Topic 2.9 Resistive Forces: model a velocity-dependent resistive force, set up and solve the equation of motion for fall with drag, and determine the terminal velocity and the exponential approach to it.

What this topic is asking

The College Board (Topic 2.9) wants you to model a velocity-dependent resistive force such as air resistance, to write Newton's second law as a differential equation, and to find the terminal velocity and the way the motion approaches it. This topic is distinctive to AP Physics C, the calculus-based course, and it is the clearest place where you set up and solve a differential equation of motion rather than use a constant-acceleration formula.

Velocity-dependent resistance

Unlike kinetic friction, which has a fixed magnitude, a resistive force depends on the speed: the faster the object moves, the harder the fluid pushes back. This velocity dependence is what makes the motion interesting, and it is why constant-acceleration kinematics cannot be used. The exam usually specifies linear drag, $F = -bv$, because it leads to a differential equation you can solve in closed form.

The equation of motion

For an object of mass $m$ falling through air with linear drag, take down as positive and apply Newton's second law. The two forces are gravity $mg$ (down, positive) and drag $bv$ (up, opposing the downward motion, negative):

This is a first-order differential equation for $v(t)$, not an algebraic equation, because the acceleration $dv/dt$ depends on the velocity. Setting it up correctly, with the right sign for the drag relative to the motion, is the first scored step on the exam. The same template, "write $\sum F = m\,dv/dt$ with a velocity-dependent term," covers a boat coasting to rest, a bead in oil, or any object slowed by drag.

Terminal velocity

The terminal velocity is the steady speed at which the drag exactly balances the driving force, so the acceleration is zero. Setting $dv/dt = 0$ in the equation of motion:

A skydiver reaches terminal velocity when air resistance grows to equal her weight; from then on she falls at constant speed. The terminal velocity grows with weight and shrinks with the drag constant, which is why a feather and a heavy ball fall at very different terminal speeds in air. You can read off $v_t$ directly from the equation of motion without solving it fully, which is often all a part of the question asks.

Solving for the motion

Solving the differential equation with the initial condition $v(0) = 0$ gives the full velocity history:

The speed starts at zero, rises steeply, and levels off at $v_t$ as the exponential term dies away. Differentiating gives the acceleration $a(t) = g\,e^{-bt/m}$, which begins at $g$ (the moment of release, when there is no drag yet) and decays to zero. The combination $\tau = m/b$ is the time constant that sets how quickly terminal velocity is reached: after a few time constants the object is effectively at $v_t$. This exponential approach is the signature result of the topic.

Try this

Q1. An object of mass $2.0$ kg falls with linear drag and reaches a terminal velocity of $40$ m/s ($g = 9.8$ m/s squared). Calculate the drag constant $b$. [2 points]

• Cue. $v_t = mg/b \Rightarrow b = mg/v_t = (2.0)(9.8)/40 = 0.49$ kg/s.

Q2. Explain why an object released from rest with linear drag never quite reaches its terminal velocity. [2 points]

• Cue. The solution $v = v_t(1 - e^{-bt/m})$ approaches $v_t$ only as $t \to \infty$; the exponential term shrinks but never becomes exactly zero.