Question

The acceleration experienced by a moving boat after its engine is cut off, is given by $ a=-k{{v}^{3}}, $ where k is a constant. If $ {{v}_{0}} $ is the magnitude of velocity at cut off, then the magnitude of the velocity at time t after the cut off is :

• A. $ \frac{{{v}_{0}}}{2ktv_{0}^{2}} $
• B. $ \frac{{{v}_{0}}}{1+2ktv_{0}^{2}} $
• C. $ \frac{{{v}_{0}}}{\sqrt{1-2kv_{0}^{2}}} $
• D. $ \frac{{{v}_{0}}}{\sqrt{1+2kt\,v_{0}^{2}}} $

Answer 1

Answer: (D)

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 ofmotion $ \int_{{{v}_{0}}}^{v}{\frac{dv}{{{v}^{3}}}=-\int_{0}^{t}{k\,dt}} $ or $ \left[ -\frac{1}{2{{v}^{2}}} \right]_{{{v}_{0}}}^{v}=-[kt]_{0}^{t} $ or $ \frac{1}{2{{v}^{2}}}-\frac{1}{2v_{0}^{2}}=kt $ or $ v=\frac{{{v}_{0}}}{\sqrt{1+2kt\,v_{0}^{2}}} $