A spy report about a suspected car reads as follows. "The car moved 2.00 km towards east, made a perpendicular left turn, ran for 500 m, made a perpendicular right turn, ran for 4.00 km and stopped". Find the displacement of the car.
Solution
The displacement of the car is represented by \[\overrightarrow{AD}\].
\[\overrightarrow{AD} = 2 \hat {i}+ 0 . 5 \hat {j} + 4 \hat {i} \]
\[ = 6 \hat {i} + 0 . 5 \hat {j}\]
Magnitude of \[\overrightarrow{AD}\] is given by
\[AD = \sqrt{{AE}^2 + {DE}^2}\]
\[ = \sqrt{6^2 + \left( 0 . 5 \right)^2}\]
\[ = \sqrt{36 + 0 . 25} = 6 . 02 km\]
Now,
\[\tan \theta = \frac{DE}{AE} = \frac{1}{12}\]
\[\Rightarrow \theta = \tan^{- 1} \left( \frac{1}{12} \right)\]
Hence, the displacement of the car is 6.02 km along the direction \[\tan^{- 1} \left( \frac{1}{12} \right)\] with positive the x-axis.
