Question

A boy walks 30m [E25°S] then 60m [E40°N]. Determine his net displacement.

Answer 1

Given,

The distances; a=30 m

b=60 m

Angles; θ=E25°S

α=E40°N

From the diagram, ∠A is given by,

`\angle A=180\degree-\theta-\alpha`

On substituting the known values,

`\begin{gathered} \angle A=180\degree-25\degree-40\degree \\ =115\degree^{} \end{gathered}`

From the cos rule,

`d^2=a^2+b^2-2ab\cos A`

On substituting the known values,

`\begin{gathered} d^2=30^2+60^2-2*30*60*\cos 115\degree \\ \Rightarrow d=\sqrt[]{30^2+60^2-2*30*60*\cos 115\degree} \\ =77.6\text{ m} \end{gathered}`

Thus the total displacement of the boy is 77.6 m