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.

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Solution

## 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,${\mathrm{V}}_{\mathrm{p}}=\mathrm{V}\mathrm{sin\theta }\stackrel{^}{\mathrm{i}}+\mathrm{V}\mathrm{cos\theta }\stackrel{^}{\mathrm{j}}$Here, ${\mathrm{V}}_{\mathrm{p}}$ is the velocity of person.Step 2: Find the velocity of imageSo, the image of the man has same distance from mirror.So, the velocity of image will be,${\mathrm{V}}_{\mathrm{I}}=-\mathrm{V}\mathrm{sin\theta }\stackrel{^}{\mathrm{i}}+\mathrm{V}\mathrm{cos\theta }\stackrel{^}{\mathrm{j}}$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, $\stackrel{\to }{{\mathrm{V}}_{\mathrm{pI}}}=\stackrel{\to }{{\mathrm{V}}_{\mathrm{p}}}-\stackrel{\to }{{\mathrm{V}}_{\mathrm{I}}}$ $=\left(\mathrm{V}\mathrm{sin\theta }\stackrel{^}{\mathrm{i}}+\mathrm{V}\mathrm{cos\theta }\stackrel{^}{\mathrm{j}}\right)-\left(-\mathrm{V}\mathrm{sin\theta }\stackrel{^}{\mathrm{i}}+\mathrm{V}\mathrm{cos\theta }\stackrel{^}{\mathrm{j}}\right)$ $=\mathrm{Vsin\theta }\stackrel{^}{\mathrm{i}}+\overline{)\mathrm{Vcos\theta }\stackrel{^}{\mathrm{j}}}+\mathrm{Vsin\theta }\stackrel{^}{\mathrm{i}}-\overline{)\mathrm{Vcos\theta }\stackrel{^}{\mathrm{j}}}$ $=2\mathrm{Vsin\theta }\stackrel{^}{\mathrm{i}}$Hence, the velocity $\mathrm{V}$ relative with which he approaches his image is $2\mathrm{Vsin\theta }$.

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