Question

A man is approaching a plane mirror with constant acceleration of 4 $m/s^2$. Initially the person is at rest. Mirror is moving towards the person with a constant velocity of 4 m/s. The correct variation of velocity of image of man (wrt ground) vs time is best given by consider leftwards as positive)

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

At any time, velocity of man is $v_0=4t$
Using ${\overrightarrow{v}}_{im}=-{\overrightarrow{v}}_{om}\Rightarrow {\overrightarrow{v}}_i-{\overrightarrow{v}}_m=-{\overrightarrow{v}}_0+{\overrightarrow{v}}_m\Rightarrow {\overrightarrow{v}}_i=-{\overrightarrow{v}}_0+{\overrightarrow{v}}_m$
With given sign convension, $v_i=-(–4t)+2\left(4\right)$
$v_i=4t+8$
Here we have to consider the things only upto time when mirror and person meet.