Question

A person walks with velocity $\mathrm{V}$ in a straight line forming an angle theta with the plane of a plane mirror. Find the velocity $\mathrm{V}$ relative with which he approaches his image.

Answer 1

Step 1. Given data:

It is given that a person walks with velocity $\mathrm{V}$ in a straight line forming an angle $\mathrm{\theta }$ with the plane of a plane mirror.

The velocity $\mathrm{V}$ has two components that is, $\mathrm{Vsin\theta }$ and $\mathrm{Vcos\theta }$.

So, Velocity of person will be,

${\mathrm{V}}_{\mathrm{p}}=\mathrm{V}\mathrm{sin\theta }\hat{\mathrm{i}}+\mathrm{V}\mathrm{cos\theta }\hat{\mathrm{j}}$

Here, ${\mathrm{V}}_{\mathrm{p}}$ is the velocity of person.

Step 2: Find the velocity of image

So, the image of the man has same distance from mirror.

So, the velocity of image will be,

${\mathrm{V}}_{\mathrm{I}}=-\mathrm{V}\mathrm{sin\theta }\hat{\mathrm{i}}+\mathrm{V}\mathrm{cos\theta }\hat{\mathrm{j}}$

Here, ${\mathrm{V}}_{\mathrm{I}}$ is the velocity of an image.

Let, ${\mathrm{V}}_{\mathrm{pI}}$ be the velocity of person with respect to image.

So,

$\overrightarrow{{\mathrm{V}}_{\mathrm{pI}}}=\overrightarrow{{\mathrm{V}}_{\mathrm{p}}}-\overrightarrow{{\mathrm{V}}_{\mathrm{I}}}$

$=\left(\mathrm{V}\mathrm{sin\theta }\hat{\mathrm{i}}+\mathrm{V}\mathrm{cos\theta }\hat{\mathrm{j}}\right)-\left(-\mathrm{V}\mathrm{sin\theta }\hat{\mathrm{i}}+\mathrm{V}\mathrm{cos\theta }\hat{\mathrm{j}}\right)$

$=\mathrm{Vsin\theta }\hat{\mathrm{i}}+\cancel{\mathrm{Vcos\theta }\hat{\mathrm{j}}}+\mathrm{Vsin\theta }\hat{\mathrm{i}}-\cancel{\mathrm{Vcos\theta }\hat{\mathrm{j}}}$

$=2\mathrm{Vsin\theta }\hat{\mathrm{i}}$

Hence, the velocity $\mathrm{V}$ relative with which he approaches his image is $2\mathrm{Vsin\theta }$.