A person walks at a velocity v in a straight line forming an angle α with the plane of a plane mirror. the velocity at which he approaches his image is:
A
v
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B
vsinα
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C
2vsinα
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D
2vcosα
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Solution
The correct option is D2vcosα
Since it is a plane mirror vertical component of velocity have no significance.
The person approaches the mirror with a horizontal speed of vcosα.
Hence his image also move towards the mirror with same velocity vcosα in the opposite direction.
Velocity at which he approaches his image is ,v′=vcosα−(−vcosα)=2vcosα