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

Explanation

From the mirror equation, Eq. `(1)/(upsilon)+(1)/(u)=(1)/(f)`
we get
`upsilon=(fu)/(u-f)`
For convex mirror, since R = 2m, f = 1 m.
Then for `u=-39m.v" "=((39)xx1)/(-39-1)`
`=(39)/(40)`
Since the jogger moves at a constant speed of `5ms^(-1)`, after 1s the position of the image `upsilon` (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)`
Therefore, the average speed of the miage when the jogger is between 39 m and 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)/(150)ms^(-1),(1)/(60)ms^(-1) and (1)/(10)ms` respectively.