An object approaches a convergent lens from the left of the lens with a uniform speed \(5 \mathrm{~m} / \mathrm{s}\) and stops at the focus, the image

From lens formula,
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
Differentiating w.r.t. time, we get
\(\begin{aligned}
0 &=-\frac{1}{v^{2}} \frac{d v}{d t}+\frac{1}{u^{2}} \frac{d u}{d t} \\
\Rightarrow \quad \frac{d v}{d t} &=\left(\frac{v^{2}}{u^{2}}\right) \frac{d u}{d t} \\
v_{i} &=m^{2} v_{O} ...(i)
\end{aligned}\)
where, \(v_{i}=\) image speed, \(v_{0}=\) object speed, and \(m=\) magnification.
Again differentiating w.r.t. time, we get
\(\begin{aligned}
\frac{d^{2} v_{i}}{d t^{2}} &=v_{0} 2 m \frac{d m}{d t} \\
\Rightarrow \quad a_{i} &=2 m v_{0} \frac{d m}{d t} ...(ii)
\end{aligned}\)
where, \(a_{i}\) is the acceleration of the image.
So, when the object approaches the lens with a uniform speed \(5 \mathrm{~m} / \mathrm{s}\) and stops at focus, then from Eq. (i) and (ii), we can say that, the image will move away from the lens with a non-uniform acceleration as \(\frac{d m}{d t}\) is variable.