Question

A convex driving mirror of focal length $20\text{cm}$, is fitted in a motor car. If the second car $2\text{m}$ broad and $1.6\text{m}$ high is $6\text{m}$ away from the first car and overtakes the first car at a relative speed of $15\text{m/s}$, then how fast relatively in $\text{m/s}$ will the image be moving? Answer upto 3 decimal places

• A. 0.017
• B. 0.015
• C. 0.016

Answer 1

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

Mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Differentiating w.r.t $t$, we get$0=-\frac{1}{v^2}\frac{dv}{dt}-\frac{1}{u^2}\frac{du}{dt}$
or $\frac{dv}{dt}=-\frac{v^2}{u^2}\frac{du}{dt}$
But
$\frac{du}{dt}=15\text{m/s}$
Now,$\frac{1}{v}=\frac{1}{20}+\frac{1}{600}$
or $v=\frac{600}{31}\text{cm}$

and speed of image $=\frac{dv}{dt}={\left(\frac{1}{31}\right)}^2\times 15=0.016\text{m/s}$

Why this question? To learn to find the image velocity when the object velocity and its relative position with respect to the mirror is known.