A particle moving in a straight line is acted by a force, which works at a constant power and changes its velocity from u to $v$ in passing over a distance x. The time taken will be:

The force acting on the particle $=\frac{\mathrm{mdv}}{\mathrm{dt}}$

Power of the force $=\left(\frac{\mathrm{mdv}}{\mathrm{dt}}\right)\mathrm{v}=\mathrm{k}$ (constant)

$\mathrm{m}\frac{{\mathrm{v}}^2}{2}=\mathrm{kt}+\mathrm{c}\dots \left(1\right)$

$\text{ at }\mathrm{t}=0,\mathrm{v}=\mathrm{u}$

$\therefore \mathrm{c}=\frac{{\mathrm{mu}}^2}{2}$

Now from (1),

$\mathrm{m}\frac{{\mathrm{v}}^2}{2}=\mathrm{kt}+\frac{{\mathrm{mu}}^2}{2}$

$\frac{1}{2}\mathrm{m}\left({\mathrm{v}}^2-{\mathrm{u}}^2\right)=\mathrm{kt}\dots \left(2\right)$

$\text{ Again }\frac{\mathrm{mdv}}{\mathrm{dt}}\mathrm{v}=\mathrm{k}$

$\mathrm{m}.\mathrm{v}\frac{\mathrm{dv}}{\mathrm{dx}}\mathrm{v}=\mathrm{k}$

${\int }_0^{\mathrm{v}} {\mathrm{mv}}^2\mathrm{dv}={\int }_0^{\mathrm{x}} \mathrm{kdx}$

Integrating,

$\frac{1}{3}\mathrm{m}\left({\mathrm{v}}^3-{\mathrm{u}}^3\right)=\mathrm{kx}\dots \left(3\right)$

From eqn (2) and (3),

$\mathrm{t}=\frac{3}{2}\left(\frac{{\mathrm{v}}^2-{\mathrm{u}}^2}{{\mathrm{v}}^3-{\mathrm{u}}^3}\right)\mathrm{x}$