Question

suppose while sitting in a park car you notice a jogger approaching towards you in this side mirror of radius 2 metre is a jogger is running at a speed of 5 metre per second how fast does the image of the jogger to move when the jogger is
9 metre Away? ​

Answer 1

Answer:

Explanation:Here, R = 2 m,

f = R/2 =2/2 = 1m

Using mirror formula, we have

1/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 = -39m

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 5 m/s, after 1 s, the position of the image for

u = -39 +5

===> -34 m

Again using Eq. (i), we get

v = 1-(-34) / -34-1

===> v = 34/35 m

Difference in apparent position of Jogger in 1 s = 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.