AP Physics C: Mechanics · Unit 2 · Topic 5 of 7 · 6 min lesson

Resistive Forces

A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.

Quiz this topic · 5 questions

The big idea

A drag force grows with speed. For linear drag, F = −bv. A falling object speeds up until drag balances gravity at the terminal speed v_T = mg/b. It approaches v_T exponentially: v = v_T(1 − e^−t/τ), with τ = m/b.

1. Falling with drag

The acceleration starts at g (no drag at v = 0) and shrinks toward zero as the speed approaches v_T.

m dv/dt = mg − bvfalling, down = +

v_T = mg/bnet force = 0

v(t) = v_T(1 − e^−t/τ)τ = m/b

Coasting: v = v₀e^−t/τdrag only

Quadratic drag cv²v_T = √(mg/c)

2. Solving the differential equation

Separate the variables: dv/(g − bv/m) = dt. Integrate from 0 to v and from 0 to t:

−(m/b) ln(1 − v/v_T) = t, which gives v = v_T(1 − e^−bt/m).

After τ: 63% of v_T. After 3τ: 95%. The object never quite reaches v_T, but gets as close as you like.

Heavier objects (same shape) fall faster: v_T ∝ m for linear drag, ∝ √m for quadratic drag.

3. Example, then try it

Example

A 0.20 kg ball falls with linear drag, b = 0.50 kg/s. Find its terminal speed and time constant.

v_T = mg/b = (0.20)(10) ÷ 0.50 = 4.0 m/s.

τ = m/b = 0.20 ÷ 0.50 = 0.40 s.

Try it · work it on paper first

For that ball, dropped from rest, find its speed at t = 0.40 s and t = 1.2 s, and its acceleration at the moment its speed is 2.0 m/s.

At t = τ = 0.40 s: v = 4.0(1 − e⁻¹) ≈ 4.0(0.63) ≈ 2.5 m/s.

At t = 3τ = 1.2 s: v = 4.0(1 − e⁻³) ≈ 4.0(0.95) ≈ 3.8 m/s.

At v = 2.0 m/s: a = (mg − bv)/m = (2.0 − 1.0) ÷ 0.20 = 5.0 m/s², half of g.

Answer: 2.5 m/s, 3.8 m/s; 5.0 m/s² at 2.0 m/s.

4. Common mistakes

“The acceleration stays g.”Drag grows with speed, so the acceleration shrinks to zero.

“It reaches v_T after τ.”Only 63%; it approaches v_T exponentially.

Mixing linear and quadratic drag.bv gives v_T = mg/b; cv² gives v_T = √(mg/c).

Check yourself

Two balls with the same size and shape fall with linear drag F = bv. One has twice the mass. How do their terminal speeds compare?

v_T = mg/b. Same shape means the same b, so doubling m doubles v_T.

Ready to practice?

5 questions on Resistive Forces, each with a hint and an explanation for every wrong answer.

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