﻿−﻿2</sup> across much of the Earth's surface.</p>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection6.1
6.1 course summaryS207_2<p>1. A coordinate system provides a systematic means of specifying the position of a particle. A system in one dimension involves choosing an origin and a positive direction in which values of the position coordinate increase. Values of the position coordinate are positive or negative numbers multiplied by an appropriate unit of length, usually the SI unit of length, the metre (m).</p><p>2. The movement of a particle along a line can be described graphically by plotting values of the particle's position <i>x</i>, against the corresponding times <i>t</i>, to produce a positiontime graph. Alternatively, by choosing an appropriate reference position <i>x</i><sub>ref</sub> and defining the displacement from that point by <i>s<sub>x</sub></i> = <i>x</i> − <i>x</i> <sub>ref</sub>, the motion may be described by means of a displacementtime graph.</p><p>3. Uniform motion along a line is characterised by a straightline positiontime graph that may be described by the equation</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_006a"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/6bc4d638/s207_1_ue006ai.gif" alt=""/></div><p>where <i>v</i><sub><i>x</i></sub> and <i>x</i><sub>0</sub> are constants. Physically, <i>v</i><sub><i>x</i></sub> represents the particle's velocity, the rate of change of its position with respect to time, and is determined by the gradient of the positiontime graph</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="eqn001_005a"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/680a03dc/s207_1_ue005i.gif" alt=""/></div><p><i>x</i><sub>0</sub> represents the particle's initial position, its position at <i>t</i> = 0, and is determined by the intercept of the positiontime graph, the value of <i>x</i> at which the plotted line crosses the <i>x</i>axis, provided that axis has been drawn through <i>t</i> = 0.</p><p>4. Nonuniform motion along a line is characterised by a positiontime graph that is not a straight line. In such circumstances the rate of change of position with respect to time may vary from moment to moment and defines the instantaneous velocity. Its value at any particular time is determined by the gradient of the tangent to the positiontime graph at that time.</p><p>5. More generally, if the position of a particle varies with time in the way described by a function <i>x</i>(<i>t</i>), then the way in which the (instantaneous) velocity varies with time will be described by the associated derived function or derivative</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="eqn001_014"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/6a0d5dc2/s207_1_ue014i.gif" alt=""/></div><p>6. The instantaneous acceleration is the rate of change of the instantaneous velocity with respect to time. Its value at any time is determined by the gradient of the tangent to the velocitytime graph at that time. More generally, the way in which the (instantaneous) acceleration varies with time will be described by the derivative of the function that describes the instantaneous velocity, or, equivalently, the second derivative of the function that describes the position:</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="eqn001_0182"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/a4361a81/s207_1_ue018_2i.gif" alt=""/></div><p>7. Results and rules relating to differentiation and the determination of derivatives are contained in <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection4.3.2#tbl001006">Table 6</a>. The derivative of a constant is zero, the derivative of <i>f</i>(<i>y</i>) = <i>Ay<sup>n</sup></i> is d<i>f</i>/d<i>y</i> = <i>nAy<sup>n</sup> </i> <sup> −1</sup>.</p><p>8. The signed area under a velocitytime graph, between specified values of time, represents the change in position of the particle during that interval.</p><p>9. Uniformly accelerated motion is a special case of nonuniform motion characterised by a constant value of the acceleration, <i>a<sub>x</sub> </i> = constant. In such circumstances the velocity is a linear function of time (<i>v</i><sub><i>x</i></sub>(<i>t</i>) = <i>u<sub>x</sub></i> + <i>a<sub>x</sub>t</i>), and the position is a quadratic function of time (<span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/2b4bb3e6/s207_1_ie077i.gif" alt="" width="156" height="27" style="maxwidth:156px;" class="oucontentinlinefigureimage"/></span>).</p><p>10. The most widely used equations describing uniformly accelerated motion are</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_028a"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/7ab379df/s207_1_ue028i.gif" alt=""/></div><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn001_028b"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/1ae6d41d/s207_1_ue028di.gif" alt=""/></div><p>11. Position <i>x</i>, displacement <i>s<sub>x</sub> </i>, velocity <i>v</i> <sub> <i>x</i> </sub>, and acceleration, <i>a<sub>x</sub> </i>, are all signed quantities that may be positive or negative, depending on the associated direction. The magnitude of each of these quantities is a positive quantity that is devoid of directional information. The magnitude of the displacement of one point from another, <i>s</i> = <i>s<sub>x</sub></i>, represents the distance between those two points, while the magnitude of a particle's velocity, <i>v</i> = <i>v</i><sub><i>x</i></sub>, represents the speed of the particle. The magnitude of the acceleration due to gravity is represented by the symbol <i>g</i>, and has the approximate value 9.81 m s<sup>−2</sup> across much of the Earth's surface.</p>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

6.2 Endofcourse questions
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection6.2
Tue, 12 Apr 2016 23:00:00 GMT
<div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_023"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 25</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Table 8 shows the atmospheric pressure <i>P</i> in pascals (Pa) at various heights <i>h</i> above the Earth's surface. Plot a graph to give a visual representation of the data in the table. Be careful to label your axes correctly. Explain why you have chosen to plot particular variables on the horizontal and vertical axes. Use your graph to find the rate of change of atmospheric pressure with height at <i>h</i> = 10 km.</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_008"><h3 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 8:</b> Data for question 25</h3><div class="oucontenttablewrapper"><table><tr><th scope="col"><i>h</i>/km</th><th scope="col"><i>P</i>/Pa</th></tr><tr><td>0</td><td class="oucontenttablemiddle ">101 325</td></tr><tr><td>5</td><td class="oucontenttablemiddle ">48 586</td></tr><tr><td>10</td><td class="oucontenttablemiddle ">23 297</td></tr><tr><td>15</td><td class="oucontenttablemiddle ">11 171</td></tr><tr><td>20</td><td class="oucontenttablemiddle ">5357</td></tr><tr><td>25</td><td class="oucontenttablemiddle ">2569</td></tr><tr><td>30</td><td class="oucontenttablemiddle ">1232</td></tr></table></div><div class="oucontentsourcereference"></div></div></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>The graph is given in Figure 43. Notice how the vertical axis is scaled by 10<sup>5</sup> in order to avoid a confusion of numbers on this axis. Since the pressure is the dependent variable, it is plotted on the vertical axis. Conversely, since height is the independent variable, it is plotted along the horizontal axis.</p><p>The rate of change of atmospheric pressure with height at 10 km is given by the gradient of the graph at this point. Using the tangent shown in Figure 43, we obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_052"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/37f4c08e/s207_1_ie060i.gif" alt=""/></div><p>(Differentiating the function that was used to produce <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection6.2#tbl001008">Table 8</a> gives a gradient of −3435 Pa km<sup> −1</sup> at 10 km.)</p><div class="oucontentfigure" style="width:393px;" id="fig001_048"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/44c26579/s207_2_043i.jpg" alt="" width="393" height="541" style="maxwidth:393px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"><b>Figure 43</b>: Graph for question 25</span></div></div></div></div></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_024"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 26</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>(a) Define the terms <i>position</i> and <i>displacement</i>, and carefully distinguish between them.</p><p>(b) The position <i>x</i> of a particle at time <i>t</i> is given in Table 9. Plot a positiontime graph for the data in Table 9.</p><p>(c) Using your graph, measure the velocity of the particle at <i>t</i> = 5 s.</p><p>(d) A second particle undergoes uniform motion and has the same position and velocity as the first particle at time <i>t</i> = 5 s. What is the displacement of the second particle from the first at time <i>t</i> = 10 s?</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_009"><h3 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 9:</b> Data for question 26</h3><div class="oucontenttablewrapper"><table><tr><th scope="col"><i>t</i>/s</th><th scope="col"><i>x</i>/km</th></tr><tr><td>0</td><td class="oucontenttablemiddle ">0</td></tr><tr><td>1</td><td class="oucontenttablemiddle ">0.443</td></tr><tr><td>2</td><td class="oucontenttablemiddle ">0.984</td></tr><tr><td>3</td><td class="oucontenttablemiddle ">1.64</td></tr><tr><td>4</td><td class="oucontenttablemiddle ">2.45</td></tr><tr><td>5</td><td class="oucontenttablemiddle ">3.44</td></tr><tr><td>6</td><td class="oucontenttablemiddle ">4.64</td></tr><tr><td>7</td><td class="oucontenttablemiddle ">6.11</td></tr><tr><td>8</td><td class="oucontenttablemiddle ">7.91</td></tr><tr><td>9</td><td class="oucontenttablemiddle ">10.1</td></tr><tr><td>10</td><td class="oucontenttablemiddle ">12.8</td></tr></table></div><div class="oucontentsourcereference"></div></div></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) The <i>position</i> of a point on a line is represented by a coordinate <i>x</i>, measured from some arbitrarily chosen origin. Such a point might represent the instantaneous position of a particle moving along the line. <i>Displacement</i> refers to the difference in position of two points, these might be the initial and final positions of a moving particle, in which case the displacement would represent the change in the particle's position.</p><p>(b) The positiontime graph for <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection6.2#tbl001009">Table 9</a> is given by the curve in Figure 44.</p><p>(c) The instantaneous velocity is given by the gradient of the tangent at <i>t</i> = 5 s. The gradient of the tangent shown is</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_053"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/b24c0d1c/s207_1_ie061i.gif" alt=""/></div><p>(This result compares with the value of 1.087, which is obtained by differentiating the function that was used to produce <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection6.2#tbl001009">Table 9</a>.)</p><p>(d) The displacementtime graph for the second particle is the tangent at <i>t</i> = 5 s, which we have already drawn in Figure 44 since for uniform motion <i>x</i> = <i>x</i><sub>0</sub> + <i>v</i><sub>x</sub><i>t</i>. From this tangent, we can see that the position of the second particle at <i>t</i> = 10 s is 8.5 km. Hence, the displacement of the second particle from the first is (8.5 − 12.8) km = −4.3 km. Note that the minus sign is an essential part of the answer.</p><div class="oucontentfigure" style="width:391px;" id="fig001_049"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/d78110d4/s207_2_044i.jpg" alt="" width="391" height="462" style="maxwidth:391px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 44:</b> Positiontime graph for question 26</span></div></div></div></div></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_025"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 27</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>(a) Define the terms <i>velocity</i> and <i>acceleration</i>.</p><p>(b) The velocity <i>v</i><sub><i>x</i></sub> of a particle moving along the <i>x</i>axis at various times <i>t</i> is given in <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection6.2#tbl001010">Table 10</a>.</p><p>(i) Assuming the particle has a constant acceleration between the given positions, draw a velocitytime graph representing the data in Table 10.</p><p>(ii) Use your graph to calculate the total displacement of the particle over the time interval given in the table.</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_010"><h3 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 10:</b> Data for Selfassessment question 27</h3><div class="oucontenttablewrapper"><table><tr><th scope="col"><i>v</i> <sub> <i>x</i> </sub>/m s<sup>−1</sup></th><th scope="col"><i>t</i>/s</th></tr><tr><td class="oucontenttablemiddle ">4</td><td class="oucontenttablemiddle ">0</td></tr><tr><td class="oucontenttablemiddle ">−4</td><td class="oucontenttablemiddle ">10</td></tr><tr><td class="oucontenttablemiddle ">0</td><td class="oucontenttablemiddle ">15</td></tr><tr><td class="oucontenttablemiddle ">2</td><td class="oucontenttablemiddle ">20</td></tr></table></div><div class="oucontentsourcereference"></div></div></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) The <i>velocity</i> of a particle is its rate of change of position. The <i>acceleration</i> of a particle is its rate of change of velocity.</p><p>(b) (i) The velocitytime graph for <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection6.2#tbl001010">Table 10</a> is given in Figure 45.</p><p>(ii) The displacement is equal to the signed area under the velocitytime graph. For the interval between 0 s and 10 s, the areas above and below the time axis cancel and consequently the displacement is zero.</p><p>For the interval between 10 s and 15 s, the displacement is given by</p><p>−(1/2) x 5 x 4 m = −10 m.</p><p>For the interval between 15 s and 20 s, the displacement is given by</p><p>(1/2) x 5 x 2 m = 5 m.</p><p>The total displacement is the sum of all three contributions:</p><p>(0 − 10 + 5) m = −5 m.</p><div class="oucontentfigure" style="width:383px;" id="fig001_050"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/47594e73/s207_2_045i.jpg" alt="" width="383" height="302" style="maxwidth:383px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 45:</b> Velocitytime graph for question 27</span></div></div></div></div></div></div></div><p><i>Comment In questions 28 to 32, you are not required to draw any graphs</i>.</p><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_026"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 28</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>The variable <i>z</i> is related to the variable <i>y</i> by the equation</p><div class="oucontentquote oucontentsbox" id="q007"><blockquote><p><i>z</i> = 3 + 2<i>y</i> + <i>y</i><sup>3</sup>.</p></blockquote></div><p>(a) Find the derivative d<i>z</i>/d<i>y</i>.</p><p>(b) Evaluate d<i>z</i>/d<i>y</i> at <i>y</i> = 2.</p><p>(c) What would be the gradient of a graph of <i>z</i> plotted against <i>y</i> for the value <i>y</i> = 2?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) d<i>z</i>/d<i>y</i> = 2 + 3y<sup>2</sup>.</p><p>(b) Substituting <i>y</i> = 2 into the derivative found in part (a) we obtain</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/74531689/s207_1_ie073i.gif" alt="" width="155" height="32" style="maxwidth:155px;" class="oucontentinlinefigureimage"/></span>.</p><p>(c) Since the gradient is equal to d<i>z</i>/d<i>y</i> evaluated at <i>y</i> = 2, the answer is 14.</p></div></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_027"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 29</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>A rocket travels vertically away from the surface of the Moon. It is still close to the Moon's surface when it jettisons an empty fuel tank. The fuel tank initially travels with the same velocity as the rocket, but falls back to the Moon, reaching the Moon's surface 50 s after being released. If the fuel tank hits the surface at a speed of 50 m s<sup>−1</sup>, calculate the speed of the fuel tank when it was jettisoned. You may assume that the magnitude of the acceleration due to gravity near the Moon's surface is 1.6 m s<sup>﻿−﻿2</sup>.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Choose an <i>x</i>axis pointing vertically upwards from the Moon's surface. Using Equation 28b</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_054"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/cd9b5ff1/s207_1_ue023i.gif" alt=""/></div><p>with <i>v<sub>x</sub> </i> = −50 m s<sup>−1</sup>, <i>a<sub>x</sub> </i> = −﻿1.6 m s<sup>−2</sup> and <i>t</i> = 50 s, we obtain</p><div class="oucontentquote oucontentsbox" id="q008"><blockquote><p><i>u<sub>x</sub></i> = <i>v<sub>x</sub></i> − <i>a<sub>x</sub>t</i> = (−50 + 1.6 × 50) m s<sup>−1</sup> = 30 m s<sup>−1</sup>.</p></blockquote></div></div></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_028"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 30</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>For time <i>t</i> greater than or equal to zero, a particle's position as it travels along the <i>x</i>axis is described by the function <i>x</i>(<i>t</i>) = <i>At</i> <sup>2</sup>, where <i>A</i> = 4.0 m s<sup>−2</sup>. Use differentiation to calculate how fast the particle is travelling at <i>t</i> = 10 s.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/d83a07ce/s207_1_ue030i.gif" alt="" width="283" height="29" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span> and therefore</p><div class="oucontentquote oucontentsbox" id="q009"><blockquote><p><i>v<sub>x</sub></i> (10 s) = (2 × 4.0 × 10) m s<sup>−1</sup> = 80 m s<sup>−1</sup>.</p></blockquote></div></div></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_029"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 31</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>A vase falls to the ground from a shelf at height 1.80 m above the floor. Neglecting air resistance, calculate the time taken for the vase to strike the floor.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Taking up as the positive direction, and using Equation 28a</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_057"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/569431a9/s207_1_ue032i.gif" alt=""/></div><p>with <i>u<sub>x</sub> </i> = 0 m s<sup>−1</sup>, <i>a<sub>x</sub> </i> = −9.81 m s<sup>−2</sup> and <i>s<sub>x</sub> </i> = −1.80m, we obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_058"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/1d912e80/s207_1_ue033i.gif" alt=""/></div></div></div></div></div><div class=" oucontentactivity oucontentsheavybox1 oucontentsbox " id="que001_030"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 32</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>(This question is more difficult than its predecessors.) A rocket is initially at rest on the Earth's surface. At time <i>t</i> = 0 s the rocket motor is fired and the rocket accelerates vertically upwards at a constant 2.00 m s<sup>−2</sup>. After a time interval of 20.0 s the motor fails completely and the rocket accelerates back to Earth with a downward acceleration of magnitude 9.81 m s<sup>−2</sup>.</p><p>(a) Calculate the height reached by the rocket at the instant the motor fails.</p><p>(b) Calculate the velocity of the rocket at the instant the motor fails.</p><p>(c) Find an equation that, for situations involving constant acceleration, gives the displacement in terms of the initial velocity, the final velocity, and the constant acceleration.</p><p>(d) Calculate the distance travelled by the rocket from the instant the motor fails until the rocket reaches its highest point. Hence, find the total height gained by the rocket. [<i>Hint</i>: The equation you obtained in part (c) may be useful here.]</p><p>(e) Find the total time taken for the rocket to fall back to Earth from its highest point.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a)Taking up as positive, and using Equation 28a</p><p>with <i>u<sub>x</sub> </i> = 0 m s<sup>−1</sup>, <i>a<sub>x</sub> </i> = 2.00 m s<sup>−2</sup> and <i>t</i> = 20.0 s, we obtain</p><div class="oucontentquote oucontentsbox" id="q010"><blockquote><p><i>s<sub>x</sub></i> = (1/2) × 2 × (20.0)<sup>2</sup> m = 400 m.</p></blockquote></div><p>This is the displacement from the initial position when the motor fails. It follows that the height at which the motor fails is <i>s</i> = <i>s<sub>x</sub> </i>  = 400 m.</p><p>(b) Using Equation 28b</p><div class="oucontentquote oucontentsbox" id="q011"><blockquote><p><i>v<sub>x</sub></i> = <i>u<sub>x</sub></i> + <i>a<sub>x</sub>t</i></p></blockquote></div><p>with <i>u<sub>x</sub> </i> = 0 m s<sup>−1</sup>, <i>a<sub>x</sub> </i> = 2.00 m s<sup>−2</sup> and <i>t</i> = 20.0 s, we obtain</p><div class="oucontentquote oucontentsbox" id="q012"><blockquote><p><i>v<sub>x</sub></i> = 0 + 2 × 20.0 m s<sup>−1</sup> = 40.0 m s<sup>−1</sup>.</p></blockquote></div><p>(c) As shown in Section 5.2, eliminating <i>t</i> from Equations 28a and 28b gives Equation 28c</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_062"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/789532a1/s207_1_ue037i.gif" alt=""/></div><p>which we can write as</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_063"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/5834c7d9/s207_1_ue038i.gif" alt=""/></div><p>(d) Substituting <i>v<sub>x</sub></i> = 0 m s<sup>−1</sup>, <i>u<sub>x</sub> </i> = 40.0 m s<sup>−1</sup> and <i>a<sub>x</sub> </i> = −9.81 m s<sup>−2</sup> in the above equation, we obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_064"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/3ac4ac75/s207_1_ue039i.gif" alt=""/></div><p>Adding this result to the displacement obtained in part (a) we obtain a total height of 482 m.</p><p>(e) Using Equation 28a</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_065"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/569431a9/s207_1_ue032i.gif" alt=""/></div><p>with <i>u<sub>x</sub></i> = 0 m s<sup>−1</sup>, <i>a<sub>x</sub> </i> = −9.81 m s<sup>−2</sup> and <i>s<sub>x</sub> </i> = −482 m, we obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_066"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/e6062a66/s207_1_ue040i.gif" alt=""/></div></div></div></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection6.2
6.2 Endofcourse questionsS207_2<div class="
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oucontentsheavybox1 oucontentsbox " id="que001_023"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 25</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>Table 8 shows the atmospheric pressure <i>P</i> in pascals (Pa) at various heights <i>h</i> above the Earth's surface. Plot a graph to give a visual representation of the data in the table. Be careful to label your axes correctly. Explain why you have chosen to plot particular variables on the horizontal and vertical axes. Use your graph to find the rate of change of atmospheric pressure with height at <i>h</i> = 10 km.</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_008"><h3 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 8:</b> Data for question 25</h3><div class="oucontenttablewrapper"><table><tr><th scope="col"><i>h</i>/km</th><th scope="col"><i>P</i>/Pa</th></tr><tr><td>0</td><td class="oucontenttablemiddle ">101 325</td></tr><tr><td>5</td><td class="oucontenttablemiddle ">48 586</td></tr><tr><td>10</td><td class="oucontenttablemiddle ">23 297</td></tr><tr><td>15</td><td class="oucontenttablemiddle ">11 171</td></tr><tr><td>20</td><td class="oucontenttablemiddle ">5357</td></tr><tr><td>25</td><td class="oucontenttablemiddle ">2569</td></tr><tr><td>30</td><td class="oucontenttablemiddle ">1232</td></tr></table></div><div class="oucontentsourcereference"></div></div></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>The graph is given in Figure 43. Notice how the vertical axis is scaled by 10<sup>5</sup> in order to avoid a confusion of numbers on this axis. Since the pressure is the dependent variable, it is plotted on the vertical axis. Conversely, since height is the independent variable, it is plotted along the horizontal axis.</p><p>The rate of change of atmospheric pressure with height at 10 km is given by the gradient of the graph at this point. Using the tangent shown in Figure 43, we obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_052"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/37f4c08e/s207_1_ie060i.gif" alt=""/></div><p>(Differentiating the function that was used to produce <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection6.2#tbl001008">Table 8</a> gives a gradient of −3435 Pa km<sup> −1</sup> at 10 km.)</p><div class="oucontentfigure" style="width:393px;" id="fig001_048"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/44c26579/s207_2_043i.jpg" alt="" width="393" height="541" style="maxwidth:393px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"><b>Figure 43</b>: Graph for question 25</span></div></div></div></div></div></div></div><div class="
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oucontentsheavybox1 oucontentsbox " id="que001_024"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 26</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>(a) Define the terms <i>position</i> and <i>displacement</i>, and carefully distinguish between them.</p><p>(b) The position <i>x</i> of a particle at time <i>t</i> is given in Table 9. Plot a positiontime graph for the data in Table 9.</p><p>(c) Using your graph, measure the velocity of the particle at <i>t</i> = 5 s.</p><p>(d) A second particle undergoes uniform motion and has the same position and velocity as the first particle at time <i>t</i> = 5 s. What is the displacement of the second particle from the first at time <i>t</i> = 10 s?</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_009"><h3 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 9:</b> Data for question 26</h3><div class="oucontenttablewrapper"><table><tr><th scope="col"><i>t</i>/s</th><th scope="col"><i>x</i>/km</th></tr><tr><td>0</td><td class="oucontenttablemiddle ">0</td></tr><tr><td>1</td><td class="oucontenttablemiddle ">0.443</td></tr><tr><td>2</td><td class="oucontenttablemiddle ">0.984</td></tr><tr><td>3</td><td class="oucontenttablemiddle ">1.64</td></tr><tr><td>4</td><td class="oucontenttablemiddle ">2.45</td></tr><tr><td>5</td><td class="oucontenttablemiddle ">3.44</td></tr><tr><td>6</td><td class="oucontenttablemiddle ">4.64</td></tr><tr><td>7</td><td class="oucontenttablemiddle ">6.11</td></tr><tr><td>8</td><td class="oucontenttablemiddle ">7.91</td></tr><tr><td>9</td><td class="oucontenttablemiddle ">10.1</td></tr><tr><td>10</td><td class="oucontenttablemiddle ">12.8</td></tr></table></div><div class="oucontentsourcereference"></div></div></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) The <i>position</i> of a point on a line is represented by a coordinate <i>x</i>, measured from some arbitrarily chosen origin. Such a point might represent the instantaneous position of a particle moving along the line. <i>Displacement</i> refers to the difference in position of two points, these might be the initial and final positions of a moving particle, in which case the displacement would represent the change in the particle's position.</p><p>(b) The positiontime graph for <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection6.2#tbl001009">Table 9</a> is given by the curve in Figure 44.</p><p>(c) The instantaneous velocity is given by the gradient of the tangent at <i>t</i> = 5 s. The gradient of the tangent shown is</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_053"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/b24c0d1c/s207_1_ie061i.gif" alt=""/></div><p>(This result compares with the value of 1.087, which is obtained by differentiating the function that was used to produce <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection6.2#tbl001009">Table 9</a>.)</p><p>(d) The displacementtime graph for the second particle is the tangent at <i>t</i> = 5 s, which we have already drawn in Figure 44 since for uniform motion <i>x</i> = <i>x</i><sub>0</sub> + <i>v</i><sub>x</sub><i>t</i>. From this tangent, we can see that the position of the second particle at <i>t</i> = 10 s is 8.5 km. Hence, the displacement of the second particle from the first is (8.5 − 12.8) km = −4.3 km. Note that the minus sign is an essential part of the answer.</p><div class="oucontentfigure" style="width:391px;" id="fig001_049"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/d78110d4/s207_2_044i.jpg" alt="" width="391" height="462" style="maxwidth:391px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 44:</b> Positiontime graph for question 26</span></div></div></div></div></div></div></div><div class="
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oucontentsheavybox1 oucontentsbox " id="que001_025"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 27</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>(a) Define the terms <i>velocity</i> and <i>acceleration</i>.</p><p>(b) The velocity <i>v</i><sub><i>x</i></sub> of a particle moving along the <i>x</i>axis at various times <i>t</i> is given in <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection6.2#tbl001010">Table 10</a>.</p><p>(i) Assuming the particle has a constant acceleration between the given positions, draw a velocitytime graph representing the data in Table 10.</p><p>(ii) Use your graph to calculate the total displacement of the particle over the time interval given in the table.</p><div class="oucontenttable oucontentsnormal oucontentsbox" id="tbl001_010"><h3 class="oucontenth3 oucontentheading oucontentnonumber"> <b>Table 10:</b> Data for Selfassessment question 27</h3><div class="oucontenttablewrapper"><table><tr><th scope="col"><i>v</i> <sub> <i>x</i> </sub>/m s<sup>−1</sup></th><th scope="col"><i>t</i>/s</th></tr><tr><td class="oucontenttablemiddle ">4</td><td class="oucontenttablemiddle ">0</td></tr><tr><td class="oucontenttablemiddle ">−4</td><td class="oucontenttablemiddle ">10</td></tr><tr><td class="oucontenttablemiddle ">0</td><td class="oucontenttablemiddle ">15</td></tr><tr><td class="oucontenttablemiddle ">2</td><td class="oucontenttablemiddle ">20</td></tr></table></div><div class="oucontentsourcereference"></div></div></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) The <i>velocity</i> of a particle is its rate of change of position. The <i>acceleration</i> of a particle is its rate of change of velocity.</p><p>(b) (i) The velocitytime graph for <a class="oucontentcrossref" href="https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection6.2#tbl001010">Table 10</a> is given in Figure 45.</p><p>(ii) The displacement is equal to the signed area under the velocitytime graph. For the interval between 0 s and 10 s, the areas above and below the time axis cancel and consequently the displacement is zero.</p><p>For the interval between 10 s and 15 s, the displacement is given by</p><p>−(1/2) x 5 x 4 m = −10 m.</p><p>For the interval between 15 s and 20 s, the displacement is given by</p><p>(1/2) x 5 x 2 m = 5 m.</p><p>The total displacement is the sum of all three contributions:</p><p>(0 − 10 + 5) m = −5 m.</p><div class="oucontentfigure" style="width:383px;" id="fig001_050"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/47594e73/s207_2_045i.jpg" alt="" width="383" height="302" style="maxwidth:383px;" class="oucontentfigureimage"/><div class="oucontentfiguretext"><div class="oucontentcaption oucontentnonumber"><span class="oucontentfigurecaption"> <b>Figure 45:</b> Velocitytime graph for question 27</span></div></div></div></div></div></div></div><p><i>Comment In questions 28 to 32, you are not required to draw any graphs</i>.</p><div class="
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oucontentsheavybox1 oucontentsbox " id="que001_026"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 28</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>The variable <i>z</i> is related to the variable <i>y</i> by the equation</p><div class="oucontentquote oucontentsbox" id="q007"><blockquote><p><i>z</i> = 3 + 2<i>y</i> + <i>y</i><sup>3</sup>.</p></blockquote></div><p>(a) Find the derivative d<i>z</i>/d<i>y</i>.</p><p>(b) Evaluate d<i>z</i>/d<i>y</i> at <i>y</i> = 2.</p><p>(c) What would be the gradient of a graph of <i>z</i> plotted against <i>y</i> for the value <i>y</i> = 2?</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a) d<i>z</i>/d<i>y</i> = 2 + 3y<sup>2</sup>.</p><p>(b) Substituting <i>y</i> = 2 into the derivative found in part (a) we obtain</p><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/74531689/s207_1_ie073i.gif" alt="" width="155" height="32" style="maxwidth:155px;" class="oucontentinlinefigureimage"/></span>.</p><p>(c) Since the gradient is equal to d<i>z</i>/d<i>y</i> evaluated at <i>y</i> = 2, the answer is 14.</p></div></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_027"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 29</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>A rocket travels vertically away from the surface of the Moon. It is still close to the Moon's surface when it jettisons an empty fuel tank. The fuel tank initially travels with the same velocity as the rocket, but falls back to the Moon, reaching the Moon's surface 50 s after being released. If the fuel tank hits the surface at a speed of 50 m s<sup>−1</sup>, calculate the speed of the fuel tank when it was jettisoned. You may assume that the magnitude of the acceleration due to gravity near the Moon's surface is 1.6 m s<sup>−2</sup>.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Choose an <i>x</i>axis pointing vertically upwards from the Moon's surface. Using Equation 28b</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_054"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/cd9b5ff1/s207_1_ue023i.gif" alt=""/></div><p>with <i>v<sub>x</sub> </i> = −50 m s<sup>−1</sup>, <i>a<sub>x</sub> </i> = −1.6 m s<sup>−2</sup> and <i>t</i> = 50 s, we obtain</p><div class="oucontentquote oucontentsbox" id="q008"><blockquote><p><i>u<sub>x</sub></i> = <i>v<sub>x</sub></i> − <i>a<sub>x</sub>t</i> = (−50 + 1.6 × 50) m s<sup>−1</sup> = 30 m s<sup>−1</sup>.</p></blockquote></div></div></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_028"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 30</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>For time <i>t</i> greater than or equal to zero, a particle's position as it travels along the <i>x</i>axis is described by the function <i>x</i>(<i>t</i>) = <i>At</i> <sup>2</sup>, where <i>A</i> = 4.0 m s<sup>−2</sup>. Use differentiation to calculate how fast the particle is travelling at <i>t</i> = 10 s.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p><span class="oucontentinlinefigure"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/d83a07ce/s207_1_ue030i.gif" alt="" width="283" height="29" style="maxwidth:283px;" class="oucontentinlinefigureimage"/></span> and therefore</p><div class="oucontentquote oucontentsbox" id="q009"><blockquote><p><i>v<sub>x</sub></i> (10 s) = (2 × 4.0 × 10) m s<sup>−1</sup> = 80 m s<sup>−1</sup>.</p></blockquote></div></div></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_029"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 31</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>A vase falls to the ground from a shelf at height 1.80 m above the floor. Neglecting air resistance, calculate the time taken for the vase to strike the floor.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>Taking up as the positive direction, and using Equation 28a</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_057"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/569431a9/s207_1_ue032i.gif" alt=""/></div><p>with <i>u<sub>x</sub> </i> = 0 m s<sup>−1</sup>, <i>a<sub>x</sub> </i> = −9.81 m s<sup>−2</sup> and <i>s<sub>x</sub> </i> = −1.80m, we obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_058"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/1d912e80/s207_1_ue033i.gif" alt=""/></div></div></div></div></div><div class="
oucontentactivity
oucontentsheavybox1 oucontentsbox " id="que001_030"><div class="oucontentouterbox"><h2 class="oucontenth3 oucontentheading oucontentnonumber">Question 32</h2><div class="oucontentinnerbox"><div class="oucontentsaqquestion"><p>(This question is more difficult than its predecessors.) A rocket is initially at rest on the Earth's surface. At time <i>t</i> = 0 s the rocket motor is fired and the rocket accelerates vertically upwards at a constant 2.00 m s<sup>−2</sup>. After a time interval of 20.0 s the motor fails completely and the rocket accelerates back to Earth with a downward acceleration of magnitude 9.81 m s<sup>−2</sup>.</p><p>(a) Calculate the height reached by the rocket at the instant the motor fails.</p><p>(b) Calculate the velocity of the rocket at the instant the motor fails.</p><p>(c) Find an equation that, for situations involving constant acceleration, gives the displacement in terms of the initial velocity, the final velocity, and the constant acceleration.</p><p>(d) Calculate the distance travelled by the rocket from the instant the motor fails until the rocket reaches its highest point. Hence, find the total height gained by the rocket. [<i>Hint</i>: The equation you obtained in part (c) may be useful here.]</p><p>(e) Find the total time taken for the rocket to fall back to Earth from its highest point.</p></div>
<div class="oucontentsaqanswer"><h3 class="oucontenth4">Answer</h3><p>(a)Taking up as positive, and using Equation 28a</p><p>with <i>u<sub>x</sub> </i> = 0 m s<sup>−1</sup>, <i>a<sub>x</sub> </i> = 2.00 m s<sup>−2</sup> and <i>t</i> = 20.0 s, we obtain</p><div class="oucontentquote oucontentsbox" id="q010"><blockquote><p><i>s<sub>x</sub></i> = (1/2) × 2 × (20.0)<sup>2</sup> m = 400 m.</p></blockquote></div><p>This is the displacement from the initial position when the motor fails. It follows that the height at which the motor fails is <i>s</i> = <i>s<sub>x</sub> </i>  = 400 m.</p><p>(b) Using Equation 28b</p><div class="oucontentquote oucontentsbox" id="q011"><blockquote><p><i>v<sub>x</sub></i> = <i>u<sub>x</sub></i> + <i>a<sub>x</sub>t</i></p></blockquote></div><p>with <i>u<sub>x</sub> </i> = 0 m s<sup>−1</sup>, <i>a<sub>x</sub> </i> = 2.00 m s<sup>−2</sup> and <i>t</i> = 20.0 s, we obtain</p><div class="oucontentquote oucontentsbox" id="q012"><blockquote><p><i>v<sub>x</sub></i> = 0 + 2 × 20.0 m s<sup>−1</sup> = 40.0 m s<sup>−1</sup>.</p></blockquote></div><p>(c) As shown in Section 5.2, eliminating <i>t</i> from Equations 28a and 28b gives Equation 28c</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_062"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/789532a1/s207_1_ue037i.gif" alt=""/></div><p>which we can write as</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_063"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/5834c7d9/s207_1_ue038i.gif" alt=""/></div><p>(d) Substituting <i>v<sub>x</sub></i> = 0 m s<sup>−1</sup>, <i>u<sub>x</sub> </i> = 40.0 m s<sup>−1</sup> and <i>a<sub>x</sub> </i> = −9.81 m s<sup>−2</sup> in the above equation, we obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_064"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/3ac4ac75/s207_1_ue039i.gif" alt=""/></div><p>Adding this result to the displacement obtained in part (a) we obtain a total height of 482 m.</p><p>(e) Using Equation 28a</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_065"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/569431a9/s207_1_ue032i.gif" alt=""/></div><p>with <i>u<sub>x</sub></i> = 0 m s<sup>−1</sup>, <i>a<sub>x</sub> </i> = −9.81 m s<sup>−2</sup> and <i>s<sub>x</sub> </i> = −482 m, we obtain</p><div class="oucontentequation oucontentequationequation oucontentnocaption" id="ueqn004_066"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/c1ea820b/e6062a66/s207_1_ue040i.gif" alt=""/></div></div></div></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

Conclusion
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection7
Tue, 12 Apr 2016 23:00:00 GMT
<p>This free course provided an introduction to studying Science. It took you through a series of exercises designed to develop your approach to study and learning at a distance, and helped to improve your confidence as an independent learner.</p>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection7
ConclusionS207_2<p>This free course provided an introduction to studying Science. It took you through a series of exercises designed to develop your approach to study and learning at a distance, and helped to improve your confidence as an independent learner.</p>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

Keep on learning
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection8
Tue, 12 Apr 2016 23:00:00 GMT
<div class="oucontentfigure oucontentmediamini"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/1b9129f0/d3c986e6/ol_skeleton_keeponlearning_image.jpg" alt="" width="300" height="200" style="maxwidth:300px;" class="oucontentfigureimage"/></div><div class="oucontentinternalsection"><h2 class="oucontenth2 oucontentinternalsectionhead">Study another free course</h2><p>There are more than <b>800 courses on OpenLearn</b> for you to choose from on a range of subjects.</p><p>Find out more about all our <span class="oucontentlinkwithtip"><a class="oucontenthyperlink" href="http://www.open.edu/openlearn/freecourses?utm_source=openlearn&utm_campaign=ol&utm_medium=ebook">free courses</a></span>.</p></div><div class="oucontentinternalsection"><h2 class="oucontenth2 oucontentinternalsectionhead">Take your studies further</h2><p>Find out more about studying with The Open University by <a class="oucontenthyperlink" href="http://www.open.ac.uk/courses?utm_source=openlearn&utm_campaign=ou&utm_medium=ebook">visiting our online prospectus</a>.</p><p>If you are new to university study, you may be interested in our <a class="oucontenthyperlink" href=" http://www.open.ac.uk/courses/doit/access?utm_source=openlearn&utm_campaign=ou&utm_medium=ebook">Access Courses</a> or <a class="oucontenthyperlink" href=" http://www.open.ac.uk/courses/certificateshe?utm_source=openlearn&utm_campaign=ou&utm_medium=ebook">Certificates</a>.</p></div><div class="oucontentinternalsection"><h2 class="oucontenth2 oucontentinternalsectionhead">What's new from OpenLearn?</h2><p><a class="oucontenthyperlink" href="http://www.open.edu/openlearn/aboutopenlearn/subscribetheopenlearnnewsletter?utm_source=openlearn&utm_campaign=ol&utm_medium=ebook">Sign up to our newsletter</a> or view a sample.</p></div><div class="oucontentbox oucontentshollowbox2 oucontentsbox oucontentsnoheading "><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>For reference, full URLs to pages listed above:</p><p>OpenLearn  <a class="oucontenthyperlink" href="http://www.open.edu/openlearn/freecourses?utm_source=openlearn&utm_campaign=ol&utm_medium=ebook">www.open.edu/<span class="oucontenthidespace"> </span>openlearn/<span class="oucontenthidespace"> </span>freecourses</a></p><p>Visiting our online prospectus  <a class="oucontenthyperlink" href="http://www.open.ac.uk/courses?utm_source=openlearn&utm_campaign=ou&utm_medium=ebook">www.open.ac.uk/<span class="oucontenthidespace"> </span>courses</a></p><p>Access Courses  <a class="oucontenthyperlink" href=" http://www.open.ac.uk/courses/doit/access?utm_source=openlearn&utm_campaign=ou&utm_medium=ebook">www.open.ac.uk/<span class="oucontenthidespace"> </span>courses/<span class="oucontenthidespace"> </span>doit/<span class="oucontenthidespace"> </span>access</a></p><p>Certificates  <a class="oucontenthyperlink" href=" http://www.open.ac.uk/courses/certificateshe?utm_source=openlearn&utm_campaign=ou&utm_medium=ebook">www.open.ac.uk/<span class="oucontenthidespace"> </span>courses/<span class="oucontenthidespace"> </span>certificateshe</a></p><p>Newsletter  <a class="oucontenthyperlink" href=" http://www.open.edu/openlearn/aboutopenlearn/subscribetheopenlearnnewsletter?utm_source=openlearn&utm_campaign=ol&utm_medium=ebook">www.open.edu/<span class="oucontenthidespace"> </span>openlearn/<span class="oucontenthidespace"> </span>aboutopenlearn/<span class="oucontenthidespace"> </span>subscribetheopenlearnnewsletter</a></p></div></div></div>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsection8
Keep on learningS207_2<div class="oucontentfigure oucontentmediamini"><img src="https://www.open.edu/openlearn/ocw/pluginfile.php/65607/mod_oucontent/oucontent/439/1b9129f0/d3c986e6/ol_skeleton_keeponlearning_image.jpg" alt="" width="300" height="200" style="maxwidth:300px;" class="oucontentfigureimage"/></div><div class="oucontentinternalsection"><h2 class="oucontenth2 oucontentinternalsectionhead">Study another free course</h2><p>There are more than <b>800 courses on OpenLearn</b> for you to choose from on a range of subjects.</p><p>Find out more about all our <span class="oucontentlinkwithtip"><a class="oucontenthyperlink" href="http://www.open.edu/openlearn/freecourses?utm_source=openlearn&utm_campaign=ol&utm_medium=ebook">free courses</a></span>.</p></div><div class="oucontentinternalsection"><h2 class="oucontenth2 oucontentinternalsectionhead">Take your studies further</h2><p>Find out more about studying with The Open University by <a class="oucontenthyperlink" href="http://www.open.ac.uk/courses?utm_source=openlearn&utm_campaign=ou&utm_medium=ebook">visiting our online prospectus</a>.</p><p>If you are new to university study, you may be interested in our <a class="oucontenthyperlink" href=" http://www.open.ac.uk/courses/doit/access?utm_source=openlearn&utm_campaign=ou&utm_medium=ebook">Access Courses</a> or <a class="oucontenthyperlink" href=" http://www.open.ac.uk/courses/certificateshe?utm_source=openlearn&utm_campaign=ou&utm_medium=ebook">Certificates</a>.</p></div><div class="oucontentinternalsection"><h2 class="oucontenth2 oucontentinternalsectionhead">What's new from OpenLearn?</h2><p><a class="oucontenthyperlink" href="http://www.open.edu/openlearn/aboutopenlearn/subscribetheopenlearnnewsletter?utm_source=openlearn&utm_campaign=ol&utm_medium=ebook">Sign up to our newsletter</a> or view a sample.</p></div><div class="oucontentbox oucontentshollowbox2 oucontentsbox
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"><div class="oucontentouterbox"><div class="oucontentinnerbox"><p>For reference, full URLs to pages listed above:</p><p>OpenLearn  <a class="oucontenthyperlink" href="http://www.open.edu/openlearn/freecourses?utm_source=openlearn&utm_campaign=ol&utm_medium=ebook">www.open.edu/<span class="oucontenthidespace"> </span>openlearn/<span class="oucontenthidespace"> </span>freecourses</a></p><p>Visiting our online prospectus  <a class="oucontenthyperlink" href="http://www.open.ac.uk/courses?utm_source=openlearn&utm_campaign=ou&utm_medium=ebook">www.open.ac.uk/<span class="oucontenthidespace"> </span>courses</a></p><p>Access Courses  <a class="oucontenthyperlink" href=" http://www.open.ac.uk/courses/doit/access?utm_source=openlearn&utm_campaign=ou&utm_medium=ebook">www.open.ac.uk/<span class="oucontenthidespace"> </span>courses/<span class="oucontenthidespace"> </span>doit/<span class="oucontenthidespace"> </span>access</a></p><p>Certificates  <a class="oucontenthyperlink" href=" http://www.open.ac.uk/courses/certificateshe?utm_source=openlearn&utm_campaign=ou&utm_medium=ebook">www.open.ac.uk/<span class="oucontenthidespace"> </span>courses/<span class="oucontenthidespace"> </span>certificateshe</a></p><p>Newsletter  <a class="oucontenthyperlink" href=" http://www.open.edu/openlearn/aboutopenlearn/subscribetheopenlearnnewsletter?utm_source=openlearn&utm_campaign=ol&utm_medium=ebook">www.open.edu/<span class="oucontenthidespace"> </span>openlearn/<span class="oucontenthidespace"> </span>aboutopenlearn/<span class="oucontenthidespace"> </span>subscribetheopenlearnnewsletter</a></p></div></div></div>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University

Acknowledgements
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsectionacknowledgements
Tue, 12 Apr 2016 23:00:00 GMT
<p>The content acknowledged below is Proprietary (see <span class="oucontentlinkwithtip"><a class="oucontenthyperlink" href="http://www.open.ac.uk/conditions">terms and conditions</a></span>) and is used under <a class="oucontenthyperlink" href="https://creativecommons.org/licenses/byncsa/4.0/">licence.</a></p><p>Course image: <a class="oucontenthyperlink" href="https://www.flickr.com/photos/eugeneb/">Eugene</a> in Flickr made available under <a class="oucontenthyperlink" href="https://creativecommons.org/licenses/byncsa/2.0/legalcode">Creative Commons AttributionNonCommercialShareAlike 2.0 Licence</a>.</p><p>Grateful acknowledgement is made to the following sources for permission to reproduce material in this course:</p><p><b><i>Figure 1</i></b>: ZARM Droptower of Bremen at the University of Bremen;</p><p><b><i>Figure 2</i></b>: LERC/NASA;</p><p><b><i>Figure 3</i></b>: Alton Towers;</p><p><b><i>Figure 27</i></b>: Paolo Fioratti/Oxford Scientific Films (swift in flight), Keren Su/Oxford Scientific Films (growth of bamboo shoot), Spectrum Colour Library (growth of a child);</p><p><b><i>Figure 29</i></b>: Science Photo Library (explosion), Popperfoto/Reuters (surfacetoair missile), NASA/Sci Mus/Sci & Soc Pic Lib (highest rocket acceleration), NASA/OSF (lunar gravity);</p><p><b><i>Figure 41</i></b>: ZARM Droptower of Bremen at the University of Bremen.</p><p><b>Don't miss out:</b></p><p>If reading this text has inspired you to learn more, you may be interested in joining the millions of people who discover our free learning resources and qualifications by visiting The Open University  <a class="oucontenthyperlink" href="http://www.open.edu/openlearn/freecourses?utm_source=openlearn&utm_campaign=ol&utm_medium=ebook">www.open.edu/<span class="oucontenthidespace"> </span>openlearn/<span class="oucontenthidespace"> </span>freecourses</a></p>
https://www.open.edu/openlearn/sciencemathstechnology/describingmotionalongline/contentsectionacknowledgements
AcknowledgementsS207_2<p>The content acknowledged below is Proprietary (see <span class="oucontentlinkwithtip"><a class="oucontenthyperlink" href="http://www.open.ac.uk/conditions">terms and conditions</a></span>) and is used under <a class="oucontenthyperlink" href="https://creativecommons.org/licenses/byncsa/4.0/">licence.</a></p><p>Course image: <a class="oucontenthyperlink" href="https://www.flickr.com/photos/eugeneb/">Eugene</a> in Flickr made available under <a class="oucontenthyperlink" href="https://creativecommons.org/licenses/byncsa/2.0/legalcode">Creative Commons AttributionNonCommercialShareAlike 2.0 Licence</a>.</p><p>Grateful acknowledgement is made to the following sources for permission to reproduce material in this course:</p><p><b><i>Figure 1</i></b>: ZARM Droptower of Bremen at the University of Bremen;</p><p><b><i>Figure 2</i></b>: LERC/NASA;</p><p><b><i>Figure 3</i></b>: Alton Towers;</p><p><b><i>Figure 27</i></b>: Paolo Fioratti/Oxford Scientific Films (swift in flight), Keren Su/Oxford Scientific Films (growth of bamboo shoot), Spectrum Colour Library (growth of a child);</p><p><b><i>Figure 29</i></b>: Science Photo Library (explosion), Popperfoto/Reuters (surfacetoair missile), NASA/Sci Mus/Sci & Soc Pic Lib (highest rocket acceleration), NASA/OSF (lunar gravity);</p><p><b><i>Figure 41</i></b>: ZARM Droptower of Bremen at the University of Bremen.</p><p><b>Don't miss out:</b></p><p>If reading this text has inspired you to learn more, you may be interested in joining the millions of people who discover our free learning resources and qualifications by visiting The Open University  <a class="oucontenthyperlink" href="http://www.open.edu/openlearn/freecourses?utm_source=openlearn&utm_campaign=ol&utm_medium=ebook">www.open.edu/<span class="oucontenthidespace"> </span>openlearn/<span class="oucontenthidespace"> </span>freecourses</a></p>The Open UniversityThe Open UniversityCoursetext/htmlenGBDescribing motion along a line  S207_2Copyright © 2016 The Open University