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You noticed a jogger approaching towards you, sitting in a parked car in the side view mirror of radius of curvature of 2m. If the jogger is running at a speed of $5 {ms}^{- 1},$ the average speed of image of jogger in 1 sec when he is initially at a distance of 9m is nearly

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$0.5 {ms}^{- 1}$

$0.2 {ms}^{- 1}$

$0.3 {ms}^{- 1}$

$0.1 {ms}^{- 1}$

answer is D.

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Detailed Solution

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v_{1}} - \frac{1}{9} = \frac{+ 1}{1}$

$\frac{1}{v_{1}} = 1 + \frac{1}{9} =\Rightarrow v_{1} = \frac{9}{10}$

$\frac{1}{v_{2}} - \frac{1}{4} = + 1 \Rightarrow v_{2} = \frac{4}{5}$

$Avg vel = \frac{v_{1} - v_{2}}{t}$

$= \frac{0.9 - 0.8}{1} = 0.1 {ms}^{- 1}$

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