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Question

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 ms−1, how fast image of the jogger appear to move when the jogger is:

1. 39 m
2. 29 m
3. 19 m
4. 9 m

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Solution

Since R=2m,f=R2=m
Given that , dudt=−5m/s
we have, 1u+1v=1f
v=fuu−f
Differentiating w.r.t t on both sides.
dvdt=(fdudt)−(dudt)(fu)(u−f)2
=−f2dudt(u−f)2
dvdt=−f2(u−f)2dudt
(A)When jogger is 39m away,
dvdt=−12(39−1)2(−5)=3.46×10−3m/s
(B)When jogger is 29m away,
dvdt=−12(29−1)2(−5)=6.38×10−3m/s
(C)When jogger is 19m away,
dvdt=−12(19−1)2(−5)=0.0154m/s
(D) when 9m away,
dvdt=0.078m/s

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