Physics - Kinematics Question with Solution | TestHub

Given, acceleration$a=-k{v}^3$

Initial velocity at cut-off,$v_1=v_0$

Initial time of cut-off,$t=0$

And final time after cut-off,$t_2=t$

Again,$a=\frac{dv}{dt}=-k{v}^3$

Or$\frac{dv}{v^3}=-kdt$

Integrating both sides, with in the condition of motion.

${\int }_{v_0}^v\frac{dv}{v^3}=-{\int }_0^{t}k dt$

Or${\left[-\frac{1}{2v^2}\right]}_{v_0}^v=-{\left[kt\right]}_0^t$

Or$\frac{1}{2v^2}-\frac{1}{2v_0^2}=kt$

Or$v=\frac{v_0}{\sqrt{1+2 kt v_0^2}}$