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\newcommand{\ds}{\displaystyle}
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\begin{document}

{\bf Question}

A body falling \emph{from rest} at $t=0$, in a medium where the
resistance is proportional to the velocity obeys the equation

$$\ds\frac{dv}{dt}=g-kv$$

where $k$ is constant. Show that eventually the body achieves a
constant (terminal) velocity. If this terminal velocity is $V$,
find, in terms of $k$, the time that elapses before a body falling
from rest attains a velocity $\ds\frac{V}{2}$.

\medskip

{\bf Answer}

$\ds\frac{dv}{dt}=g-kv$

Variables separable

$\ds\int\ds\frac{dv}{g-kv}=\ds\int \,dt \Rightarrow
-\ds\frac{1}{k}\ln(g-kv)=t+c$

What is $c$? Particle falls from rest ($v=0$) at $t=0$.

Thus, $-\ds\frac{1}{k}\ln(g-0)=0+c \Rightarrow -\ds\frac{1}{k}\ln
g=c$

Thus

$\begin{array} {crcl} &amp; -\ds\frac{1}{k}\ln(g-kv) &amp; = &amp;
t-\ds\frac{1}{k}\ln g\\ \rm{or} &amp; \ds\frac{1}{k}\ln
g-\ds\frac{1}{k}\ln(g-kv) &amp; = &amp; t\\ \Rightarrow &amp;
\ds\frac{1}{k}\ln\left(\ds\frac{g}{g-kv}\right) &amp; = &amp; t\\ \rm{or}
&amp; \ds\frac{g}{g-kv} &amp; = &amp; e^{kt}\\ &amp; ge^{-kt} &amp; = &amp; g-kv
\end{array}$

$\Rightarrow \un{v=\ds\frac{g}{k}(1-e^{-kt})}$

Now as $t \to \infty$,\ $v \to \ds\frac{g}{k}(1-e^{-\infty})$

But $e^{-\infty}=0 \Rightarrow \un{v \to \ds\frac{g}{k}}$ the
terminal velocity

Thus $V=\ds\frac{g}{k}$.

\newpage
Now when $v=\ds\frac{V}{2}$ we have

$\begin{array} {crcl} &amp; \ds\frac{V}{2} &amp; = &amp;
\ds\frac{g}{k}(-e^{-kt})\\ or &amp; \ds\frac{g}{2k} &amp; = &amp;
\ds\frac{g}{k}(1-e^{-kt})\\ \Rightarrow &amp; \ds\frac{1}{2} &amp; = &amp;
1-e^{-kt}\\ \Rightarrow &amp; e^{-kt} &amp; = &amp; \ds\frac{1}{2}\\
\Rightarrow -kt=\ln\left(\ds\frac{1}{2}\right)\\ \Rightarrow
kt=\ln 2 \end{array}$

$\ \Rightarrow \un{t=\ds\frac{1}{k}\ln 2}$

\end{document}
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