Drag Forces & Terminal Velocity

You’ll be able to

Write Newton’s second law as a differential equation when a resistive force depends on speed
Determine terminal velocity from the condition that net force and dv/dt vanish
Describe qualitatively how velocity approaches its terminal value over time

Resistive forces grow with speed

Unlike friction on a surface, a drag force from a fluid gets stronger the faster you move. Two models appear on the AP exam: a linear law f = bv (good for slow motion or thick fluids) and a quadratic law f = cv² (fast motion through air). Because the force depends on the very quantity v you are solving for, ΣF = ma is no longer algebra — it becomes a differential equation for v(t).

Falling object with linear drag

m dv/dt = mg − bv

Down is positive. Weight mg pulls down; drag bv pushes up and grows as v grows. This first-order differential equation governs the whole descent.

Terminal velocity is where the acceleration dies

Start from rest and the drag is zero, so the object accelerates at g. As v climbs, the drag bv climbs with it, shrinking the net force and thus the acceleration. Eventually drag exactly balances weight: the net force is zero, dv/dt = 0, and the speed stops changing. That steady speed is the terminal velocity v_t. Setting dv/dt = 0 in the equation of motion, mg − bv_t = 0, so v_t = mg/b.

Terminal velocity

v_t = mg / b (linear drag) v_t = √(mg / c) (quadratic drag)

Found by setting dv/dt = 0 so the resistive force equals the weight. No calculus needed for v_t itself — only the balance condition.

Worked example

A 2 kg object falls from rest through a fluid that exerts a linear drag f = bv with b = 4 N·s/m. Using g = 10 m/s², find (a) the terminal velocity and (b) the acceleration at the instant its speed is 2.5 m/s.

1.Write the equation of motion (down positive): m dv/dt = mg − bv.
2.Terminal velocity is where dv/dt = 0: mg = b v_t → v_t = mg/b = (2 × 10)/4 = 5 m/s.
3.For the acceleration at v = 2.5 m/s, solve the equation for dv/dt: dv/dt = g − (b/m)v.
4.Substitute: dv/dt = 10 − (4/2)(2.5) = 10 − 5 = 5 m/s². It is already down to half of g, and it will fall to zero as v approaches 5 m/s.

Answer: v_t = mg/b = 5 m/s; at v = 2.5 m/s the acceleration is g − (b/m)v = 5 m/s²

Checkpoint

An object falls through air and reaches terminal velocity, where it experiences a drag force f = bv opposing its motion. Which statement is then true?

Checkpoint

A 3 kg object falls under gravity with a linear drag force f = bv, where b = 6 N·s/m. Using g = 10 m/s², what is its terminal speed?

Watch out

Do not treat drag problems with the constant-acceleration kinematic equations — the acceleration is not constant, it decreases as speed builds. Start from ΣF = m dv/dt. For terminal velocity you only need the balance dv/dt = 0; for the full v(t) you would separate variables and integrate.

Answer the 2 checkpoints as you read.