Resistive Forces

Drop a marble in air and it speeds up at $9.8\ \text{m/s}^2$; drop it in a jar of honey and within a moment it sinks at a steady crawl. The difference is a force that depends on how fast the marble moves: the faster it goes, the harder the fluid pushes back. Because that force changes as the speed changes, the acceleration isn't constant, the constant-acceleration equations from Unit 1 fail, and you need calculus.

A force that depends on velocity

A resistive force depends on the object's velocity and points opposite to it. Air resistance, drag in water, and the braking force of a magnet on a moving metal plate (eddy-current braking) are all examples. The simplest model, and the one AP builds its calculus on, makes the force proportional to the velocity:

The linear model fits small objects moving slowly through thick fluids (a bead in oil, a dust grain in air) and magnetic braking. Larger, faster objects in air, like a falling person or a baseball, feel a drag closer to $bv^2$. The method below works for any resistive force; only the integral changes.

Newton's second law gives a differential equation

Since $a = \frac{dv}{dt}$ and the force depends on $v$, Newton's second law becomes an equation relating $v$ to its own derivative, a differential equation. You solve it by separation of variables: get every $v$ on one side with $dv$ and every $t$ on the other with $dt$, then integrate both sides between matching limits (the initial time with the initial velocity, a general time $t$ with the velocity $v$ at that time).

Coasting with only a resistive force. An object moving along $+x$ with initial speed $v_0$ and nothing but $-kv$ acting on it:

Once you have $v(t)$, differentiate for acceleration and integrate for position, using the initial conditions:

The speed and acceleration decay toward zero, but never reach it in any finite time, and the position approaches a horizontal asymptote, $x_{\max} = \frac{mv_0}{k}$: the object coasts a finite distance in infinite time. The combination $\tau = \frac{m}{k}$, in seconds, is the time constant. Every $\tau$, the speed falls by a factor of $e$; after one $\tau$ it is at $37\%$ of $v_0$, after $3\tau$ at $5\%$.

Falling against a resistive force: terminal velocity

Now add a constant force in the other direction: gravity on an object released from rest. Take down as positive, so gravity is $+mg$ and the resistive force is $-kv$:

At the start, $v = 0$, so the object accelerates at $g$. As it speeds up, drag grows, the net force shrinks, and so does the acceleration. The speed keeps rising but more and more slowly, approaching the speed at which drag equals weight. That limit is the terminal velocity: the largest speed reached when a constant force and a resistive force act in opposite directions, reached when the net force is zero.

To get the whole motion, separate and integrate from $(0, 0)$ to $(t, v)$:

Then $a(t) = \frac{dv}{dt} = g\,e^{-t/\tau}$, and integrating $v$ from $y = 0$ gives $y(t) = v_T\left[t - \tau\left(1 - e^{-t/\tau}\right)\right]$. The graphs below show the shapes: $v$ rises toward the asymptote $v_T$ (reaching $63\%$ of it at $t = \tau$), and $a$ decays from $g$ toward the asymptote $0$. The position graph starts as a parabola and straightens into a line of slope $v_T$. Which asymptotes appear is set by the forces ($v_T = \frac{mg}{k}$) and the initial conditions ($v_0$, and where the object starts).

Worked examples

Where students lose points

On the exam

Free-response questions ask you to write Newton's second law as a differential equation, separate variables, integrate with limits, and sketch $v(t)$ or $a(t)$ with the asymptote labeled. An experimental question may ask how to tell whether drag is linear or quadratic: drop objects of different mass with the same shape (stacked coffee filters, say), measure each terminal speed, and graph. A straight line for $v_T$ against $m$ means $v_T \propto m$, linear drag; a straight line for $v_T^2$ against $m$ means quadratic drag.