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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?
Solution
(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.
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