22.1: Model I F ∝ v

In this first model, we model the resistive force \(F_{R}\) through a fluid as being proportional to the first power of the velocity \(v\) :

\[F_{R}=-b v\]

where \(b\) is the constant of proportionality; the minus sign shows that the resistive force is always opposite the direction of motion.

Under this model, the net downward force on the falling body is \(m g+F_{R}=m g-b v\). Then by Newton's second law,

\[
\begin{align}
F & =m a \\
m g-b v & =m \frac{d v}{d t}
\end{align}
\]

Dividing through by \(m\), we have

\[\frac{d v}{d t}+\frac{b}{m} v=g\]

This is a first-order differential equation, which you will learn to solve for \(v(t)\) in a course on differential equations. But briefly, for a differential equation of the form

\[\frac{d y}{d t}+p(t) y=q(t),\]

the solution \(y(t)\) is found to be (Ref. [2])

\[y(t)=\frac{1}{\mu(t)}\left[\int \mu(t) q(t) d t+C\right]\]

where \(C\) is a constant that depends on the initial conditions, and \(\mu(t)\) is an integrating factor, given by

\[\mu(t)=\exp \left[\int p(t) d t\right]\]

Since this is a first-order differential equation, there will be one arbitrary constant of integration, and it is the constant \(C\) in Eq. \(\PageIndex{6}\).

Comparing Eq. \(\PageIndex{4}\) with Eq. \(\PageIndex{5}\), we have

\[
\begin{align}
& y(t)=v(t) \\
& p(t)=b / m \\
& q(t)=g .
\end{align}
\]

Then the integrating factor \(\mu(t)\) is, from Eq. \(\PageIndex{7}\),

\[
\begin{align}
\mu(t) & =\exp \left[\int p(t) d t\right] \\
& =\exp \left[\int \frac{b}{m} d t\right] \\
& =A e^{b t / m},
\end{align}
\]

where \(A\) is a constant of integration. The solution to Eq. \(\PageIndex{4}\) is then given by Eq. \(\PageIndex{6}\):

\[
\begin{align}
v & =\frac{e^{-b t / m}}{A}\left[\int A e^{b t / m} g d t+C\right] \\
& =e^{-b t / m}\left[\frac{m g}{b} e^{b t / m}+C^{\prime}\right] \\
& =\frac{m g}{b}+C^{\prime} e^{-b t / m} .
\end{align}
\]

To find the constant \(C^{\prime}\), we use the initial condition: if we release the body at time zero, then \(v=0\) when \(t=0\); Eq. \(\PageIndex{16}\) then becomes at \(t=0\)

\[0=\frac{m g}{b}+C^{\prime}\]

and so

\[C^{\prime}=-\frac{m g}{b}\]

Therefore, from Eq. \(\PageIndex{16}\), the solution is

\[v=\frac{m g}{b}-\frac{m g}{b} e^{-b t / m},\]

or

\[v=\frac{m g}{b}\left(1-e^{-b t / m}\right)\]

This is the solution we're after: it gives the falling object's velocity \(v\) at any time \(t\), assuming that it's dropped from rest at time \(t=0\).

Now let's examine what happens to the solution (Eq. \(\PageIndex{20}\)) as \(t \rightarrow \infty\). In this case, the exponential term approaches zero, and the falling object’s velocity approaches the limiting value

\[v_{\infty}=\frac{m g}{b}\]

This is called the terminal velocity, and is a general feature of bodies falling through resistive fluids: at some point the resistive force balances the downward gravitational force, and the body stops accelerating and moves at a constant velocity. \({ }^{2}\) Sky divers experience this phenomenon: some time after jumping out of an airplane, a sky diver will reach a terminal velocity of roughly 100 miles per hour, and will not change speed further until the parachute is deployed.

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\({ }^{2}\) A simpler way to arrive at Eq. \(\PageIndex{21}\) is to simply set the acceleration \(d v / d t=0\) in Eq. (19.5), which immediately gives \(v_{\infty}=\) \(m g / b\).