is the hypotenuse of the right-angled triangle formed by
,
and the resultant
, so from Pythagoras’ theorem its magnitude is given by
This gives
So the magnitude of is 170.29 (to 2 d.p.).
Angles and
are alternate angles, so they are equal. We can find
using the tangent function, which is given by
So
therefore
This gives (to 1 d.p.) and this is the direction of the vector
.
After 30 seconds we have
and
So the block is travelling at a speed of approximately and has travelled a distance of approximately 76.5 m.
After 60 seconds we have
and
So the block is travelling at a speed of approximately and has travelled a distance of approximately 306 m.
First we need to determine the size of angle . We can use the angles 24° and 47° to calculate
because they sit on the same straight line as
, so
Now, using the cosine rule, we can calculate the length of edge :
So
Using the sine rule, we get
so
Using the inverse sine function, we get
Newton’s second law gives
so acceleration is given by
therefore
The direction of the acceleration is the same as the direction of , and this is
measured clockwise from the positive
-axis.
The magnitude of the acceleration is
So the block accelerates at (to 2 d.p.) in a direction that is
measured clockwise from the positive
-axis.
The magnitude of the vertical component is given by
so
The magnitude of the horizontal component is given by
so
The horizontal displacement is and the vertical displacement is
, so
.
The horizontal displacement is and the vertical displacement is
, so
.
The horiztonal displacment is and the vertical displacement is
, so
.
The component form of the vector is given by
The component form of the vector is given by
The magnitude of is
and the direction is given by
The calculator value for is
, but looking at a drawing of the vector below, shows that this is not the correct angle. Instead, we are looking for a value of
that is greater than
and less than
.
By considering the graph of , we can identify the angles that are within the range, and using the periodicity of the tangent function we can say that the value for
is
The vector has no vertical component, and a negative horizontal component. So it points in the negative
-direction and its magnitude is
The magnitude of is
and the direction is given by
The calculator value for is
. Looking at a drawing of the vector below shows that this is the correct value for
, because it is greater than
and less than
.
and
.
Rearranging gives
So .
Expand the brackets by using property 4:
Simplify by using property 3 to give
Simplify further by using property 2:
First let’s use the components of and
to find
,
and
. We have
Using these we can calculate :
So
The angle between the vectors is 27° (to the nearest degree).
OpenLearn - Part 2: Chapter 4 Applications of vectors Except for third party materials and otherwise, this content is made available under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 Licence, full copyright detail can be found in the acknowledgements section. Please see full copyright statement for details.