Q. The retardation experienced by a moving motor boat, after its engine is cut off, is given by $\frac{du}{dt} = - kv^{3},$ where $k$ is a constant. If $v_{0}\ $is the magnitude of the velocity at cut-off, the magnitude of the velocity at time $t$ after the cut-off is
A) $v_{0}$
B) $\frac{v_{0}}{2}$
C) $v_{0}e^{- kt}$
D) $\frac{v_{0}}{\sqrt{2v_{0}^{2}kt + 1}}$
Explanation
(d) $\int_{v_{0}}^{v}{\frac{dv}{kv^{3}} = \int_{0}^{t}{dt}}$
$- \frac{1}{k}\int_{v_{0}}^{v}{v^{- 3}dv = t}$ or $- \frac{1}{k}\left| \frac{v^{- 3 + 1}}{- 3 + 1} \right|_{v_{0}}^{v} = t$
or $\frac{1}{2k}\left\lbrack \frac{1}{v^{2}} - \frac{1}{v_{0}^{2}} \right\rbrack = t$ or $\frac{1}{v^{2}} - \frac{1}{v_{0}^{2}} = 2kt$
or $\frac{1}{v^{2}} = \frac{1}{v_{0}^{2}} + 2kt$ or $\frac{1}{v^{2}} = \frac{1 + 2v_{0}^{2}kt}{v_{0}^{2}}$
or $v = \frac{v_{0}}{\sqrt{2v_{0}^{2}kt + 1}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.