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Question:

A spy report about a suspected car reads as follows. "The car moved $2.00 \mathrm{~km}$ towards east, made a perpendicular left turn, ran for $500 \mathrm{~m}$, made a perpendicular right turn, ran for $4.00 \mathrm{~km}$ and stopped." Find the displacement of the car.

Solution:

$A B=2 i+0.5 j+4 i=6 i+0.5 j$

As the car went forward, took a left and then a right. So, $\mathrm{AB}=\left(6^{2}+0.5^{2}\right)^{1 / 2}=6.02 \mathrm{~km}$

And $\phi=\tan ^{-1}(\mathrm{BE} \backslash \mathrm{AE})=\tan ^{-1}(0.5 / 6)=\tan ^{-1}(1 / 12)$