question_answer

A person walks at a velocity v in a straight line forming an angle a with the plane of a plane mirror. Determine the velocity \[{{v}_{rel}}\] at which he approaches his image, assuming that the object and its image are symmetric relative to the plane of the mirror.     

A) \[2v\,\text{sin}\,\alpha \]

B) \[2v\cos \alpha \]

C) \[v\sin \alpha \]

D) \[v\cos \alpha \]

Correct Answer: A

Solution :

[a] We resolve the velocity vector \[\vec{v}\] of the person into two components, one parallel to the mirror,  \[{{\vec{v}}_{||}}\] and the other perpendicular to the mirror, \[{{\vec{v}}_{\bot }},\]i.e.\[\vec{v}={{\vec{v}}_{||}}+{{\vec{v}}_{\bot }}\] (figure). The velocity of the image will   obviously be\[\vec{v}={{\vec{v}}_{||}}-{{\vec{v}}_{\bot }}\] . Therefore the velocity at which the person approaches his image is defined as his velocity relative to the image. From the formula \[{{v}_{rel}}=2{{v}_{\bot }}=2v\sin \alpha .\]