A Skydiver's Terminal Velocity

Gravity pulls a skydiver down while air drag pushes back harder the faster they fall. The equation never says what the speed is, only how fast it changes. Draw that rate as tiny slopes everywhere and every possible fall appears at once.

Click or drag anywhere to start a new solution there.

How it works

Differential equations and direction fields

In one sentence

A differential equation relates an unknown function to its derivatives; $v' = g - (c/m)v$ is a first-order linear equation whose solutions all approach the equilibrium $v_T = mg/c$ .

General formula

{\htmlData{sym=0}{m}}{\htmlData{sym=4}{\frac{dv}{dt}}} = {\htmlData{sym=0}{m}}{\htmlData{sym=2}{g}} - {\htmlData{sym=3}{c}}{\htmlData{sym=1}{v}} \quad\Rightarrow\quad {\htmlData{sym=4}{\frac{dv}{dt}}} = {\htmlData{sym=2}{g}} - \frac{\htmlData{sym=3}{{c}}}{\htmlData{sym=0}{{m}}}\,{\htmlData{sym=1}{v}}

The right side is the acceleration: gravity's $g$ minus drag's $(c/m)v$ . Where the two cancel, the slope marks lie flat.

What the symbols mean

$m$
Mass
The skydiver's mass in kilograms.
$v$
Velocity
Falling speed in meters per second: the unknown function of time.
$g$
Gravity
9.8 m/s², the acceleration gravity alone would give.
$c$
Drag coefficient
How strongly the air resists; bigger for a spread-out body or an open parachute.
$\frac{dv}{dt}$
Acceleration
How fast the speed is changing right now.

Two forces act: gravity $mg$ pulls down, and air drag $cv$ pushes up, growing with speed.
Newton's second law turns that into an equation for the unknown function $v(t)$ : its derivative depends on $v$ itself.
At the start drag is zero, so $v$ rises at $g = 9.8$ m/s². As $v$ grows the slope flattens.
Where $g - \frac{c}{m}v = 0$ the speed stops changing: the terminal velocity $v_T = \frac{mg}{c}$ .

Learn the math

Differential Equations: DefinitionsDiff Eq · Lesson 1, Basic Concepts Direction FieldsDiff Eq · Lesson 3, Basic Concepts