An object is moving towards a concave mirror of focal length $24 \mathrm{cm}$. When it is at a distance of $60 \mathrm{cm}$ from the mirror its speed is $9 {\mathrm{cms}}^{-1}$. The speed of its image at that instant, is

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\Rightarrow v^{-1}+u^{-1}=f^{-1}=$ constant

Taking derivative w.r.t. time on both sides

 $\frac{d}{dt}\left(v^{-1}\right)+\frac{d}{dt}\left(u^{-1}\right)=0$

$\frac{d}{dt}\left(v^{-1}\right)+\frac{d}{dt}\left(u^{-1}\right)=0$

when object moves towards mirror u decreases with passage of time and hence $\frac{du}{dt}=-9{\mathrm{cms}}^{-1}$

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

$\Rightarrow \frac{dv}{dt}=-{\left(\frac{-24}{-24+60}\right)}^2\frac{du}{dt}$

$\Rightarrow \frac{dv}{dt}=-{\left(\frac{2}{3}\right)}^2\frac{du}{dt}$

$\Rightarrow \frac{dv}{dt}=-\frac{4}{9}(-9)$

$\Rightarrow \frac{dv}{dt}=+4{\mathrm{cms}}^{-1}$

Positive value of  $\frac{dv}{dt}$ indicates that $\mathrm{v}$ increases with the passage of time i.e. image must be going away from mirror.