A classical mechanics problem by sukhpreet saini

We have by lens formula that $\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$ Plugging in value $f=0.3$ and $u=-0.4$ we get $\boxed{v=1.2m}$ Again using lens formula $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$ Now diffrentiating w.r.t. time and denoting time by ' $t$ '. $0=-\dfrac{dv}{v^{2}dt}+\dfrac{du}{u^{2}dt}$ Rearranging the terms $\dfrac{dv}{dt} = \dfrac{v^{2}du}{u^{2}dt}$ Again plugging in values we get $\boxed{\dfrac{dv}{dt}=0.09ms^{-1}}$

By Lens formula, $\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$ Differentiate it wrt time and keep f as constant(because focal length does not change)

$\frac{dv}{dt}$ = $\frac{v^{2}}{u^{2}}$ . $\frac{du}{dt}$

from lens formula v=1.2 m hence, $\frac{dv}{dt}$ =0.09 m/s