Unit 2 · Topic 2.9

2.9 Resistive Forces

Resistive forces like air drag point opposite the velocity and grow with speed. Modeling drag as $\vec{F}_r = -k\vec{v}$ and putting it into Newton's second law gives a differential equation, which you solve by separating variables. The solutions are exponentials that level off, approaching a terminal velocity where the forces balance.

Key terms

resistive force
drag
terminal velocity
differential equation
separation of variables
exponential function

What a resistive force is

A resistive force acts on an object moving through a fluid, like air or water, and points opposite the object's velocity. Unlike kinetic friction, its size depends on speed: the faster you go, the harder it pushes back.

In this course the usual model is a force proportional to velocity: $\vec{F}_r = -k\vec{v}$ , where k (in kg/s) depends on the object's shape and size and the fluid. The minus sign keeps the force opposite the velocity. Real drag on fast objects is closer to proportional to v², but the linear model is the one you'll solve.

Terminal velocity

Drop an object from rest. At first it's slow, so drag is tiny and it accelerates at about g. As it speeds up, drag grows and the net force shrinks, so the acceleration shrinks. Eventually drag equals the weight, the net force is zero, and the object falls at a constant terminal velocity. For $F_r = kv$ :

$kv_T = mg \quad\Rightarrow\quad v_T = \frac{mg}{k}$

You can find terminal velocity just by setting the net force to zero. No calculus is needed for that part. At any moment, the acceleration is $a = g - \frac{k}{m}v$ , so at half the terminal speed, the acceleration is g/2.

Solving the differential equation

For a falling object (down positive), Newton's second law is $m\frac{dv}{dt} = mg - kv$ . This is a differential equation: it relates v to its own derivative. Separate the variables, putting everything with v on one side and t on the other, and integrate from the starting conditions:

$\int_0^v \frac{dv'}{g - \frac{k}{m}v'} = \int_0^t dt$

Working through the logarithm and solving for v gives:

$v(t) = \frac{mg}{k}\left(1 - e^{-kt/m}\right)$

The quantity m/k has units of seconds and is called the time constant. After one time constant the object reaches $1 - e^{-1}$ , about 63%, of its terminal velocity, and after several it is essentially at $v_T$ .

With drag as the only horizontal force, as for a boat whose engine shuts off, $m\frac{dv}{dt} = -kv$ gives $v(t) = v_0e^{-kt/m}$ . The speed decays toward zero, and the total distance it coasts is finite: $\frac{mv_0}{k}$ .

Graphs to recognize

Falling from rest: v rises steeply at first (initial slope g), then curves over and levels off at $v_T$ , approaching but never quite reaching it.
Its acceleration starts at g and decays exponentially toward zero.
Coasting to a stop: v decays exponentially toward zero; the position approaches a maximum value.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Falling with linear drag
A 0.20 kg ball is dropped from rest. Air resistance is $F_r = kv$ with k = 0.050 kg/s. Use g = 9.8 m/s². Find (a) the terminal speed, (b) the speed after 4.0 s and (c) the acceleration when the ball moves at half its terminal speed.
Show the solutionHide the solution
1. Step 1: (a) Set net force to zero: $v_T = \frac{mg}{k} = \frac{(0.20)(9.8)}{0.050} = 39.2$ m/s.
2. Step 2: (b) The time constant is $\frac{m}{k} = \frac{0.20}{0.050} = 4.0$ s. $v(4.0) = 39.2(1 - e^{-1}) \approx 24.8$ m/s.
3. Step 3: (c) $a = g - \frac{k}{m}v = g - \frac{k}{m}\cdot\frac{v_T}{2} = g - \frac{g}{2} = 4.9$ m/s².
Answer: (a) 39.2 m/s; (b) about 24.8 m/s; (c) 4.9 m/s².
Example 2Calculator allowed
Coasting to a stop
A 200 kg boat moving at 8.0 m/s shuts off its engine. Water resistance is $F_r = kv$ with k = 50 kg/s. Derive v(t), find when the speed reaches 2.0 m/s, and find how far the boat coasts in total.
Show the solutionHide the solution
1. Step 1: Newton's second law: $m\frac{dv}{dt} = -kv$ . Separate: $\frac{dv}{v} = -\frac{k}{m}dt$ .
2. Step 2: Integrate from $v_0$ to v and from 0 to t: $\ln\frac{v}{v_0} = -\frac{k}{m}t$ , so $v = v_0e^{-kt/m} = 8.0e^{-t/4.0}$ m/s.
3. Step 3: Speed 2.0 m/s: $e^{-t/4.0} = 0.25$ , so $t = 4.0\ln 4 \approx 5.5$ s.
4. Step 4: Total distance: $\int_0^\infty 8.0e^{-t/4.0}\,dt = (8.0)(4.0) = 32$ m.
Answer: $v = 8.0e^{-t/4.0}$ m/s; about 5.5 s; 32 m.

Common mistakes

Using constant-acceleration equations with drag. The acceleration changes as the speed changes.
Saying the object stops accelerating as soon as it's dropped, or that it reaches terminal velocity at a definite time. It approaches $v_T$ gradually.
Losing the sign in the drag force. Drag always points opposite the velocity, so write $-kv$ relative to the direction of motion.
Forgetting the limits when integrating, which leaves an unknown constant instead of the starting speed.

On the exam

Expect a derivation of v(t) by separating variables, plus a sketch of v or a against t. Points come from writing the correct differential equation, separating, integrating with limits, and showing the curve leveling off at $\frac{mg}{k}$ .
Questions also ask for terminal velocity directly. Set the net force to zero; a full derivation isn't needed.

Connected topics

Videos

Motion as a function of time: Resistive force example (part 1) | AP Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Topic 2.9 - Resistive Forces
Lessons With LondotWatch on YouTube (opens in a new tab)
Deriving Motion Equations with Drag Force
Flipping PhysicsWatch on YouTube (opens in a new tab)
AP Physics C Mechanics - Unit 2 - Lesson 12C - Drag Forces
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Motion as a function of time: Resistive force example (part 2) | AP Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Physics: Retarding and Drag Forces
Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)

Check yourself

4 questions on 2.9 Resistive Forces. Pick an answer to see if you got it, and why.

A small 0.50 kg ball is released from rest and falls through a thick liquid. The liquid exerts a resistive force $\vec{F}_r = -k\vec{v}$ on it, with k = 0.25 kg/s. Ignore buoyancy and use g = 10 m/s². Take down as positive.

Described situation

Question 1 of 4Calculator allowed

What is the ball's terminal speed?

Question 2 of 4Calculator allowed

What is the ball's acceleration at the instant it is released?

Question 3 of 4Calculator allowed

Which equation correctly applies Newton's second law to the ball while it falls?

Question 4 of 4Calculator allowed

How fast is the ball moving at t = 2.0 s?

0 of 4 answered