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

$$\therefore AD=\sqrt{{AF}^2+{DF}^2}=\sqrt{{30}^2+{40}^2}$$

$$=50\text{ m }$$

In ∆AED,

$$\tan \theta =\frac{DE}{AE}=\frac{30}{40}$$
$$\Rightarrow \theta ={\tan }^{-1}\left(\frac{3}{4}\right)$$

Displacement from the house to the field = 50 m in the direction $${\tan }^{-1}\left(\frac{3}{4}\right)$$ north to east.