Question

A motor car is fitted with a convex driving mirror of focal length 20cm. A second motor car 2.8m behind the first car is overtaking speed of 15 ${ms}^{-1}$. The speed of the image of the second car as seen in the mirror of the first is :

• A. $\frac{1}{10}{ms}^{-1}$
• B. $\frac{1}{15}{ms}^{-1}$
• C. $10{ms}^{-1}$
• D. $15{ms}^{-1}$

Answer 1

Answer: (B)

$\frac{1}{15}{ms}^{-1}$

Given

focal length $f=20cm$

Object distance, $u=2.8m=280cm$

Speed of object, $v_0=15m/s$

from mirror formula

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}....\left(1\right)$

differentiating on b/s wrt $t$

$0=-\frac{1}{v^2}\frac{dv}{dt}-\frac{1}{u^2}\frac{du}{dt}$

$\Rightarrow$ $\frac{dv}{dt}=-\left[\frac{v^2}{u^2}\frac{du}{dt}\right]...\left(2\right)$

Also from (1)

$\frac{1}{20}=\frac{1}{v}-\frac{1}{280}$

$\frac{1}{v}=\frac{15}{280}$

$v=\frac{280}{15}cm$

Putting value in (2)

$\Rightarrow$ $\frac{dv}{dt}=v_i,\frac{du}{dt}=v_0$

$\Rightarrow$ $v_i=-{\left[\frac{280}{15\times 280}\right]}^{2}15m/s$

$v_i=-\frac{1}{15}m/s$

Speed $=\frac{1}{15}m/s$