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

5 $m s^{- 1}$

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15 $m s^{- 1}$

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20 $m s^{- 1}$

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25 $m s^{- 1}$

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Solution

The correct option is C
20 $m s^{- 1}$

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

To find:
Speed of image

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

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

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

Distance moved by image in time $t$ is:
$s_{i} = x_{i f} - x_{i i}$
$s_{i} = - 25 t$

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 $25 m s^{- 1}$

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