\subsection{Drag Forces and Terminal Velocity}

This subsection uses the AP linear-drag model, in which the resistive force depends on the object's instantaneous velocity. The local-first viewpoint is Newton's second law with a velocity-dependent force.

\dfn{Linear drag and terminal velocity}{Let an object move through a fluid with velocity $\vec{v}$ relative to the fluid, and let $b>0$ denote the linear-drag coefficient. In the linear-drag model, the drag force exerted by the fluid on the object is
\[
\vec{F}_D=-b\vec{v}.
\]
The negative sign means that the drag force always points opposite the velocity.

If an object falls vertically through the fluid and eventually moves with constant velocity, that steady velocity is called the \emph{terminal velocity}. At terminal velocity, the net force is zero, so the acceleration is zero.}

\wc{Terminal velocity is not zero speed}{Terminal velocity is a \emph{nonzero constant speed}: the object is still moving, but its acceleration is zero because drag balances the driving force. The object approaches $v_T$ asymptotically as $t\to\infty$; it never stops.}

\wc{Drag force is not kinetic friction}{Drag $F_D=bv$ or $F_D=\tfrac12 C\rho A v^2$ is velocity-dependent and acts in fluids. Kinetic friction $f_k=\mu_k N$ is approximately velocity-independent and acts at solid-solid contacts. They are physically distinct interactions.}

\thm{Vertical-fall ODE and terminal-speed result}{Choose the vertical axis positive downward. Let $v(t)$ denote the downward velocity of an object of mass $m$ at time $t$, let $g$ denote the gravitational field strength, and let $b>0$ denote the linear-drag coefficient. Then the vertical equation of motion is
\[
m\frac{dv}{dt}=mg-bv.
\]
The terminal speed $v_T$ is the steady-state value obtained by setting the net force equal to zero:
\[
mg-bv_T=0,
\]
so
\[
v_T=\frac{mg}{b}.
\]

If the initial velocity is $v(0)=v_0$, then
\[
v(t)=v_T+(v_0-v_T)e^{-bt/m}.
\]
In particular, if the object is released from rest, then
\[
v(t)=v_T\left(1-e^{-bt/m}\right).
\]}

\nt{The linear drag model $F_D=bv$ is valid for low-Reynolds-number flow (small objects, slow speeds, viscous fluids). For most everyday situations (baseballs, skydivers, cars at highway speed), drag is approximately quadratic: $F_D=\tfrac12 C\rho A v^2$, where $C$ is the drag coefficient, $\rho$ is the fluid density, and $A$ is the cross-sectional area. The AP exam typically uses the linear model for analytical tractability, but recognizes the quadratic model qualitatively.}

\pf{Short derivation of the velocity function}{Start with Newton's second law for vertical fall in the downward-positive direction:
\[
m\frac{dv}{dt}=mg-bv.
\]
Define the terminal speed by
\[
v_T=\frac{mg}{b}.
\]
Then the differential equation becomes
\[
\frac{dv}{dt}=\frac{b}{m}(v_T-v).
\]
Separate variables:
\[
\frac{dv}{v_T-v}=\frac{b}{m}\,dt.
\]
Integrating gives
\[
-\ln|v_T-v|=\frac{b}{m}t+C.
\]
Therefore
\[
v_T-v=Ce^{-bt/m}
\]
for some constant $C$, so
\[
v=v_T-Ce^{-bt/m}.
\]
Now use the initial condition $v(0)=v_0$:
\[
v_0=v_T-C.
\]
Thus $C=v_T-v_0$, and
\[
v(t)=v_T-(v_T-v_0)e^{-bt/m}=v_T+(v_0-v_T)e^{-bt/m}.
\]
If $v_0=0$, this reduces to
\[
v(t)=v_T\left(1-e^{-bt/m}\right).
\]
As $t\to\infty$, the exponential term approaches $0$, so $v(t)\to v_T$.}

\ex{Illustrative example}{Choose downward as positive. A ball of mass $m=0.20\,\mathrm{kg}$ falls through air with linear drag coefficient $b=0.50\,\mathrm{N\cdot s/m}$. Find the terminal speed and the acceleration when the downward speed is $v=2.0\,\mathrm{m/s}$.

The terminal speed is
\[
v_T=\frac{mg}{b}=\frac{(0.20\,\mathrm{kg})(9.8\,\mathrm{m/s^2})}{0.50\,\mathrm{N\cdot s/m}}=3.92\,\mathrm{m/s}.
\]
When $v=2.0\,\mathrm{m/s}$, Newton's second law gives
\[
m\frac{dv}{dt}=mg-bv,
\]
so the acceleration is
\[
a=g-\frac{b}{m}v=9.8-\frac{0.50}{0.20}(2.0)=4.8\,\mathrm{m/s^2}.
\]
Since this value is positive in the downward-positive coordinate system, the ball is still accelerating downward.}

\qs{Worked example}{A small package of mass $m=0.40\,\mathrm{kg}$ is dropped from rest and falls vertically through air. Choose downward as positive. Let $v(t)$ denote the package's downward velocity at time $t$, let $g=9.8\,\mathrm{m/s^2}$, and let the drag force be modeled by
\[
\vec{F}_D=-b\vec{v}
\]
with $b=0.80\,\mathrm{N\cdot s/m}$.

\begin{enumerate}[label=(\alph*)]
\item Write the differential equation for $v(t)$.
\item Find the terminal speed.
\item Find an explicit formula for $v(t)$.
\item Find the package's velocity and acceleration at $t=1.0\,\mathrm{s}$.
\end{enumerate}}

\sol The forces on the package are its weight $\vec{W}$ downward and the drag force $\vec{F}_D$ upward because the package is moving downward. Since downward is chosen as positive, the scalar force equation is
\[
mg-bv=m\frac{dv}{dt}.
\]

For part (a), the differential equation is therefore
\[
m\frac{dv}{dt}=mg-bv.
\]
Substitute $m=0.40\,\mathrm{kg}$ and $b=0.80\,\mathrm{N\cdot s/m}$:
\[
(0.40)\frac{dv}{dt}=(0.40)(9.8)-0.80v.
\]
So an equivalent form is
\[
\frac{dv}{dt}=9.8-2.0v.
\]

For part (b), terminal speed occurs when the acceleration is zero, so
\[
\frac{dv}{dt}=0.
\]
Then
\[
mg-bv_T=0,
\]
which gives
\[
v_T=\frac{mg}{b}=\frac{(0.40)(9.8)}{0.80}=4.9\,\mathrm{m/s}.
\]

For part (c), because the package is dropped from rest, the initial condition is
\[
v(0)=0.
\]
Using the linear-drag result for release from rest,
\[
v(t)=v_T\left(1-e^{-bt/m}\right).
\]
Substitute the values:
\[
v(t)=4.9\left(1-e^{-(0.80/0.40)t}\right).
\]
Therefore,
\[
v(t)=4.9\left(1-e^{-2.0t}\right)\,\mathrm{m/s}.
\]

For part (d), evaluate this at $t=1.0\,\mathrm{s}$:
\[
v(1.0)=4.9\left(1-e^{-2.0}\right)\,\mathrm{m/s}.
\]
Since
\[
e^{-2.0}\approx 0.135,
\]
we get
\[
v(1.0)\approx 4.9(0.865)=4.24\,\mathrm{m/s}.
\]
So after $1.0\,\mathrm{s}$ the package is moving downward at about
\[
4.24\,\mathrm{m/s}.
\]

Now find the acceleration. From the differential equation,
\[
a=\frac{dv}{dt}=9.8-2.0v.
\]
At $t=1.0\,\mathrm{s}$,
\[
a=9.8-2.0(4.24)=1.32\,\mathrm{m/s^2}.
\]
This is positive in the downward-positive coordinate system, so the acceleration is still downward.

Therefore the differential equation is
\[
\frac{dv}{dt}=9.8-2.0v,
\]
the terminal speed is
\[
4.9\,\mathrm{m/s},
\]
the velocity function is
\[
v(t)=4.9\left(1-e^{-2.0t}\right)\,\mathrm{m/s},
\]
and at $t=1.0\,\mathrm{s}$ the package has velocity
\[
4.24\,\mathrm{m/s}
\]
downward and acceleration
\[
1.32\,\mathrm{m/s^2}
\]
downward.
