Question

. In a children’s game the instructions are to start from the school flag pole and go 30 paces north, 25 paces at 30o south of west, and 10 paces due east. How far is the student from the flag pole? What is the student’s compass bearing?

Answer 1

Apologies for the badly drawn diagram!

Here, we need to use vector analysis to add up all the x- and y-components of each of the three separate movements, from which we can determine the magnitude and direction of the total displacement vector. In the graph, we have set up a coordinate scale, and the angle of all the vectors will be with respect to the positive x-axis (i.e. 0 degrees on the positive x-axis, increasing in an anticlockwise direction until we return to the same axis at 360 degrees)

1: 30 paces north
We can split this vector up into its x- and y-components by using its magnitude (30) and the angle the vector makes with the positive x-axis (here, it is 90 degrees).
So, the x-component is given by 30cos(90) = 0 (from a calculator).
The y-component will be 30sin(90) = 30

2: 25 paces at 30 degrees south of west
Although the student is turning 30 degrees south of west from where they are standing, they are actually now at an angle of 210 degrees with respect to the positive x-axis ( if we were to bring the vector down on the x-axis on its own, this is easier to see).
So, x-component is 25cos(210) = -21.65 (approximately)
y-component is 25sin(210) = -12.5

3: 10 paces east
As in (2) above, if we were to superimpose this vector onto the x-axis, we would see that it made an angle of 0 degrees with respect to the POSITIVE x-axis.
So, x-component is 10cos(0) = 10
y-component is 10sin(0) = 0

Adding up all the x-components we get 0 + (-21.65) + 10 = -11.65
Adding up all the y-components we get 30 + (-11.5) + 0 = 17.5

This gives us the x - and y- components of the overall displacement vector from the flagpole at the origin to the end point.
We can use Pythagoras' Theorem to find the overall vector


So, if D is the number of paces from the flag pole to the end point, then
\[D^{2} = (-11.65)^{2} +(17.5) ^{2}\]
So \[D = \sqrt{(-11.65)^{2} +(17.5) ^{2}}\] = 21.02 paces

The student's compass bearing can be found by the angle between the x-component and the total displacement vector. We'll call this angle A.
So, tan(A) = y-component/x-component
This means that A=tan^-1(y-component/x-component)
=tan^-1(17.5/-11.65) = -56.34 degrees (or approximately a south -east bearing).

Answer 2

@Ciarán95 thank you so much!