Question

The acceleration experienced by a moving boat after its engine is cut-off, of 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. $\frac{v_0}{2kt{v}_0^2}$
• B. $\frac{v_0}{1+2kt{v}_0^2}$
• C. $\frac{v_0}{\sqrt{1-2kt{v}_0^2}}$
• D. $\frac{v_0}{\sqrt{1+2kt{v}_0^2}}$

Answer 1

The correct option is D $\frac{v_0}{\sqrt{1+2kt{v}_0^2}}$

• The regarding acceleration
  

$a=-k{v}^3$

which can be written as

$\frac{dv}{dt}=-K{v}^3$

$\frac{dv}{v^3}=-Kat$

integrating both side

$\frac{-1}{2v_0^2}=-Kt+C$

$C=\frac{-1}{2v_0^2}$

so the equation becomes

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

$\frac{1}{2v^2}=Kt+\frac{1}{2v_0^2}$

$2v^2=\frac{1}{Kt+1/2v_0^2}=\frac{2v_0^2}{2kt{v}_0^2+1}$

$v=\frac{v_0}{\sqrt{1+2kt{v}_0^2}}$

Hence(D) is the correct answer