---
title: "Terminal Speed — AP Physics C Mechanics Definition"
description: "Terminal speed is the constant speed where drag equals gravity and acceleration hits zero. Learn how to derive v_T from Newton's second law for the AP exam."
canonical: "https://fiveable.me/ap-physics-c-mechanics/key-terms/terminal-speed"
type: "key-term"
subject: "AP Physics C: Mechanics"
unit: "Unit 2"
---

# Terminal Speed — AP Physics C Mechanics Definition

## Definition

Terminal speed is the constant speed a falling object reaches when the resistive (drag) force grows to equal the gravitational force, making the net force and acceleration zero; for linear drag F = -bv it equals mg/b, and for quadratic drag F = -bv² it equals √(mg/b).

## What It Is

Terminal speed is what happens when drag finally catches up to gravity. Drop an object through air or oil and the [resistive force](/ap-physics-c-mechanics/unit-2/9-resistive-forces/study-guide/pXbIz3a4RtJYP8Gq "fv-autolink") depends on speed, so it starts at zero and grows as the object speeds up. Gravity stays constant at mg. The object keeps accelerating, just less and less, until the drag force exactly balances the [weight](/ap-physics-c-mechanics/key-terms/weight "fv-autolink"). At that point the net force is zero, the acceleration is zero, and the speed stops changing. That constant speed is the terminal speed, v_T.

The fastest way to find it is to skip the calculus entirely. Set the magnitudes equal. For linear drag (F = bv), set bv_T = mg and get v_T = mg/b. For quadratic drag (F = bv²), set bv_T² = mg and get v_T = √(mg/b). One subtlety the exam loves to test: an object dropped from rest *approaches* terminal speed asymptotically but never quite reaches it. After one time constant τ = m/b (linear drag), the object is moving at v_T(1 - e⁻¹), about 63% of terminal speed, not 100%.

## Why It Matters

Terminal speed lives in **Topic 2.9 Resistive Forces** in [Unit 2](/ap-physics-c-mechanics/unit-2 "fv-autolink") (Force and Translational Dynamics), and it's the payoff concept for the whole topic. Resistive forces are where [AP Physics C](/ap-physics-c-mechanics "fv-autolink") stops being plug-and-chug. Because drag depends on velocity, Newton's second law becomes a differential equation, m(dv/dt) = mg - bv, and terminal speed is the equilibrium solution of that equation. The exam uses it to check two skills at once. First, can you reason qualitatively about a velocity-dependent force (drag grows, net force shrinks, acceleration goes to zero)? Second, can you handle the math, either by setting the net force to zero for v_T or by separating variables to get v(t) = v_T(1 - e^(-bt/m))? It also shows up in graph analysis. A velocity-time graph for a falling object with drag is concave down and levels off at a horizontal asymptote, and that asymptote is v_T.

## Connections

### [Drag coefficient (Unit 2)](/ap-physics-c-mechanics/key-terms/drag-coefficient)

The [drag coefficient](/ap-physics-c-mechanics/key-terms/drag-coefficient "fv-autolink") b sets how strong the resistive force is at a given speed, so it directly controls terminal speed. Bigger b means more drag at every speed, which means the balance point with gravity happens sooner, at a lower v_T. In both formulas, v_T = mg/b and v_T = √(mg/b), the coefficient sits in the denominator.

### Newton's second law as a differential equation (Unit 2)

Terminal speed is the steady-state solution of m(dv/dt) = mg - bv. When you separate variables and integrate, you get exponential approach toward v_T with time constant τ = m/b. This is the same math pattern you'll see again with RC circuits in E&M, which is exactly why Physics C drills it here.

### Velocity-time graphs and kinematics (Unit 1)

With drag, the v-t graph isn't a straight line like constant-acceleration [kinematics](/ap-physics-c-mechanics/unit-1 "fv-autolink"). It curves, with slope starting at g and bending toward zero as the speed flattens at the asymptote v_T. If a question hands you a graph that levels off, the horizontal asymptote is the terminal speed.

### [Translational equilibrium (Unit 2)](/ap-physics-c-mechanics/key-terms/translational-equilibrium)

At terminal speed the object is in equilibrium even though it's moving fast. Zero [net force](/ap-physics-c-mechanics/key-terms/net-force "fv-autolink") means zero acceleration, not zero velocity. Terminal speed is the cleanest example of a moving object in equilibrium, the same idea behind an object sliding at constant velocity with friction.

## On the AP Exam

Terminal speed shows up in both MCQs and FRQs, and the questions almost always test one of three moves. First, derive v_T by setting drag equal to gravity. Second, compare terminal speeds when mass changes. With linear drag (F = -bv), tripling the mass triples v_T because v_T = mg/b. With quadratic drag (F = -bv²), tripling the mass only multiplies v_T by √3 because of the square root. Practice questions hit this scaling comparison constantly, so check which drag model the problem gives you before you scale anything. Third, connect v_T to the time-dependent solution. A classic stem asks for the speed at t = τ = m/b, and the answer is v_T(1 - e⁻¹), roughly 0.63v_T, not v_T itself. On the FRQ side, the 2024 exam (Q2) gave a sphere dropped from rest with drag modeled as F_drag = bv and asked you to work with that model, the kind of question where identifying v_T = mg/b and sketching the asymptotic v-t graph earns points. Expect to justify your answer with Newton's second law, not just quote the formula.

## terminal speed vs Terminal speed with linear drag vs. quadratic drag

Both models give a terminal speed, but the formulas and scaling are different, and mixing them up is the most common point-loser. Linear drag (F = -bv) gives v_T = mg/b, so terminal speed scales directly with mass. Quadratic drag (F = -bv²) gives v_T = √(mg/b), so terminal speed scales with the square root of mass. A 3m skydiver with the same drag coefficient falls at √3 times v_T under quadratic drag, but 3 times v_T under linear drag. Always read the force model in the problem statement before computing.

## Key Takeaways

- Terminal speed is the constant speed where the drag force equals the gravitational force, so the net force and acceleration are both zero.
- Find v_T by setting magnitudes equal, which gives v_T = mg/b for linear drag (F = -bv) and v_T = √(mg/b) for quadratic drag (F = -bv²).
- Mass scaling depends on the drag model. Tripling the mass triples v_T under linear drag but only multiplies it by √3 under quadratic drag.
- An object dropped from rest approaches terminal speed asymptotically and never exactly reaches it. At t = τ = m/b it's moving at v_T(1 - e⁻¹), about 63% of v_T.
- On a velocity-time graph, terminal speed is the horizontal asymptote, and the initial slope of the curve is g.
- An object at terminal speed is in equilibrium while moving, which is a direct application of Newton's second law with zero net force.

## FAQs

### What is terminal speed in AP Physics C?

Terminal speed is the constant [speed](/ap-physics-c-mechanics/unit-1/1-scalars-and-vectors/study-guide/rVQeOgdT8itcgCoV "fv-autolink") a falling object reaches when the velocity-dependent drag force grows large enough to equal the gravitational force mg. With zero net force, acceleration is zero and speed stays constant. For linear drag it's v_T = mg/b; for quadratic drag it's v_T = √(mg/b).

### Does an object actually reach terminal speed?

Mathematically, no. The solution v(t) = v_T(1 - e^(-bt/m)) approaches v_T asymptotically but never equals it exactly. In practice the object gets extremely close after a few time constants, but on the exam, the speed at t = τ = m/b is v_T(1 - e⁻¹) ≈ 0.63v_T, not v_T.

### How is terminal speed different for F = -bv versus F = -bv² drag?

Linear drag gives v_T = mg/b, so v_T is proportional to mass. Quadratic drag gives v_T = √(mg/b), so v_T is proportional to the square root of mass. A skydiver system with triple the mass and the same b falls at 3v_T under linear drag but only √3·v_T under quadratic drag.

### Is acceleration zero at terminal speed?

Yes. At terminal speed the drag force exactly cancels gravity, so the net force is zero and Newton's second law gives zero acceleration. The object is in equilibrium even though it's moving, which is a classic MCQ trap (zero acceleration does not mean zero velocity).

### How does terminal speed show up on the AP Physics C exam?

It appears in MCQs asking you to derive or scale v_T and in FRQs built around the differential equation m(dv/dt) = mg - bv. The 2024 FRQ Q2 used exactly this setup, a sphere dropped from rest with drag modeled as F_drag = bv. Expect to set net force to zero, separate variables, or sketch the asymptotic v-t graph.

## Related Study Guides

- [2.9 Resistive Forces](/ap-physics-c-mechanics/unit-2/9-resistive-forces/study-guide/pXbIz3a4RtJYP8Gq)

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