Question

The acceleration experienced by a boat after the engine is cut off, is given by (dv/dt)=-kv^3,where k is a constant. if ''v(1)'' is magnitude of the velocity at cut off, find the magnitude of the velocity at time 't' after the cut off.

Answer 1

\[\frac{dv}{dt}=-kv ^{3}=f(v)\]
\[\frac{dv}{f(v)}=dt\]
\[\int\limits_{}^{}\frac{dv}{f(v)}=t+C\]
\[\int\limits_{}^{}\frac{1}{-kv ^{3}}dv=t+C\]
\[-\frac{1}{k} \times -\frac{v ^{-2}}{2}=t+C\]
\[\frac{1}{2kv ^{2}}=t+C\ ..................(1)\]
\[v=v _{1}\ when\ t=0\]
Substituting into (1) we get:
\[\frac{1}{2kv _{1}^{2}}=C\ .................(2)\]
Subtracting (2) from (1) we get:
\[\frac{1}{2kv ^{2}}-\frac{1}{2kv _{1}^{2}}=t\ ................(3)\]

Answer 2

Dividing both sides of (3) by t, we get:
\[\frac{1}{2ktv ^{2}}-\frac{1}{2ktv _{1}^{2}}=1\]
Rearranging gives:
\[\frac{1}{2ktv ^{2}}=1+\frac{1}{2ktv _{1}^{2}}\]
Multiplying both sides by 2kt we get:
\[\frac{1}{v ^{2}}=2kt+\frac{1}{v _{1}^{2}}=\frac{1+2ktv _{1}^{2}}{v _{1}^{2}}\]
Inverting both sides gives:
\[v ^{2}=\frac{v _{1}^{2}}{1+2ktv _{1}^{2}}\]
Taking the square root of both sides gives:
\[v=\frac{v _{1}}{\sqrt{1+2ktv _{1}^{2}}}\]