Question

A man has to go 50 m due north, 40 m due east and 20 m due south to reach a field. (a) What distance he has to walk to reach the field? (b) What is his displacement from his house to the field?

Answer 1

(a) Distance travelled by the man = AB + BC + CD = 50 + 40 + 20 = 110 m

(b) AF = AB − BF = 50 − 20 = 30 m

Displacement = Final position − Initial position = AD
$$\begin{gathered} \therefore \mathrm{AD}=\sqrt{{\mathrm{AF}}^2+{\mathrm{DF}}^2}=\sqrt{{30}^2+{40}^2} \\ =50\mathrm{m} \end{gathered}$$

In ∆AED,
$$\begin{gathered} \tan \mathrm{\theta }=\frac{\mathrm{DE}}{\mathrm{AE}}=\frac{30}{40} \\ \Rightarrow \mathrm{\theta }={\tan }^{-1}\left(\frac{3}{4}\right) \end{gathered}$$
Displacement from the house to the field = 50 m in the direction ${\tan }^{-1}\left(\frac{3}{4}\right)$ north to east.