Question

A bike is fitted witha convex mirror of $f=40cm$. Another bike $5.6m$ behind is coming with a relative speed of $15m/s$. What is the speed of image of second bike ?

Answer 1

we know form mirror formula

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$------(1)

Thus image distance v

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{u}-\frac{1}{f}\text{(as focal length for convex mirror is-)}=\frac{1}{5.6}-\frac{1}{0.4}=\frac{-13}{5.6}=0.43m$

Now differentiating (1) with respect to time we get

$\frac{-1}{u^2}\frac{du}{dt}+\left(\frac{-1}{v^2}\right)\frac{dv}{dt}=0\frac{\dot{u}}{-u^2}+\left(\frac{\dot{v}}{-v^2}\right)=0\text{where}\dot{u}and\dot{v}\text{are object and image velocity respectively}\dot{v}=\left(\frac{\dot{u}}{u^2}\right)v^2=\left(\frac{15}{{5.6}^2}\right){0.43}^2=0.088m/s$