Suppose while sitting in a parked car,you notice a jogger approaching towards you in the rear view mirror of R=2 m.if the jogger running in a speed of 5 m/s,how fast is the image of the jogger moving,when the jogger is (a) 39 m (b) 29 m (c) 19 m and (d) 9 m away

Answers (1)
S safeer

V_{i m}=-m^{2} V_{O M}

\implies V_{i }-V_{m}=-m^{2} (V_{O }-V_{m})

Here mirror is at rest and f=1m

v = \frac{u}{u-1}

therefore

\\\frac{dv}{dt}=-\frac{1}{(u-1)^2}\frac{du}{dt}\\

du/dt=-5m/s

a)  u = -39 m

\frac{dv}{dt} = -\frac{1}{40^2}*5 m/s=-0.003125 m/s

b) u = -29 m
\frac{dv}{dt} = -\frac{1}{30^2}*5 m/s=-0.0055 m/s

c) u = -19 m
\frac{dv}{dt} = -\frac{1}{20^2}*5 m/s=-0.0125m/s

d) u = -9 m
\frac{dv}{dt} = -\frac{1}{10^2}*5 m/s=-0.05m/sec

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