Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of R= 2m. If the jogger is running at a speed of 5ms^-1 , how fast the image of the jogger appear to move when the jogger is 39m away?

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Solution

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

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Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of R= 2m. If the jogger is running at a speed of 5ms-1 , how fast the image of the jogger appear to move when the jogger is 39m away?

Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of R= 2m. If the jogger is running at a speed of 5ms^-1 , how fast the image of the jogger appear to move when the jogger is 39m away?