Motion with Drag and Terminal Velocity, differential equations lesson

A small ball dropped into oil, or a skydiver before the parachute opens, speeds up at first and then falls at an almost constant speed. Why does the speed level off, and at what value? Newton’s second law with a drag force gives a first-order equation for the velocity, and the limiting speed is an equilibrium of that equation. This lesson fixes the signs of the forces, solves the linear and quadratic drag models, and separates the approach to the limiting speed from the speed itself. Throughout, $t$ is time in seconds and a prime means $d/dt$ .

Key terms

Sign convention for forces

Definition Choose one positive direction along the line of motion, here downward. Let $x(t)$ be the position measured in that direction and $v = x'$ the velocity, so $v > 0$ means moving down and $v < 0$ means moving up. Every force is a signed number, positive when it points down. For an object of constant mass $m$ under a constant gravitational acceleration $g > 0$ , Newton’s second law reads

\begin{gathered} mv' = mg + F_{\text{drag}}(v), \\[0.4em] x' = v, \end{gathered}

where the drag $F_{\text{drag}}(v)$ always points against the motion. It has the sign opposite to $v$ and is zero when $v = 0$ . The model assumes a medium at rest and leaves out buoyancy and every other force.

In plain words Pick a direction once and give every velocity and every force a sign relative to it. Gravity keeps the same sign throughout, while drag changes sign whenever the motion changes direction.

Example With downward positive, a ball moving up at 3 meters per second has $v = -3$ . The drag on it points down, so $F_{\text{drag}} > 0$ adds to gravity, and the ball slows more quickly than under gravity alone. Falling at 3 meters per second, the same ball has $v = 3$ and a negative drag force. Choosing upward as positive instead gives $mv' = -mg + F_{\text{drag}}(v)$ , with drag again opposite to $v$ . It describes the same motion, with the velocity and every force changing sign.

Linear drag

Definition In the linear drag model $F_{\text{drag}} = -bv$ with a constant $b > 0$ , measured in kilograms per second so that $bv$ is a force in newtons. The minus sign makes the drag oppose the motion for both signs of $v$ . Newton’s law becomes the linear first-order equation

v' + \frac{b}{m}\,v = g,

and its solution with $v(0) = v_0$ is defined for all $t$ ,

\begin{gathered} v(t) = v_T + (v_0 - v_T)\,e^{-bt/m}, \\[0.4em] v_T = \frac{mg}{b}. \end{gathered}

It follows because the difference $v - v_T$ satisfies $(v - v_T)' = -(b/m)(v - v_T)$ . The model suits small, slow objects in a viscous fluid, where drag is close to proportional to speed.

In plain words Drag here is proportional to the velocity and points the other way. The equation is linear, and the gap between the velocity and $mg/b$ shrinks exponentially.

Example Take $m = 2$ kilograms, $b = 4$ kilograms per second and, for round numbers, $g = 10$ meters per second squared, with the object released from rest at $x(0) = 0$ . Then $v' = 10 - 2v$ , the terminal velocity is $mg/b = 5$ meters per second, and

v(t) = 5\bigl(1 - e^{-2t}\bigr).

As a check, $v' = 10e^{-2t}$ and $10 - 2v = 10e^{-2t}$ , and $v(0) = 0$ . Integrating the velocity from 0 to $t$ gives the position

x(t) = 5t + \tfrac{5}{2}\bigl(e^{-2t} - 1\bigr),

where the term $-\tfrac{5}{2}$ comes from the lower limit. After 1 second the velocity is about $4.32$ meters per second and the object has fallen about $2.84$ meters. The velocity levels off at 5, while the position keeps growing.

Quadratic drag

Definition In the quadratic drag model

F_{\text{drag}} = -c\,\lvert v \rvert\,v

with a constant $c > 0$ , measured in kilograms per meter so that $cv^2$ is a force in newtons. The absolute value keeps the force opposite to the motion in both directions. For $v \ge 0$ , moving down, the force is $-cv^2$ and

v' = g - \frac{c}{m}\,v^2,

while for $v < 0$ , moving up, the force is $+cv^2$ and $v' = g + (c/m)v^2$ . Each branch is a nonlinear separable equation. The full right side $g - (c/m)\lvert v \rvert v$ is continuously differentiable in $v$ , so the velocity of a ball thrown upward is unique and passes from one branch to the other at $v = 0$ . Using $-cv^2$ in both directions would make the drag push an upward moving object farther up. The model suits larger objects moving fast through air, where drag grows roughly like the square of the speed.

In plain words Drag grows with the square of the speed, but it must still point against the motion. The absolute value takes care of that, and on any stretch of motion in one direction the sign can be written out.

Example Keep $m = 2$ and $g = 10$ , take $c = \tfrac{4}{5}$ kilogram per meter, and release the object from rest. It starts moving down, so the downward branch applies and $v' = 10 - \tfrac{2}{5}v^2$ . With $w = v/5$ this becomes $w' = 2(1 - w^2)$ with $w(0) = 0$ . Separating on $0 \le w < 1$ gives

\frac{1}{2}\ln\frac{1 + w}{1 - w} = 2t,

so $(1 + w)/(1 - w) = e^{4t}$ and

\begin{aligned} v(t) &= 5\,\frac{e^{4t} - 1}{e^{4t} + 1} \\[0.4em] &= 5\tanh(2t). \end{aligned}

Since $(\tanh u)' = 1 - \tanh^2 u$ , the derivative is $v' = 10 - 10\tanh^2(2t) = 10 - \tfrac{2}{5}v^2$ , and $v(0) = 0$ . The solution satisfies $0 \le v < 5$ for all $t \ge 0$ , which confirms the branch and shows that the division by $1 - w^2$ was safe.

Terminal velocity

Definition A terminal velocity is a constant velocity $v_T$ at which the net force is zero,

mg + F_{\text{drag}}(v_T) = 0.

It is an equilibrium solution of the velocity equation, not an equilibrium of the position, which keeps changing at the rate $v_T$ . With downward positive, linear drag gives $v_T = mg/b$ and quadratic drag gives $v_T = \sqrt{mg/c}$ , both positive. In both models the net force is positive below $v_T$ and negative above it, so on the phase line $v_T$ is asymptotically stable and every solution approaches it as $t \to \infty$ . A solution that does not start at $v_T$ never reaches it in finite time, because two solutions of the velocity equation cannot meet.

In plain words Terminal velocity is the speed at which drag exactly cancels gravity. Both forces still act, but they balance, so the velocity stops changing while the object keeps moving.

Example The two examples above share the terminal velocity $mg/b = 20/4 = 5$ and $\sqrt{mg/c} = \sqrt{20/0.8} = 5$ meters per second. A start above it also leads there. An object thrown downward at 8 meters per second under the same linear drag has $v = 5 + 3e^{-2t}$ , which decreases toward 5 because the drag then outweighs gravity. Indeed $v' = -6e^{-2t}$ and $10 - 2v = -6e^{-2t}$ .

Transient velocity

Definition The transient velocity is the difference $v(t) - v_T$ between the velocity and its terminal value, the part of the motion that dies out. For linear drag it is exactly

v - v_T = (v_0 - v_T)\,e^{-bt/m},

an exponential decay with time constant $\tau = m/b$ , so each time constant shrinks it by the factor $e^{-1} \approx 0.37$ . For quadratic drag starting from rest the velocity is $v = v_T\tanh(gt/v_T)$ , and the transient is

v - v_T = -\frac{2v_T}{e^{2gt/v_T} + 1},

which for large $t$ is close to $-2v_T e^{-2gt/v_T}$ . In both models the transient tends to zero, and the drag law sets how fast.

In plain words The transient measures how far the velocity still is from its terminal value. It shrinks with time, and the drag law decides how quickly.

Example From rest in the two examples, linear drag gives the transient $-5e^{-2t}$ and quadratic drag gives $-10/(e^{4t} + 1)$ . Both start at $-5$ with the same initial acceleration $g = 10$ , since drag is zero at rest. The quadratic transient is smaller in size at every $t > 0$ , because below 5 meters per second the quadratic drag $0.8v^2$ is weaker than the linear drag $4v$ . The linear model reaches 4 meters per second when $e^{-2t} = \tfrac{1}{5}$ , at $t = \tfrac{1}{2}\ln 5 \approx 0.80$ seconds, and the quadratic model when $\tanh(2t) = \tfrac{4}{5}$ , that is $e^{4t} = 9$ , at $t = \tfrac{1}{2}\ln 3 \approx 0.55$ seconds. After 1 second the two transients are about $-0.68$ and $-0.18$ meters per second.

Check yourself

An object of mass 4 kilograms falls from rest with linear drag coefficient $b = 2$ kilograms per second. Using $g = 10$ meters per second squared, find its terminal velocity, its time constant and the time at which it reaches 90 percent of the terminal velocity.

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The terminal velocity is $v_T = mg/b = 40/2 = 20$ meters per second and the time constant is $\tau = m/b = 2$ seconds. From rest

v(t) = 20\bigl(1 - e^{-t/2}\bigr).

Ninety percent of $v_T$ means $e^{-t/2} = 0.1$ , so $t = 2\ln 10 \approx 4.61$ seconds.

A ball of mass 2 kilograms is thrown straight up at 6 meters per second through air with quadratic drag coefficient $c = \tfrac{4}{5}$ kilogram per meter. With downward positive and $g = 10$ , a student uses $v' = g - (c/m)v^2$ from the moment of the throw. What is the correct acceleration just after the throw, and what does the student’s equation predict instead?

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Just after the throw the velocity is $v = -6$ . The ball moves up, so the drag points down, and the signed drag force is

\begin{aligned} -c\,\lvert v \rvert\,v &= -\tfrac{4}{5}(6)(-6) \\[0.4em] &= 28.8 \end{aligned}

newtons. The correct equation on this branch is $v' = g + (c/m)v^2$ , which gives $v' = 10 + 0.4 \cdot 36 = 24.4$ meters per second squared, downward, so the upward speed drops quickly. The student’s equation gives $10 - 14.4 = -4.4$ , an upward acceleration that would make the ball rise faster, as if the drag pushed it along. The form $-cv^2$ is correct only on the branch $v \ge 0$ , after the ball has reached its highest point and started to fall.

With the quadratic drag example values $m = 2$ , $g = 10$ and $c = \tfrac{4}{5}$ , an object is thrown downward at 10 meters per second. Find $v(t)$ and check that the branch used stays valid.

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The motion is downward, so the branch $v \ge 0$ applies. With $w = v/5$ the equation is again $w' = 2(1 - w^2)$ , now with $w(0) = 2$ . On $w > 1$ separation gives

\begin{aligned} \frac{1}{2}\ln\frac{w + 1}{w - 1} &= 2t + K, \\[0.4em] K &= \tfrac{1}{2}\ln 3, \end{aligned}

so $(w + 1)/(w - 1) = 3e^{4t}$ and

v(t) = 5\,\frac{3e^{4t} + 1}{3e^{4t} - 1}.

As a check, $w = 1 + 2/(3e^{4t} - 1)$ gives $w' = -24e^{4t}/(3e^{4t} - 1)^2$ , which equals $2(1 - w^2)$ , and $v(0) = 5 \cdot 4/2 = 10$ . For $t \ge 0$ the denominator is at least 2, so $v > 5 > 0$ throughout. The branch $v \ge 0$ holds, and the velocity decreases toward the terminal velocity of 5 meters per second from above.

Motion with Drag and Terminal Velocity

Key terms

Sign convention for forces

Linear drag

Quadratic drag

Terminal velocity

Transient velocity

Check yourself

How this lesson connects

Builds on