A car ‘A’ moves due north at a speed of 40 km/hr, while another car ‘B’ moves due east at a speed of 30 km/hr. Find the velocity of car B relative to car A (both in magnitude and direction).

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40 km/hr, at an angle $\tan^{- 1} (\frac{3}{5}) east of south$

50 km/hr, at an angle $\tan^{- 1} (\frac{3}{5}) east of south$

40 km/hr, at an angle $\tan^{- 1} (\frac{3}{4}) east of south$

50 km/hr, at angle $\tan^{- 1} (\frac{3}{4}) east of south$

answer is D.

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Detailed Solution

$\vec{v_{A}} = 40 \hat{j}, \vec{v_{B}} = 30 \hat{i}$

$\vec{v_{B / A}} = \vec{v_{B}} - \vec{v_{A}} = 30 \hat{i} - 40 \hat{j}$

$| \vec{v_{B / A}} | = \sqrt{30^{2} + 40^{2}} = 50 km / h$

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