Question

A plane mirror and an object approach each other with speeds of $5m{s}^{-1}$ and $10m{s}^{-1}$ respectively. What will be the speed of the image w.r.t. the stationary observer?

• A. $5m{s}^{-1}$
• B. $15m{s}^{-1}$
• C. $20m{s}^{-1}$
• D. $25m{s}^{-1}$

Answer 1

The correct option is D

$25m{s}^{-1}$

Given:
Speed of object, $u_o=5m{s}^{-1}$
Speed of mirror, $u_m=10m{s}^{-1}$
Mirror and object are approaching each other.

To find:
Speed of image

Procedure:
Let initial position of object be at $x_{oi}=0$
Let initial position of mirror be at $x_{mi}=u$
Object distance equals the image distance. Hence, initial position of image is $x_{ii}=2u$

Distance is the product of speed and time.
After time $t$,
Position of object will be:
$x_{of}=x_{oi}+u_{o}t=5t$
Position of mirror will be:
$x_{mf}=x_{mi}-u_{f}t=u-10t$

Let the final position of image be $x_{if}$
Using object distance equals image distance after time $t$.
$x_{if}-x_{mf}=x_{mf}-x_{of}$
$x_{if}=2u-25t$

Distance moved by image in time $t$ is:
$s_i=x_{if}-x_{ii}$
$s_i=-25t$

Speed is the ratio of distance and time. Speed of the image is:
$v_i=\frac{s_i}{t}=-25$

Negative sign appears because image is moving leftwards.
Hence, speed of image is $25m{s}^{-1}$