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Question

A person walks with velocity V in a straight line forming an angle theta with the plane of a plane mirror. Find the velocity 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 V in a straight line forming an angle θ with the plane of a plane mirror.

The velocity V has two components that is, Vsinθ and Vcosθ.

So, Velocity of person will be,

Vp=Vsinθi^+Vcosθj^

Here, Vp 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,

VI=-Vsinθi^+Vcosθj^

Here, VI is the velocity of an image.

Let, VpI be the velocity of person with respect to image.

So,

VpI=Vp-VI

=Vsinθi^+Vcosθj^--Vsinθi^+Vcosθj^

=Vsinθi^+Vcosθj^+Vsinθi^-Vcosθj^

=2Vsinθi^

Hence, the velocity V relative with which he approaches his image is 2Vsinθ.


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