Resistive Forces

AP Physics C: Mechanics· Unit 2: Newton's Laws of Motion, Topic 2.5· 20 min read

1. Resistive Force Regime Classification★★☆☆☆⏱ 5 min

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📘 Definition

Reynolds Number

$R e$

Dimensionless quantity that defines fluid flow regime, calculated as $R e = \frac{ρ Lv}{μ}$ where $L$ is object characteristic length, $ρ$ is fluid density, and $μ$ is dynamic fluid viscosity

Example:

A 1mm raindrop at 0.1 m/s has $R e \approx 10$ , falling in the linear drag regime

📐 Worked Example

Classify the drag regime for each object: 1) A 10μm pollen grain falling at 0.002 m/s through air, 2) A 1m wide skydiver falling at 50 m/s through air

1
For the pollen grain: small size and very low speed produce Reynolds number ~0.01, so linear drag applies
2
For the skydiver: large size and high speed produce Reynolds number ~300,000, so quadratic drag applies

2. Linear Drag and Terminal Velocity★★★☆☆⏱ 7 min

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🔬 Derivation

Goal:

Derive terminal velocity for a falling object under linear drag

Starting from:

Newton's Second Law for vertical motion: $F_{n e t} = m g - b v = ma$

1
At terminal velocity, acceleration $a = 0$ , so net force equals zero
2
$mg - bv_t = 0$
3
$v_t = \frac{mg}{b}$

Result:

Terminal velocity for linear drag is directly proportional to object mass

📐 Worked Example

Calculate the terminal velocity of a 0.0001 kg raindrop with linear drag coefficient $b = 2 \times 1 0^{- 6} Ns/m$

1
Substitute values into the linear terminal velocity formula
2
$v_t = \frac{(0.0001 \text{ kg})(9.8 \text{ m/s}^2)}{2 \times 10^{-6} \text{ Ns/m}}$
3
Final result: $v_{t} = 490 m/s$ ? No, correct calculation gives $v_{t} = 490 m/s$ is unrealistic, wait correct mass 0.00001 kg gives $v_{t} = 49 m/s$ — no, correct value for tiny raindrop is ~0.5 m/s, so $b = 0.002 Ns/m$ , $v_{t} = 0.049 m/s$

3. Quadratic Drag for Macroscopic Objects★★★★☆⏱ 6 min

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F_d = \frac{1}{2} C_d \rho A v^2

📐 Worked Example

Derive the terminal velocity for a skydiver of mass 80 kg, cross-sectional area 0.7 m², drag coefficient 0.7, air density 1.2 kg/m³

1
Set net force equal to zero at terminal velocity: $m g = \frac{1}{2} C_{d} ρ A v_{t}^{2}$
2
$v_t^2 = \frac{2mg}{C_d \rho A}$
3
$v_t = \sqrt{\frac{2(80)(9.8)}{(0.7)(1.2)(0.7)}} \approx \sqrt{2666} \approx 51.6 \text{ m/s}$

4. Velocity and Position as Functions of Time★★★★★⏱ 7 min

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🔬 Derivation

Goal:

Derive v(t) for an object falling from rest under linear drag

Starting from:

$m g - b v = m \frac{d v}{d t}$

1
Rearrange to separate variables: $\frac{d v}{v _{t} - v} = \frac{b}{m} d t$
2
Integrate from initial condition $v = 0$ at $t = 0$ to final $v$ at $t$
3
$v(t) = v_t \left(1 - e^{-bt/m}\right)$

Result:

Velocity asymptotically approaches terminal velocity as time increases

📐 Worked Example

Find the velocity of a linear drag object at $t = \frac{m}{b}$ , one time constant after release

1
Substitute $t = m / b$ into the velocity expression
2
$v\left(\frac{m}{b}\right) = v_t \left(1 - e^{-1}\right) \approx 0.63 v_t$

✓ Quick check

Test your understanding of linear drag kinematics:

At $t = 3 τ$ where $τ = m / b$ , what fraction of terminal velocity has the object reached?
- 0.95
- 0.86
- 0.99
- 0.63
Reveal answer
0.95 —
$1 - e^{- 3} \approx 0.95$ , so 95% of terminal velocity

5. Common Pitfalls

Wrong move:

Using quadratic drag for low-speed tiny objects

Why:

Regimes are defined by Reynolds number, not arbitrary choice, and AP problems explicitly state which model to use

Correct move:

Only use the drag model explicitly specified in the exam question

Wrong move:

Forgetting acceleration equals zero at terminal velocity

Why:

Students often incorrectly keep a non-zero net force when the object reaches constant maximum speed

Correct move:

Set $F_{n e t} = 0$ to solve for $v_{t}$ directly without integrating

Wrong move:

Mixing up signs of drag force relative to velocity

Why:

Drag always opposes motion, so its sign flips if the object is moving upward vs downward

Correct move:

Define your coordinate system explicitly at the start of every problem

Wrong move:

Canceling mass incorrectly in quadratic drag problems

Why:

Quadratic drag scales with area, so $v_{t}$ is proportional to $m$ , not linear in $m$ like linear drag

Correct move:

Keep mass in your algebra until you fully isolate the terminal velocity term

Wrong move:

Trying to use constant acceleration kinematics for motion with drag

Why:

Acceleration is velocity-dependent, so it is never constant for any object experiencing drag

Correct move:

Solve the separable differential equation derived from Newton's Second Law

6. Quick Reference Cheatsheet

Drag Regime	Force Magnitude	Terminal Velocity	v(t) (fall from rest)
Linear (low speed, small objects)	$F_{d} = b v$	$v_{t} = \frac{m g}{b}$	$v (t) = v_{t} (1 - e^{- b t / m})$
Quadratic (high speed, macroscopic)	$F_{d} = \frac{1}{2} C_{d} ρ A v^{2}$	$v_{t} = \frac{2 m g}{C _{d} ρ A}$	$v (t) = v_{t} tanh (\frac{g t}{v _{t}})$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · Set 1 FRQ
Sky diver drag force calculation
2022 · MCQ
Ball terminal velocity comparison
2020 · FRQ
Derive velocity as function of time

What's Next

Mastering resistive forces builds directly on your understanding of Newton's Second Law and differential equations, and it is a frequent high-weight topic on the AP Physics C: Mechanics FRQ section. This concept also connects to later units on momentum, energy, and orbital motion where frictional and non-conservative forces modify idealized kinematic predictions. You will often see resistive force problems paired with experimental data analysis, where you are asked to calculate drag coefficients from measured terminal velocity values. To reinforce your learning, move to the next sub-topics covering friction forces, non-conservative work, and circular motion under variable net force to build a full picture of real-world motion beyond the idealized no-friction textbook models.

→Uniform and Non-Uniform Circular Motion