June 2, 2026

Motion with Derivatives & Integrals

« Back to AP Physics Guide / Unit C1: Kinematics (Mechanics) / 1D Motion with Calculus

Conceptual visualization of a tangent line on a position-time graph translating into the exact value of a velocity-time graph.

In AP Physics C, motion is no longer limited to constant acceleration. Calculus is required to understand a continuously changing world.

Welcome to AP Physics C: Mechanics. While Algebra-based physics courses limit you to scenarios where acceleration is perfectly constant (allowing the use of the “Big 4” kinematic equations), the real world is rarely that simple. If an object is attached to a spring, or falling through air resistance, its acceleration is constantly changing. To analyze these systems, we must use the mathematics of change: Calculus.

1. Kinematic Quantities as Derivatives

Instantaneous velocity is the rate of change of the object’s position with respect to time, and instantaneous acceleration is the rate of change of the object’s velocity[cite: 2]. If you are given a position function x(t), you can find the velocity and acceleration functions by taking the derivative with respect to time t.

    \[v_x = \frac{dx}{dt}\]

    \[a_x = \frac{dv_x}{dt} = \frac{d^2x}{dt^2}\]

Where x is position, v_x is instantaneous velocity, and a_x is instantaneous acceleration[cite: 2].
Diagram showing a vertical ladder: Position is at the top, velocity in the middle, and acceleration at the bottom. An arrow pointing down is labeled 'Derivative (Slope)' and an arrow pointing up is labeled 'Integral (Area)'.

The Calculus Kinematics Ladder: Differentiate to go down, Integrate to go up.

Concept First: The derivative represents the slope of the tangent line on a graph. The slope of a Position vs. Time graph at any specific instant is the velocity at that instant. The slope of a Velocity vs. Time graph is the acceleration[cite: 2].

⚙️ Interactive Derivative Simulator

Explore the motion of a particle governed by the cubic position function: x(t) = t^3 - 6t^2 + 9t. Drag the time slider to see how the mathematical derivatives determine the physical velocity (blue vector) and acceleration (red vector)!

Position x(t)
0.00 m
Velocity v(t)
9.00 m/s
Acceleration a(t)
-12.00 m/s²

2. Kinematic Quantities as Integrals

If you are given an acceleration function a(t) and need to find velocity or position, you must work backwards up the ladder using Integration. The displacement of an object is equal to the area under the curve of a velocity-time graph, and the change in velocity is equal to the area under the curve of an acceleration-time graph[cite: 2].

    \[\Delta v_x = \int_{t_1}^{t_2} a_x(t) dt\]

    \[\Delta x = \int_{t_1}^{t_2} v_x(t) dt\]

Do not forget your constants of integration! When evaluating an indefinite integral, you must add + C. In kinematics, this constant represents the initial velocity (v_0) or initial position (x_0).

3. Quick AP Practice

📚 Unit C1 Mastery Challenge

1. The net force F exerted on an object that moves along a straight line is given as a function of time t by F(t) = At^2 + B, where A=1 \text{ N/s}^2 and B=1 \text{ N}. What is the change in momentum of the object from t = 0 to t = 3 \text{ s}?[cite: 2]

Check Answer From the impulse-momentum theorem, change in momentum is the integral of Force with respect to time: \Delta p = \int F dt.

\Delta p = \int_{0}^{3} (1t^2 + 1) dt
\Delta p = [\frac{1}{3}t^3 + t]_0^3
\Delta p = (\frac{1}{3}(27) + 3) - (0) = 9 + 3 = \mathbf{12 \text{ kg} \cdot \text{m/s}}[cite: 2].

2. An object’s position is given by x(t) = 4t^2 - 12t. At what time t does the object instantaneously reverse its direction of motion?

Check Answer An object reverses direction when its velocity crosses zero. First, find the velocity function by taking the derivative of position:
v(t) = \frac{dx}{dt} = 8t - 12.

Set velocity equal to zero:
0 = 8t - 12 \Rightarrow 8t = 12 \Rightarrow t = \mathbf{1.5 \text{ s}}.