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 = 2 m, f = 1 m. Then
for `u=-39m, v=((-39)xx1)/(-39-1)=(39)/(40)m`
Since the jogger moves at a constant speed of `5 m s^(-1)`, after 1 s the position of the image v `("for "u = 39 + 5 = 34)` is `(34//35 )m`.
The shift in the position of image in 1 s 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 and 34 m from the mirror, is `(1//280) m s^(-1)`
Similarly, it can be seen that for `u = -29 m, -19 m and -9 m`, the speed with which the image appears to move is
`(1)/(150)ms^(-1),(1)/(60)ms^(-1) and (1)/(10)ms^(-1)`, respectively.
Although the jogger has been moving 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 case of moving vehicles, a similar phenomenon could be observed if the vehicle in the rear is moving closer with a constant speed.