question_answer

A car moves 2.00 km towards east, makes a perpendicular left turn and travels 500m, makes a perpendicular right turn and travels 4.00 km and finally stops. Find the displacement of the car.

Answer:

Let the car start at A and move 2 km towards east and reach B. From B, it takes a perpendicular turn and travels 500 m and reaches C. From C, it takes a perpendicular right, turn and travels 4 km and reach D. From the figure, we need to find out displacement \[\overline{AD}\] Now, join BE and DE such that BE = 4km and DE = 500m or 0.5km \[AD=\sqrt{{{6}^{2}}+{{0.5}^{2}}}=6.02\,km\] \[\tan \theta =\frac{1/2}{6}=\frac{1}{12}\] \[\theta ={{\tan }^{-1}}\left( \frac{1}{12} \right)\] Therefore, the displacement of the car is 6.02 km, along the direction \[{{\tan }^{-1}}\left( \frac{1}{12} \right)\]with the positive x-axis.