This AP Physics C lesson on calculus kinematics notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on calculus kinematics and check yourself with the quiz.
Quick answer: In calculus-based kinematics, velocity is the derivative of position and acceleration is the derivative of velocity. Going the other way, velocity is the integral of acceleration and position the integral of velocity, with constants set by initial conditions. These methods handle variable acceleration, where the constant-acceleration equations fail.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 1
Calculus kinematics notes: the key ideas
How are position, velocity and acceleration linked by calculus?
Velocity is the rate of change of position, and acceleration is the rate of change of velocity. In calculus form, v = dx/dt and a = dv/dt = d²x/dt². Graphically, the slope of an x–t graph at an instant is the velocity, and the slope of a v–t graph is the acceleration. This works even when the acceleration is not constant, which is where the usual constant-acceleration equations fail.
Instantaneous velocity is the limit of Δx/Δt as Δt approaches zero, which is the derivative dx/dt.
v = dx/dt a = dv/dt = d²x/dt² a = v dv/dx
x is position (m), v is velocity (m/s), a is acceleration (m/s²) and t is time (s). The last form lets you work with acceleration that depends on position.
A particle moves with x = 2t³ − 9t² + 12t (m, with t in s). Method: v = dx/dt = 6t² − 18t + 12 and a = 12t − 18. The particle is at rest when 6(t − 1)(t − 2) = 0, so at t = 1 s and t = 2 s. Positions: x(1) = 5 m and x(2) = 4 m. Over 0 to 3 s, x goes 0 → 5 → 4 → 9 m. Answer: displacement 9 m but distance travelled 11 m.
| t (s) | x (m) | v (m/s) | a (m/s²) |
|---|---|---|---|
| 0 | 0 | 12 | −18 |
| 1 | 5 | 0 | −6 |
| 2 | 4 | 0 | 6 |
| 3 | 9 | 12 | 18 |
How do you go from acceleration back to velocity and position?
Integration reverses differentiation. The change in velocity is the area under the a–t graph, and the change in position is the area under the v–t graph. The constants of integration are fixed by the initial conditions, v₀ and x₀.
v = v₀ + ∫a dt x = x₀ + ∫v dt. The area under a v–t graph gives displacement, and area below the axis counts as negative.
A cart has a = 4t (m/s²), v₀ = 3 m/s and x₀ = 2 m. Method: v = 3 + ∫4t dt = 3 + 2t², then x = 2 + ∫(3 + 2t²) dt = 2 + 3t + (2/3)t³. At t = 3 s: v = 3 + 18 = 21 m/s and x = 2 + 9 + 18 = 29 m.
Using v = v₀ + at or x = x₀ + v₀t + ½at² when a changes with time. These equations come from integrating a constant. Another slip is to forget the constants v₀ and x₀ after integrating.
What if the acceleration depends on position?
Sometimes a is given as a function of x, such as a spring-like force. Use the chain rule: a = dv/dt = (dv/dx)(dx/dt) = v dv/dx. Separate the variables and integrate: ∫v dv = ∫a dx.
A mass has a = −4x (with x in m) and passes x = 0 at 6 m/s. Method: ∫v dv = ∫−4x dx, so v²/2 = −2x² + C. With v = 6 at x = 0, C = 18, so v² = 36 − 4x². It stops when 4x² = 36. Answer: x = 3 m, where the acceleration is −12 m/s².
Check the units after each derivative or integral. Differentiating divides the unit by seconds, and integrating multiplies it by seconds. If a result for velocity ends in m/s², you lost a time factor.
- v = dx/dt and a = dv/dt = d²x/dt²; slopes of x–t and v–t graphs.
- v = v₀ + ∫a dt and x = x₀ + ∫v dt; areas under a–t and v–t graphs.
- If a depends on x, use a = v dv/dx and integrate ∫v dv = ∫a dx.
- Displacement is the integral of v; distance is the integral of |v|.
- Constant-acceleration equations only work when a is constant.
Motion with derivatives and integrals: frequently asked questions
What is the difference between displacement and distance in calculus?
Displacement is the integral of velocity and can be negative. Distance is the integral of speed |v| and is never negative. To find distance, split the motion where velocity changes sign and add the pieces.
How do I find when a particle is at rest?
Set v = dx/dt equal to zero and solve for t. Then check the position and the acceleration at those times. A particle can be momentarily at rest while still accelerating, such as at the top of a throw.
When does speed increase?
Speed increases when velocity and acceleration have the same sign, and decreases when they have opposite signs. Check both signs at the instant, not just whether acceleration is positive.
Why is the constant of integration important?
Each integral gives a family of functions that differ by a constant. The initial conditions v₀ and x₀ select the one that matches the real motion, so skipping them gives wrong values at every time.
What does a = v dv/dx mean?
It is the chain rule applied to acceleration. It lets you solve for velocity as a function of position when a is given as a function of position, without first finding x(t).
