Suppose while sitting in a parked car, you notice a jogger approaching towards you in the rear view mirror of `R = 2 m`. If the jogger is running at a speed of `5 ms^-1`, how fast is the image of the jogger moving, when the jogger is
(a) 39 m
(b) 29 m
19 m and
(d) 9 m. away ?

Explanation

From the mirror equation. Eq. 9.7, we get
`v=(fu)/(u-f)`
For convex mirror, since R=2m, f=1m. Then
u`=-39m, v=(-39xx1)/(-39-1) = 39/40m`
Since the jogger moves at a constant speed of `5ms^(-1)`, after 1 s the position of the image v(for u=-39+5=-34) is `(34/35)` m.
The shit in the position of image in 1s is
`39/40-34/35=(1365-1360)/(1400)=5/1400=1/280`m
Therefore, the average speed of the image when the jogger is between 39 m nd 34 m from the mirror, is `(1//280)ms^(-1)`
Similarly, it can be seen that for u=`-29m, -19m` and `-9m`, the speed with which the image appears to move is
`1/150ms^(-1), 1/60ms^(-1)` and `1/10ms^(-1)`, repectively.
Although the jogger has been with a constant speed, the speed of his/her image appears to increase substantially as he/she moves closer to the mirror. This phenomenon can be noticed by any person sitting in a stationary car or a bus. In a case of moving vehicles, a similar phenomenon could be observed if the vehicle in the rear is moving closer with a constant speed.