Part 2: Chapter 4 Applications of vectors
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# Solutions to activities in Chapter 4

## Activity 4.1

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.).

## Activity 4.2

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 .

## Activity 4.3

1. 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.

2. 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.

## Activity 4.4

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

## Activity 4.5

Using the sine rule, we get

so

Using the inverse sine function, we get

## Activity 4.6

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.

## Activity 4.7

The magnitude of the vertical component is given by

so

The magnitude of the horizontal component is given by

so

## Activity 4.8

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 .

## Activity 4.9

## Activity 4.10

1. The component form of the vector is given by

2. The component form of the vector is given by

## Activity 4.11

1. 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

2. The vector has no vertical component, and a negative horizontal component. So it points in the negative -direction and its magnitude is

3. 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 .

## Activity 4.12

## Activity 4.13

and .

## Activity 4.14

## Activity 4.15

## Activity 4.16

## Activity 4.17

## Activity 4.18

2. Rearranging gives

   So .

## Activity 4.19

## Activity 4.20

## Activity 4.21

Expand the brackets by using property 4:

Simplify by using property 3 to give

Simplify further by using property 2:

## Activity 4.22

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).

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OpenLearn - Part 2: Chapter 4 Applications of vectors
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