Question

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).

• A. $40$ km/hr, at an angle $ta{n}^{-1}\left(\frac{3}{5}\right)$ east of south
• B. 50 km/hr, at an angle $ta{n}^{-1}\left(\frac{3}{5}\right)$ east of south
• C. $40$ km/hr, at an angle $ta{n}^{-1}\left(\frac{3}{4}\right)$ east of south
• D. $50$ km/hr, at an angle $ta{n}^{-1}\left(\frac{3}{4}\right)$ east of south

Answer 1

Answer: (D)

${\vec{v}}_A=40\hat{j},{\vec{v}}_B=30\hat{i}$

${\vec{v}}_{B/A}={\vec{v}}_B-{\vec{v}}_A=30\hat{i}-40\hat{j}$

$\left|{\vec{v}}_{B/A}\right|=\sqrt{30^2+40^2}=50Km/h$