Question

A man walks 2 km east and then 4 km at 60° north of east. What is the magnitude and direction of the resultant displacement?

Answer 1

The total displacement is about `5.29` km in a direction about `40.9°` north of east.

Explanation:

Let us solve it with trigonometry.

The man starts at point A and moves `2` km to B, so we have a triangle with side AB = `2` km. Then he moves to point C with BC = `4` km. Since the man turned `60°` through B the angle inside triangle ABC at that corner is the supplementary value, `120°`.

Then the third side of the triangle, AC, is given by the Law of Cosines:

`AC^2=AB^2+BC^2-2(AB)(BC)\cos (\angle B)`

`=2^2+4^2-(2×4×(-1/2))=28`

So the magnitude of his displacement is `2\sqrt{7}` km, about `5.29` km.

The displacement is north of east by an angle equal to angle A in the triangle. This is an acute angle whose sine is given by the Law of Sines:

`{\sin (\angle A)}/{BC}={\sin (\angle B)}/{AC}`

`\sin (\angle A) = 4×(\sqrt{3}/2)/(2\sqrt{7})={3\sqrt{21}}/{7}`

Taking the inverse sine with a calculator gives about `40.9°`.