Velocity of image in convex lens $V _{ i }=( f /( f + u ))^{2} V_o$
and image will move from focus to infinity (i.e. away from the lens).
From the formula, velocity of image
$\left( dV _{ i }\right) / dt =\left( d V _{ i } / du \right) \times( du / dt )$
$=2( f / f + u ) f \ln ( f + u ) \times 5$ (because $\left.du / dt =5 m / s \right)$
$\Rightarrow \left( dV _{ i }\right) / dt =10 f ^{2} \ln ( f + u ) /( f + u )$
We can see that acceleration of image $\propto \ln (f+u)$,
i.e varies with distance of object
So the image moves away from the lens with a non-uniform acceleration