Question

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

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 = 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

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

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

===> 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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