Question

A car $A$ is going north-east at $80km/h$ and another car $B$ is going south-east at $60km/h$. Then the direction of the velocity of $A$ relative to $B$ makes with the north an $\alpha$ angle such that $\tan \alpha \text{is}$ :

• A. $\frac{1}{7}$
• B. $\frac{3}{4}$
• C. $\frac{4}{3}$
• D. $\frac{3}{5}$

Answer 1

The correct option is A $\frac{1}{7}$

Refer image 1.

$V_A=80$ Km/h north-east= $\frac{80}{\sqrt{2}}\hat{i}+\frac{80}{\sqrt{2}}\hat{j}$

$V_B=60$ Km/h south-east= $\frac{60}{\sqrt{2}}\hat{i}-\frac{60}{\sqrt{2}}\hat{j}$

Velocity of $A$ relative to $B\Rightarrow V_{AB}=(\frac{80}{\sqrt{2}}\hat{i}+\frac{80}{\sqrt{2}}\hat{j})-(\frac{60}{\sqrt{2}}\hat{i}+\frac{60}{\sqrt{2}}\hat{j})$

$=\frac{20}{\sqrt{2}}\hat{i}+\frac{140}{\sqrt{2}}\hat{j}$

Refer image 2.

$\tan \propto =\frac{\frac{20}{\sqrt{2}}}{\frac{140}{\sqrt{2}}}=\frac{1}{7}$

Hence, $\tan \propto =\frac{1}{7}$