Question

A person crossing the a road with a certain velocity due north sees a car moving towards east. The velocity of the car with respect to the person is $\sqrt{2}$ times that of velocity of the person. The angle made by the relative velocity with the east is

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is B ${45}^{\circ }$

$\overrightarrow{V}$ person $=v\hat{j}$

$\overrightarrow{V}$ car $=u\hat{i}$

$\overrightarrow{V}cp=\overrightarrow{V}car\to person=u\hat{i}-v\hat{j}$

$\Rightarrow |\overrightarrow{V}cp|=\sqrt{u^2+v^2}$

given that $|\overrightarrow{V}cp|=\sqrt{2}V$

$\Rightarrow \sqrt{u^2+v^2}=\sqrt{2}x\Rightarrow u^2+v^2=2v^2$

$\Rightarrow v^2=u^2\Rightarrow v=u$

$\Rightarrow \overrightarrow{V}cp=v\hat{i}-v\hat{j}$

$\theta ={\tan }^{-1}\frac{v}{v}\Rightarrow \theta ={45}^o$