question_answer

The deceleration experienced by a moving motor boat after its engine is cut off is given by, \[\frac{dv}{dt}=-\text{k}{{\text{v}}^{\text{3}}}\], where k is a constant. If \[{{\text{v}}_{0}}\]is the magnitude of the velocity at cut off, find the magnitude of the velocity at time t after the engine is cut off.

Answer:

                Here, \[\frac{dv}{dt}=-\text{k}{{\text{v}}^{\text{3}}}\] or \[\frac{dv}{{{v}^{3}}}=-\text{kdt}\] Integrating it within the conditions of motion, we have                                 \[\int\limits_{{{v}_{0}}}^{v}{\frac{dv}{{{v}^{3}}}}=\int\limits_{0}^{t}{-k\,\,dt}\]  or            \[-\left( \frac{1}{2{{v}^{2}}} \right)_{{{v}_{0}}}^{v}=-k\left( t \right)_{0}^{t}\]      or            \[\frac{1}{{{v}^{2}}}-\frac{1}{v_{0}^{2}}=2kt\] or            \[\frac{1}{{{v}^{2}}}=2kt+\frac{1}{v_{0}^{2}}=\frac{2kt\,\,v_{0}^{2}+1}{v_{0}^{2}}\]        or                \[v=\frac{{{v}_{0}}}{\sqrt{2kt\,\,v_{0}^{2}+1}}\]