June 2, 2026

Velocity-Dependent Drag

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Visualization of a falling sphere surrounded by glowing air resistance vectors pushing upward, canceling out the gravity vector pulling downward.

As an object falls faster, the air pushes back harder. Eventually, these forces balance perfectly.

In classical introductory physics, objects fall with a constant acceleration of g (9.8 \text{ m/s}^2). In AP Physics C, we model the real world. As an object moves through a fluid like air or water, it experiences a resistive drag force that increases as the object’s speed increases.

1. The Drag Equation

For relatively slow-moving or small objects, the drag force (F_d) is directly proportional to the instantaneous velocity (v). The negative sign indicates that the force always acts in the direction exactly opposite to the velocity.

    \[\vec{F}_d = -b\vec{v}\]

Where b is the drag coefficient (which depends on the fluid’s density and the object’s size and shape) and v is velocity. Note: For very fast or large objects, drag is often proportional to v^2 (quadratic drag), but linear drag is the primary focus for AP calculus derivations.

2. Terminal Velocity (v_T)

Imagine dropping a ball from a helicopter. Initially, v=0, so there is no drag, and the ball accelerates at 9.8 \text{ m/s}^2. As it speeds up, the upward drag force (bv) grows larger and larger. Because the net force is decreasing, the acceleration decreases.

Eventually, the upward drag force becomes exactly equal to the downward force of gravity (mg). At this exact moment, the net force is zero. Acceleration drops to zero, and the object stops speeding up. It falls at a constant, maximum speed called Terminal Velocity.

    \[mg - b v_T = 0 \quad \Rightarrow \quad v_T = \frac{mg}{b}\]

3. The Calculus Derivation

How do we get the mathematical equation for the curve shown in the graph above? We must set up and solve a First-Order Separable Differential Equation.

Start with Newton’s Second Law, substitute a with dv/dt, separate the variables (v on one side, t on the other), and integrate!

Step 1: Newton’s 2nd Law
mg - bv = ma \quad \Rightarrow \quad mg - bv = m\frac{dv}{dt}

Step 2: Separate Variables
dt = \frac{m}{mg - bv} dv

Step 3: Integrate Both Sides (from t=0, v=0 to t, v)
\int_{0}^{t} dt = m \int_{0}^{v} \frac{1}{mg - bv} dv

Step 4: Solve (using u-substitution)
t = -\frac{m}{b} \ln\left(\frac{mg - bv}{mg}\right)

Step 5: Isolate v (using exponentials)
v(t) = \frac{mg}{b} \left(1 - e^{-\frac{bt}{m}}\right)

Concept First: Look at the final equation v(t) = \frac{mg}{b} \left(1 - e^{-\frac{bt}{m}}\right). As time (t) approaches infinity, e^{-\infty} approaches zero. The term in the parentheses becomes (1 - 0), leaving exactly v = \frac{mg}{b}, which is our Terminal Velocity!

4. Quick AP Practice

📚 Unit C2 Mastery Challenge

1. An object is dropped from a great height and experiences a drag force F_d = -cv^2. What is the terminal velocity of the object?

Check Answer At terminal velocity, the net force is zero, meaning gravity equals drag.
mg = c(v_T)^2
Divide by c and take the square root:
v_T = \sqrt{\frac{mg}{c}}.

2. Look at the graph simulator above. If you increase the mass (m) of the falling object, but keep the drag coefficient (b) exactly the same, what happens to the terminal velocity?

Check Answer Terminal velocity increases. Because v_T = mg/b, a heavier mass requires a larger drag force to balance its weight. To get a larger drag force with the same b value, the object must fall at a faster speed.