Question

A man running at a speed $ 5 \,m/s $ is viewed in the side view mirror of radius of curvature $ R = 2\, m $ of a stationary car. Calculate the speed of image when the man is at a distance of $ 9 \,m $ from the mirror

• A. $ 0.3 \,m/s $
• B. $ 0.2 \,m/s $
• C. $ 0.1\, m/s $
• D. $ 0.05 \,m/s $

Answer 1

Answer: (C)

$ d=+\ln $
$u=+9 m $
By mirror form Jea :-
$ \frac{1}{v}-\frac{1}{v} =\frac{1}{7} $
$\frac{1}{v} =\frac{1}{7}+\frac{1}{v}$
$=\frac{1}{1}+\frac{1}{9} $
$ V=\frac{9 \times 1}{9+1}$
$V=\frac{9}{10} $
After $18 sc$
$ v=9-5=4 .$
$\frac{1}{v}=\frac{1}{v}+\frac{1}{1} $
$v=\frac{u \times 1}{v+1}=\frac{4}{5} $
speed $=0.1 \,m / s $