Question

A thief is running away in a car with velocity of 20 m/s. A police jeep is following him, which is sighted by a thief in his rear view mirror, of focal length 10m. He observes that image of jeep is moving towards him with a velocity of 1cm/s If the magnification of the mirror for the jeep at that time is $\frac{1}{10}$, then at some instant of time (Assume jeep to be point object moving on the axes of mirror)

• A. Actual speed of the jeep is 21 m/s
• B. Distance of jeep from pole of mirror is 90 m
• C. Rate at which magnification is changing is $1\times {10}^{-3}/sec$
• D. Rate at which magnification is changing is $2\times {10}^{-3}/sec$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Actual speed of the jeep is 21 m/s
B Distance of jeep from pole of mirror is 90 m
C Rate at which magnification is changing is $1\times {10}^{-3}/sec$

$m=-\left(\frac{q}{p}\right)\Rightarrow \frac{1}{10}=-\frac{q}{p}\Rightarrow \frac{-p}{10}\dots \left(1\right)\frac{1}{p}+\frac{1}{q}=\frac{1}{f}\Rightarrow -\frac{10}{p}+\frac{1}{p}=\frac{-1}{10}\left(asfoeconcave\right)\Rightarrow p=90mAlso{V}_{1m}=-m^2\left(v_{om}\right)\Rightarrow -1=\frac{1}{100}(V_0-20)\Rightarrow V_0=21m/s\left(b\right)\frac{dm}{dt}=-\frac{(p\frac{dq}{dt}-q\left(\frac{dp}{dt}\right))}{p^2}=\frac{-\left[90\right(-1\times {10}^{-2})+9)\left(1\right)]}{90\times 90}=1\times {10}^{-3}/sec$