Question

The acceleration experienced by a moving boat after is engine is cut-off, is given by $a=-kv^{3},$ 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. $\frac{v_{0}}{2 k t v_{0}^{2}}$
• B. $\frac{v_{0}}{1 + 2 \, k t v_{0}^{2}}$
• C. $\frac{v_{0}}{\sqrt{1 - 2 \, k \, v_{0}^{2}}}$
• D. $\frac{v_{0}}{\sqrt{1 + 2 \, k t v_{0}^{2}}}$

Answer 1

Answer: (D)

Given, acceleration $a=-kv^{3}$
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, $a=\frac{d v}{d t}=-kv^{3}$
Or $ \, \, \frac{d v}{v^{3}}=-kdt$
Integrating both sides, with in the condition of motion.
$ \, \, \, \int\limits _{v_{0}}^{v} \frac{d v}{v^{3}}=-\int\limits _{0}^{t} k \, d t$
Or $ \, \left[- \frac{1}{2 v^{2}}\right]_{v_{0}}^{v}=-\left[k t\right]_{0}^{t}$
Or $ \, \frac{1}{2 v^{2}}-\frac{1}{2 v_{0}^{2}}=kt$
Or $ \, \, \, v=\frac{v_{0}}{\sqrt{1 + 2 \, k t \, v_{0}^{2}}}$