This AP Physics C lesson on terminal velocity notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on terminal velocity and check yourself with the quiz.
Quick answer: Terminal velocity is the steady speed at which a resistive force balances the weight, so the net force is zero. For linear drag it is mg/b and the speed follows v = vt(1 − e−t/τ); for quadratic drag it is √(mg/c) and v = vt tanh(gt/vt).
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 2
Terminal velocity notes: the key ideas
What changes when a force depends on velocity?
In earlier mechanics most forces were constant or depended on position. A resistive force is different: it depends on how fast the object moves, so the acceleration keeps changing during the motion. Newton’s second law then becomes a differential equation, m dv/dt = F(v), and calculus is the natural tool to solve it.
A resistive (drag) force is a force exerted by a fluid that opposes the velocity of an object moving through it. Its direction is always opposite to v, and its magnitude grows with speed.
Two models cover the AP Physics C: Mechanics course. At low speed in a viscous fluid (a ball bearing in syrup) the force is linear: FD = bv. At higher speed in air (a skydiver, a cricket ball) it is closer to quadratic: FD = cv².
How do you solve motion with linear drag?
Take down as positive for a falling object released from rest. The only forces are weight and drag, so m dv/dt = mg − bv. Separate the variables and integrate from v = 0 at t = 0:
∫ dv′ / (mg − bv′) = ∫ dt / m, so −(1/b) ln[(mg − bv)/mg] = t/m.
Solving for v gives an exponential approach to a limiting speed.
v(t) = vt(1 − e−t/τ), with vt = mg/b and τ = m/b. Here m is mass (kg), b the drag coefficient (kg/s), vt the terminal speed (m/s) and τ the time constant (s).
Terminal speed does not need the full solution. When the speed stops changing, dv/dt = 0, so mg − bvt = 0. Differentiating the solution gives a = g e−t/τ, so the acceleration starts at g and decays to zero.
Ananya in Pune drops a steel ball of mass 0.050 kg into a tall jar of syrup, with b = 0.40 kg/s. Find the terminal speed and the speed after 0.25 s. vt = mg/b = (0.050 × 9.81)/0.40 = 1.23 m/s. τ = m/b = 0.050/0.40 = 0.125 s, so t = 0.25 s is 2τ. v = 1.226 × (1 − e−2) = 1.226 × 0.8647 = 1.06 m/s.
How does quadratic drag differ?
With FD = cv² the equation is m dv/dt = mg − cv². Setting dv/dt = 0 gives vt = √(mg/c). Integrating ∫ dv′/(1 − v′²/vt²) = gt gives an inverse hyperbolic tangent, so v = vt tanh(gt/vt). The graph below compares both models when they share the same vt.
| Feature | Linear drag, F = bv | Quadratic drag, F = cv² |
|---|---|---|
| Typical situation | Small, slow object in a viscous liquid | Larger, faster object in air |
| Terminal speed | mg/b (proportional to m) | √(mg/c) (proportional to √m) |
| Speed against time | vt(1 − e−t/τ) | vt tanh(gt/vt) |
| Units of coefficient | b in kg/s | c in kg/m |
Kofi in Accra models a skydiver of mass 80 kg with c = 0.20 kg/m. Find vt and the speed after 5.0 s. vt = √(mg/c) = √(784.8/0.20) = √3924 = 62.6 m/s. Then gt/vt = (9.81 × 5.0)/62.6 = 0.783 and tanh(0.783) = 0.655, so v = 62.6 × 0.655 = 41 m/s.
Terminal velocity does not mean there is no force. It means the net force is zero: drag has grown until it equals the weight. The speed is constant, but the object is still moving fast, and drag and weight are both large.
To get a terminal speed, skip the integration: set dv/dt = 0 and solve the algebra. Use the full integral only when the question asks for speed at a given time.
Which signs and units should you check?
Choose one positive direction and keep it. If down is positive, drag is subtracted. If you launch an object upward, drag points down and adds to gravity. Check units: bv must be in newtons, so b is in kg/s, and cv² in newtons means c is in kg/m.
- Linear drag: m dv/dt = mg − bv, vt = mg/b, τ = m/b
- Quadratic drag: m dv/dt = mg − cv², vt = √(mg/c)
- At t = τ the speed is 63 % of vt; the acceleration is g e−t/τ
- Units: b in kg/s, c in kg/m
- At terminal speed dv/dt = 0, so drag equals weight
Drag forces and terminal velocity with calculus: frequently asked questions
What is terminal velocity?
It is the constant speed reached when the drag force equals the weight, so the net force and acceleration are zero. For linear drag it equals mg/b; for quadratic drag it equals √(mg/c).
Why is the approach to terminal speed exponential for linear drag?
Drag grows in proportion to speed, so the net force falls in proportion to the remaining gap vt − v. A quantity whose rate of change is proportional to itself decays exponentially, with time constant τ = m/b.
How do you solve m dv/dt = mg − bv?
Separate the variables to get dv/(mg − bv) = dt/m, integrate both sides, then rearrange. The result is v = (mg/b)(1 − e−bt/m).
When is drag quadratic rather than linear?
Quadratic drag applies to larger objects moving fast through air, where the fluid must be pushed aside. Linear drag applies to small objects moving slowly through a viscous liquid.
What does the time constant τ mean?
It is the time for the speed to reach about 63 % of terminal speed, and after about 5τ the speed is within 1 % of it. A larger mass or smaller b gives a longer τ.
