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.

Here, R = 2 m,f = R/2 =2/2 = 1mUsing mirror formula, we have1/v +1 /u = 1/f====> 1/v = 1/f -1/u====> 1/v = u-f/fu ===> fu/u-f -----(i)When jogger is 39 m away, then u = -39mUsing Eq. (i), we getv = fu/u-f = 1(-39) /-39-1or v = 39/40 mAs the Jogger is running at a constant speed 5 m/s, after 1 s, the position of the image foru = -39 +5===> -34 mAgain using Eq. (i), we getv = 1-(-34) / -34-1===> v = 34/35 mDifference in apparent position of Jogger in 1 s = 39/40-34/35 = 1365-1360/1400 = 1/280 mAverage speed of Jogger’s image = 1/280 m/sSimilarly, for u = -29 m, -19 m and -9 m, average speed of Jogger image is 1/150 m/s, 1/60 m/s, 1/10 m/s, respectively.The speed increases as the Jogger approaches the car.This can be experienced by the person in the car.