AP Physics C: Mechanics · Unit 2: Force and Translational Dynamics · Lesson 2.9

Deep Dive: Resistive Forces

🔬 Deep Dive

This is your textbook for this topic. Take your time. Read it more than once.

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What Is a Resistive Force?

A resistive force is a velocity-dependent force that opposes an object's velocity — the faster the object moves, the harder the resistive force pushes back. The simplest model is direct proportionality:

F(r) = −kv

🔑This is fundamentally different from every other force in this unit. Gravity, spring force, even friction — none of them depend on how fast something is moving. A resistive force does, and that single fact changes everything about how you solve for motion.

Apply Newton's second law to an object under a resistive force, and something new happens: acceleration is dv/dt, but the force itself contains v. You get a genuine differential equation — an equation relating a function to its own derivative:

m(dv/dt) = −kv

⚠️You cannot just integrate both sides directly here — v is a function of t, and it appears on the side you'd normally integrate with respect to t. This is exactly the situation that calls for separation of variables.

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Solving for Velocity and Position

Separation of variables means getting every v term on one side and every t term on the other, so each side can be integrated independently.

dv/v = −(k/m) dt

Integrating both sides — the left from v₀ to v, the right from 0 to t — and solving for v gives:

v(t) = v₀ e^(−kt/m)

🔑This is an exponential decay — velocity falls off quickly at first, then more and more slowly, approaching (but mathematically never quite reaching) zero. The quantity m/k sets the timescale: a larger m/k means a slower decay.

Once you have v(t), you can integrate again — using the initial position as another condition — to find x(t). For the simple case above, the result is just as surprising as the velocity function:

x(t) = x₀ + (v₀m/k)(1 − e^(−kt/m))

Explore both functions together below.

An object moving through a resistive medium with no other force acting on it — like a boat drifting after its engine cuts out. Watch velocity decay exponentially, and notice the surprising result for position.

v₀ (m/s)20

k/m0.50

v(t) = v₀e−(k/m)tx(t) = (v₀/(k/m))(1 − e−(k/m)t)Position limit as t→∞: 40.0 m

The object's velocity mathematically never reaches exactly zero, but its position approaches a fixed limiting value — it travels a finite total distance in infinite time.

ExampleGuided Example — Separating Variables

Starting from m(dv/dt) = −kv, walk through the full separation-of-variables process to arrive at v(t) = v₀e^(−kt/m).

Step 1 — Rewrite dv/dt as separate differentials

m(dv/dt) = −kv → m dv = −kv dt

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Terminal Velocity

Now add a constant force back into the picture — gravity pulling an object down, while a resistive force pushes back up against its increasing speed. Early on, gravity wins and the object accelerates. But as speed builds, the resistive force grows too, eating away at the net force. Eventually the two forces balance exactly, acceleration hits zero, and the object continues at a constant terminal velocity — the maximum speed it will ever reach.

🔑You don't need to solve the full differential equation just to find terminal velocity — set acceleration equal to zero directly in Newton's second law:

mg − kv(term) = 0  ⟹  v(term) = mg/k

Solving the full differential equation (mg − kv = m dv/dt, starting from rest) gives the complete velocity function, which confirms this same limiting value:

v(t) = (mg/k)(1 − e^(−kt/m))

Explore this directly below — adjust mass and the resistive constant, and watch velocity climb asymptotically toward its terminal value.

An object falling from rest through a resistive medium — gravity pulling down, resistance pushing back. Adjust mass and the resistive constant k, and watch velocity climb toward (but never quite touch) terminal velocity.

mass (kg)2.0

k (N·s/m)4.0

v(term) = mg/k = (2.0)(10)/4.0 = 5.00 m/s

Notice v(term) depends only on m, g, and k — not on how the object started. Every object with these properties settles toward the same terminal speed no matter its initial velocity.

ExampleWorked Example — Finding Terminal Velocity Two Ways

A 3 kg object falls through a fluid with resistive constant k = 6 N·s/m. Find its terminal velocity using the shortcut method, and confirm using the full velocity function at a large value of t.

← Back to Lesson 2.9Ready for 2.10? Circular Motion closes out Unit 2 — force laws applied to motion along a curve.