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.
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