Suppose while sitting in a parked car, you notice a Jogger approaching towards you in the rear view mirror having R=2m. If the Jogger is running at a speed of 5 m/s, how fast is the image of Jogger moving, when the Jogger is (i) 39 m (i) 29 m (iii) 19 m and (iv) 9 m away?

Explanation

Here, R=2 m, `f =R/2 =2/2=1 m`
Using mirror formula, we have
`1/v+1/u=1/f implies 1/v=1/f-1/u`
`implies 1/v=(u-f)/(fu) implies v=(fu)/(u-f)` ...(i)
When Jogger is 39 m away, then u= -39 m
Using Eq. (i), we get
`v=(fu)/(u-f)=(1(-39))/(-39-1)` or `v=(39)/(40)m`
As, the Jogger is running at a constant speed of 5 `m//s`, after 1 s, the position of the image (v) for
u= -39 + 5
`implies` u= -34 m
Again using Eq. (i), we get
`implies v=(1(-34))/(-34-1) implies v=(34)/(35)m`
Difference in apparent position of Jogger in 1s
`=(39)/(40)-(34)/(35)=(1365-1360)/(1400)=(1)/(280)m`
Average speed of Jogger.s image = `(1)/(280) m//s`
Similarly, 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.