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Introducing vectors for engineering applications
Introducing vectors for engineering applications

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1.2 Models of motion

We have calculated that Alice and Bob’s combined force causes the block of ice to accelerate at 0.17 times m s super negative two , but what does this mean? We can put this in context by comparing it to other common magnitudes of acceleration, as shown in Table 1. Our calculated value is less than the magnitude of acceleration of a high-speed train, but within the same order of magnitude, and if we think about how slowly a train accelerates as it initially begins to move, then this comparison sounds about right.

Table 1 Approximate magnitudes of acceleration
Object Approximate magnitude of acceleration (m s−2)
High-speed train 0.25
Executive car 4.3
Sprinter (pulling away from start line) 9.2
Gravity on Earth at sea-level standard 9.8
Saturn V moon rocket (just after launch) 11.2
Mid-engined sports car 15.2
Space shuttle (maximum during launch) 29
Formula One car (maximum under heavy braking) 49
F-16 aircraft (pulling out of dive) 79
Explosive seat ejection from aircraft 147
Automobile crash (100 km h−1 into wall) 982
Football struck by foot 2946
Baseball struck by bat 29 460
Closing jaws of a trap-jaw ant 1 000 000
Jellyfish stinger 53 000 000

We can also consider what the calculated acceleration means, by considering how it converts to motion. If we assume that acceleration is constant and in a straight line, then we can calculate speed and distance travelled using the following equations of motion.

Equations of linear motion

For linear motion under constant acceleration, the following equations relate distance ( s ), time ( t ), initial speed ( u ), final speed ( v ) and magnitude of acceleration ( a ).

Initial speed:

u equals v minus a times t full stop

Final speed:

v equals u plus a times t full stop

Finding displacement using initial and final speed:

s equals one divided by two times left parenthesis u plus v right parenthesis times t full stop

Finding displacement using initial speed and acceleration:

s equals u times t plus one divided by two times a times t squared full stop

Example 2 Calculating speed and distance

A stationary block of ice has a mass of a metric ton ( one multiplication 10 cubed times kg ). If Bob applies a force of 130 N to the left face of the block, while Alice applies a force of 110 N to the bottom face, use the equations of motion to find how far the block will move and how fast it will be moving after 10 seconds.

Solution

The block of ice is initially stationary so we have an initial speed of

u equals zero full stop

Also, in Example 1 we calculated that the magnitude of acceleration of the block of ice is 0.17 times m s super negative two (to 2 d.p.), so we have

a equals 0.17 full stop

We can calculate the final speed using

v equals u plus a times t

and distance travelled using

s equals u times t plus one divided by two times a times t squared full stop

So after 10 seconds we have

equation sequence part 1 v equals part 2 zero plus left parenthesis 0.17 multiplication 10 right parenthesis equals part 3 1.7 times m s super negative one

and

equation sequence part 1 s equals part 2 left parenthesis zero multiplication 10 right parenthesis plus left parenthesis one divided by two multiplication 0.17 multiplication 10 squared right parenthesis equals part 3 one divided by two multiplication 0.17 multiplication 100 equals part 4 8.5 m full stop

The block is travelling at a speed of approximately 1.7 times m s super negative one and has travelled a distance of approximately 8.5 m.

Activity 3

Now use the equations of motion to find how far the block will move and how fast it will be moving after the following times.

  • a.30 seconds

  • b.1 minute

How good is the model?

Do the answers calculated in Example 2 and Activity 3 seem reasonable to you?

  • After 10 seconds we calculated that the block is travelling with a speed of approximately 1.7 times m s super negative one . This is an average walking speed, and sounds pretty reasonable.
  • After 30 seconds the block is travelling at a speed approximately three times as fast. This is starting to get quite speedy, and it would be a challenge for Alice and Bob to maintain this pace while pushing a block of ice.
  • After one minute the block is travelling at a very fast speed, comparable to the world record speed of 12.4 times m s super negative one reached by Usain Bolt during the 100 m sprint final at the 2009 World Championships in Berlin, and it is clear that Alice and Bob are very unlikely to reach such a speed while pushing a block of ice.

So what has gone wrong? The problem lies in the underlying assumption in the equations of motion that we used. We assumed that acceleration is constant, but this is not realistic. Bob and Alice are unlikely to maintain the same force on the block once it has started moving, and it is much more realistic to assume that they will reduce the force that they apply once the block has reached a comfortable speed. This would result in a reduction in acceleration, and to maintain a constant speed, an acceleration of zero would be required.

For example, look at the velocity and acceleration profiles of a 100 m sprinter in the graphs in Figure 7. Certain assumptions are made that simplify the graphs; in particular, it is assumed that the acceleration at the start of the race is immediately at the maximum value, and it is also assumed that there is no deceleration towards the end of the race. These assumptions are not realistic, but since these graphs are for the purposes of illustration, they are perfectly acceptable. The graphs are similar in shape to what we would expect to see if we plotted the profiles for Alice and Bob pushing the block of ice.

Described image
Figure 7 (a) Velocity and (b) acceleration profiles for a 100 m sprint

Consider the velocity profile first. In Figure 7(a), we see that the sprinter quickly increases velocity from 0 to above 10 times m s super negative one in the first two seconds of the race. Once maximum velocity is reached, after about 4 seconds, the sprinter maintains this for the rest of the race. The velocity profile is also reflected in the acceleration profile in Figure 7(b). We see a high initial acceleration, above 10 times m s super negative two , which after around 2 seconds quickly reduces to zero. There is a strong relationship between these two graphs: where the velocity looks like a straight line, the acceleration is flat, i.e. constant, and where the velocity is flat, acceleration is zero. This is because acceleration is the rate of change of velocity, and the slope of the velocity profile at a specific value of t is equal to the acceleration for that value of t . This relationship is at the heart of calculus..

Finally, note that for small values of t , the graph of the velocity profile is approximately linear, and this gives a flat segment in the acceleration profile, where the acceleration is constant. This means that for small values of t an assumption of constant acceleration is not far from wrong, and this is why in Example 2 our calculation for 10 seconds seems reasonable, but for 30 and 60 seconds our calculations seem to be far away from reality. The statistician George Box is quoted as saying ‘all models are wrong, but some are useful’ (Box and Draper, 1987) and this perfectly sums up what we have found here.