Problem 1.
An object (its mass is $m$) falls through the air toward Earth. Assuming that the only forces acting on the object are gravity ($g$ is the gravity constant) and air resistance (proportional to the the speed of the object).(a) Determine the speed $v(t)$ of the object.
(b) Show that $\displaystyle{\lim_{t\to\infty}}v(t)=v_{*}$ independently of the initial speed $v_0$, where $v_{*}$ is the equilibrium speed.
(a) Applying Newton's second law we obtain the differential equation
\[ mv'(t)=mg-kv(t), \] where $m,g,k>0$ are constants and $v(0)=v_0\geq 0$. The equilibrium speed $v_{*}$ is \[ v_{*}=\frac{mg}{k}. \] Separating the variables in the differential equation we have \begin{equation} \label{eq:diffeqlin} \frac{m}{kv{_*}-kv}\,dv=1\,dt. \end{equation} Integrating in \eqref{eq:diffeqlin} both sides we obtain
Solving for $v$ \[
v(t)=v_{*}+ce^{-kt/m}.
\] From the initial condition we find \[
c=v_0-v_{*}.
\] Substituting this value of $c$ we obtain
\begin{equation}
\label{eq:sollin}
v(t)=v_{*}+(v_0-v_{*})e^{-kt/m}. \end{equation} (b) From \eqref{eq:sollin} it follows
\[
\lim_{t\to\infty}v(t)=v_{*}. \qquad\blacksquare \]
\[ mv'(t)=mg-kv(t), \] where $m,g,k>0$ are constants and $v(0)=v_0\geq 0$. The equilibrium speed $v_{*}$ is \[ v_{*}=\frac{mg}{k}. \] Separating the variables in the differential equation we have \begin{equation} \label{eq:diffeqlin} \frac{m}{kv{_*}-kv}\,dv=1\,dt. \end{equation} Integrating in \eqref{eq:diffeqlin} both sides we obtain
\begin{align*}
&\Step{1B}{-\frac{m}{k}\ln\,\left| v_{*}-v \right| =t+c,}\\
&\Step{2B}{\ln\,\left| v_{*}-v\right|=-\frac{kt}{m}+c,} \\
&\Step{3B}{v_{*}-v=ce^{-kt/m}.} \\
\end{align*}
Solving for $v$ \[
v(t)=v_{*}+ce^{-kt/m}.
\] From the initial condition we find \[
c=v_0-v_{*}.
\] Substituting this value of $c$ we obtain
\begin{equation}
\label{eq:sollin}
v(t)=v_{*}+(v_0-v_{*})e^{-kt/m}. \end{equation} (b) From \eqref{eq:sollin} it follows
\[
\lim_{t\to\infty}v(t)=v_{*}. \qquad\blacksquare \]
