A car is fitted with a convex sideview mirror of focal length 20 cm. A second car 2.8 m behind the first car is over taking the first car at a relative speed of 15 m/s the speed of the image of the second car as seen in the mirror of the first one is

From mirror formula $\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}$

Differentiating above equation we get 
 $\begin{array}{l}\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{1}{\mathrm{f}}\right)=\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{1}{\mathrm{v}}\right)+\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{1}{\mathrm{u}}\right)\\ 0=-\frac{1}{{\mathrm{v}}^2}\frac{\mathrm{dv}}{\mathrm{dt}}-\frac{1}{{\mathrm{u}}^2}\frac{\mathrm{du}}{\mathrm{dt}}\end{array}$

$\begin{array}{l}\frac{1}{{\mathrm{v}}^2}\frac{\mathrm{dv}}{\mathrm{dt}}=-\frac{1}{{\mathrm{u}}^2}\frac{\mathrm{du}}{\mathrm{dt}}\\ \frac{\mathrm{dv}}{\mathrm{dt}}=-\frac{{\mathrm{v}}^2}{{\mathrm{u}}^2}\frac{\mathrm{du}}{\mathrm{dt}}\end{array}$

And also we get 

$\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{f}}{\mathrm{u}-\mathrm{f}}$

$\frac{\mathrm{dv}}{\mathrm{dt}}=-{\left(\frac{\mathrm{f}}{\mathrm{u}-\mathrm{f}}\right)}^2\times \frac{\mathrm{du}}{\mathrm{dt}}$

$\begin{array}{l}\mathrm{f}=20\mathrm{cm}=0.2\mathrm{m}\\ \frac{\mathrm{du}}{\mathrm{dt}}=-15{\mathrm{ms}}^{-1}\end{array}$

$\mathrm{u}=-2.8\mathrm{m}$

$-\frac{\mathrm{dv}}{\mathrm{dt}}={\left[\frac{0.2}{-2.8-0.2}\right]}^2\times 15$

$={[\frac{10}{10}\times \frac{0.2}{-3}]}^2\times 15$

$={\left[\frac{2}{-30}\right]}^2\times 15$

$=\frac{1}{15}\times \frac{1}{15}\times 15$

$\frac{-dv}{dt}=\frac{1}{15}m{s}^{-1}$