Calculus-based Kinematics — AP Physics C: Mechanics Phys C Mech Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: Derivative definitions of velocity and acceleration, integration to solve for velocity and position, variable acceleration motion, calculus-based projectile motion, and 2D vector kinematics for AP Physics C: Mechanics.

You should already know: Strong calculus (concurrent OK), AP Physics 1 helpful.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: Mechanics style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Calculus-based Kinematics?

Calculus-based kinematics is the study of object motion using differential and integral calculus, unlike the algebra-only kinematics of AP Physics 1 which is limited to constant acceleration scenarios. It is the first unit in the AP Physics C: Mechanics Course and Exam Description, accounting for 14-20% of your total exam score, and forms the foundation for all later dynamics, work-energy, and rotational motion topics. Common synonyms include analytical kinematics and kinematics with derivatives/integrals.

2. $v=\frac{dx}{dt},a=\frac{dv}{dt}$

This pair of differential definitions is the core foundation of calculus kinematics, describing instantaneous rates of change of motion quantities rather than average values over time intervals. First, position $x(t)$ is a function that describes an object’s location along a 1D axis at time $t$ (units: meters, m, for $x$, seconds, s, for $t$). Instantaneous velocity $v(t)$ is the rate of change of position, equal to the first derivative of position with respect to time: $$v(t)=\frac{dx}{dt}$$ Velocity has units of meters per second (m/s). Instantaneous acceleration $a(t)$ is the rate of change of velocity, equal to the first derivative of velocity, or the second derivative of position: $$a(t)=\frac{dv}{dt}=\frac{d^{2}x}{d{t}^2}$$ Acceleration has units of meters per second squared (m/s²).

Worked Example

A remote-controlled car moves along a straight track with position function $x(t)=2t^3-5t+3$ (units: t in s, x in m). Find its velocity and acceleration at $t=2$ s.

1. Differentiate position to get velocity: $v(t)=\frac{dx}{dt}=6t^2-5$. Substitute $t=2$: $v(2)=6(2{)}^2-5=19$ m/s.
2. Differentiate velocity to get acceleration: $a(t)=12t$. Substitute $t=2$: $a(2)=24$ m/s².

Exam tip: Examiners regularly test that you distinguish instantaneous velocity (derivative) from average velocity ($\Delta x/\Delta t$), which are only equal if acceleration is constant.

3. Integrating to find velocity / position

Integration reverses the differentiation process, letting you recover velocity from acceleration, and position from velocity, even when acceleration is not constant. We start by rearranging the differential definition of acceleration: $$a=\frac{dv}{dt}⟹dv=a(t)dt$$ Integrate both sides with bounds matching initial conditions (velocity $v_0$ at $t=0$, velocity $v(t)$ at time $t$): $${\int }_{v_0}^{v(t)}d{v}^{\prime }={\int }_0^{t}a(t^{\prime })d{t}^{\prime }⟹v(t)=v_0+{\int }_0^{t}a(t^{\prime })d{t}^{\prime }$$ Repeat the process for position, rearranging the velocity differential: $$v=\frac{dx}{dt}⟹dx=v(t)dt$$ Integrate with initial position $x_0$ at $t=0$: $${\int }_{x_0}^{x(t)}d{x}^{\prime }={\int }_0^{t}v(t^{\prime })d{t}^{\prime }⟹x(t)=x_0+{\int }_0^{t}v(t^{\prime })d{t}^{\prime }$$

Worked Example

A particle starts at rest ($v_0=0$) at $x=2$ m, with acceleration $a(t)=4t+1$ m/s². Find its position at $t=3$ s.

1. Solve for velocity: $v(t)=0+{\int }_0^t(4t^{\prime }+1)d{t}^{\prime }=2t^2+t$
2. Solve for position: $x(t)=2+{\int }_0^t(2t^{\prime 2}+t^{\prime })d{t}^{\prime }=2+\frac{2}{3}{t}^3+\frac{1}{2}{t}^2$
3. Substitute $t=3$: $x(3)=2+\frac{2}{3}(27)+\frac{1}{2}(9)=24.5$ m

Exam tip: Never forget constants of integration! They correspond to initial conditions, which are always provided in exam problems, and forgetting them costs 1-2 marks per question.

4. Variable acceleration

A key difference between AP Physics 1 and AP Physics C is regular testing of motion with variable acceleration, where acceleration changes with time, position, or velocity. The standard constant-acceleration kinematic equations ($v=v_0+at$, $x=x_0+v_{0}t+\frac{1}{2}a{t}^2$) are only valid for constant acceleration, so you must use calculus for variable acceleration problems.

For acceleration given as a function of position $a(x)$, use the chain rule to avoid integrating with respect to time: $$a=\frac{dv}{dt}=\frac{dv}{dx}\cdot \frac{dx}{dt}=v\frac{dv}{dx}$$ Rearrange to separate variables: $$vdv=a(x)dx$$ Integrate both sides with position and velocity bounds to solve for unknown quantities.

Worked Example

A particle moves along the x-axis with acceleration $a(x)=-2x$ m/s². It starts at $x=1$ m with velocity $v=0$ m/s. Find its velocity when it reaches $x=0$ m.

1. Set up the integral with bounds $v_0=0$ at $x_0=1$, $v$ at $x=0$: $${\int }_0^v{v}^{\prime }d{v}^{\prime }={\int }_1^0(-2x^{\prime })d{x}^{\prime }$$
2. Evaluate left side: $\frac{1}{2}{v}^2$
3. Evaluate right side: $-2{\left[\frac{x^{\prime 2}}{2}\right]}_1^0=-(0-1)=1$
4. Solve for $v$: $\frac{1}{2}{v}^2=1⟹v=\pm \sqrt{2}\approx \pm 1.41$ m/s. Since the particle moves left from $x=1$ to $x=0$, $v=-1.41$ m/s.

5. Projectile motion with calculus

Ideal projectile motion (no air resistance) matches the algebra-based results from AP Physics 1, but calculus lets you solve for motion with non-constant forces like air resistance, a common AP Physics C exam question. Always separate projectile motion into independent x (horizontal) and y (vertical) components.

For ideal projectile motion, $a_x=0$ and $a_y=-g$ ($g=9.8$ m/s², upward as positive). Integrating gives the familiar results: $v_x(t)=v_{0x}$, $v_y(t)=v_{0y}-gt$, $x(t)=x_0+v_{0x}t$, $y(t)=y_0+v_{0y}t-\frac{1}{2}g{t}^2$.

For projectile motion with air resistance proportional to velocity, acceleration becomes $a_x=-k{v}_x$ and $a_y=-g-k{v}_y$, where $k$ is a positive drag constant. These are separable differential equations solvable via integration.

Worked Example

A projectile is launched horizontally with initial velocity $v_0=10$ m/s from height $y=100$ m, with drag constant $k=0.1$ s⁻¹. Find its horizontal velocity after 2 seconds.

1. Separate variables for the x-component acceleration: $\frac{d{v}_x}{v_x}=-0.1dt$
2. Integrate with bounds $v_{0x}=10$ at $t=0$: $${\int }_{10}^{v_x}\frac{d{v}^{\prime }}{v^{\prime }}={\int }_0^2-0.1d{t}^{\prime }$$
3. Evaluate: $\ln \left(\frac{v_x}{10}\right)=-0.2⟹v_x=10e^{-0.2}\approx 8.19$ m/s.

6. Vector kinematics in 2D

All kinematic quantities are vectors, so you can represent them in 2D using unit vectors $\hat{i}$ (positive x-direction) and $\hat{j}$ (positive y-direction). You differentiate or integrate each component separately to get velocity and acceleration vectors.

• Position vector: $\vec{r}(t)=x(t)\hat{i}+y(t)\hat{j}$
• Velocity vector: $\vec{v}(t)=\frac{d\vec{r}}{dt}=\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}=v_x(t)\hat{i}+v_y(t)\hat{j}$
• Acceleration vector: $\vec{a}(t)=\frac{d\vec{v}}{dt}=\frac{d{v}_x}{dt}\hat{i}+\frac{d{v}_y}{dt}\hat{j}=a_x(t)\hat{i}+a_y(t)\hat{j}$

The magnitude of velocity (speed) is a scalar quantity: $|\vec{v}|=\sqrt{v_x^2+v_y^2}$, and its direction is $\theta =\arctan \left(\frac{v_y}{v_x}\right)$ measured from the positive x-axis.

Worked Example

The position vector of a particle is $\vec{r}(t)=(3t^2-2t)\hat{i}+(4t+1)\hat{j}$ (units: t in s, r in m). Find the magnitude of its acceleration at $t=1$ s.

1. Differentiate position to get velocity: $\vec{v}(t)=(6t-2)\hat{i}+4\hat{j}$
2. Differentiate velocity to get acceleration: $\vec{a}(t)=6\hat{i}+0\hat{j}$
3. Acceleration is constant 6 m/s² in the x-direction, so its magnitude is 6 m/s² at all times, including $t=1$ s.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Using constant-acceleration kinematic equations for variable acceleration problems. Why: Students carry over habits from AP Physics 1 where constant acceleration was the default. Correct move: Always verify if acceleration is constant before using the standard 3 kinematic equations; if not, use derivatives or integration.
• Wrong move: Forgetting constants of integration when solving for velocity or position. Why: Students treat indefinite integrals as complete without accounting for initial conditions. Correct move: Always add $+C$ after indefinite integration, then use given initial position/velocity to solve for $C$, or use definite integrals with bounds matching initial conditions.
• Wrong move: Mixing up x and y components in 2D kinematics. Why: Students forget that horizontal and vertical motion are independent, so they apply vertical acceleration to horizontal velocity or vice versa. Correct move: Separate all vector quantities into x and y components first, solve each component separately, then combine only when asked for total speed or displacement.
• Wrong move: Ignoring the sign of velocity/acceleration when solving for $a(x)$ or $a(v)$ problems. Why: Students only take the positive root when solving for $v$ from a square root, without considering direction of motion. Correct move: Check problem context to determine the correct sign; e.g., a falling object has negative y-velocity if upward is defined as positive.
• Wrong move: Confusing average velocity with instantaneous velocity. Why: Students use $\Delta x/\Delta t$ when the question asks for velocity at a specific time. Correct move: Use the derivative of position for instantaneous velocity at a single time, use $\Delta x/\Delta t$ only for average velocity over a time interval.

8. Practice Questions (AP Physics C: Mechanics Style)

Question 1

A particle moves along the x-axis with position function $x(t)=t^4-4t^3+6t^2-24t+10$, where t is in seconds and x is in meters. (a) Find the instantaneous velocity and acceleration at $t=2$ s. (b) At what positive time(s) is the particle at rest? (c) What is the average acceleration between $t=0$ s and $t=3$ s?

Solution

(a) Velocity: $v(t)=\frac{dx}{dt}=4t^3-12t^2+12t-24$. At $t=2$: $v(2)=4(8)-12(4)+12(2)-24=-16$ m/s. Acceleration: $a(t)=\frac{dv}{dt}=12t^2-24t+12$. At $t=2$: $a(2)=12(4)-24(2)+12=12$ m/s². (b) Particle at rest when $v(t)=0$: $4t^3-12t^2+12t-24=0⟹t^3-3t^2+3t-6=0$. The only real positive root is approximately $t=2.77$ s (you do not need to solve cubics analytically on the AP exam, unless a rational root exists). (c) Average acceleration = $\Delta v/\Delta t$. $v(0)=-24$ m/s, $v(3)=4(27)-12(9)+12(3)-24=12$ m/s. $\Delta v=12-(-24)=36$ m/s, so average $a=36/3=12$ m/s².

Question 2

A particle has acceleration as a function of position given by $a(x)=3x^2+2$ m/s². It starts at $x=0$ m with velocity $v=1$ m/s. What is its velocity when it is at $x=2$ m?

Solution

Use the chain rule relation $vdv=a(x)dx$. Set up the integral: $${\int }_1^v{v}^{\prime }d{v}^{\prime }={\int }_0^2(3x^{\prime 2}+2)d{x}^{\prime }$$ Left side: $\frac{1}{2}{v}^2-0.5$. Right side: $[x^{\prime 3}+2x^{\prime }{]}_0^2=8+4=12$. Solve: $\frac{1}{2}{v}^2=12.5⟹v^2=25⟹v=\pm 5$ m/s. Since acceleration is positive for all x and initial velocity is positive, the particle moves in the positive x-direction, so $v=5$ m/s.

Question 3

A projectile is launched with initial velocity vector ${\vec{v}}_0=20\hat{i}+30\hat{j}$ m/s from the origin. Air resistance gives acceleration components $a_x=-0.05v_x$ m/s² and $a_y=-9.8-0.05v_y$ m/s². What is the x-component of the particle's position after 4 seconds?

Solution

First solve for $v_x(t)$: separate variables $\frac{d{v}_x}{v_x}=-0.05dt$. Integrate: $${\int }_{20}^{v_x}\frac{d{v}^{\prime }}{v^{\prime }}={\int }_0^4-0.05d{t}^{\prime }⟹\ln \left(\frac{v_x}{20}\right)=-0.2⟹v_x(t)=20e^{-0.05t}$$ Integrate $v_x(t)$ to get position: $$x(4)={\int }_0^{4}20e^{-0.05t^{\prime }}d{t}^{\prime }=20{\left[\frac{e^{-0.05t^{\prime }}}{-0.05}\right]}_0^4=-400(e^{-0.2}-1)\approx 72.5$$ m.

9. Quick Reference Cheatsheet

Note: Standard constant-acceleration kinematic equations are only valid when $a(t)=\text{constant}$; do not use for variable acceleration.

10. What's Next

Calculus-based kinematics is the foundational unit for all subsequent AP Physics C: Mechanics content. You will use the derivative and integration techniques you learned here to analyze forces and acceleration in Newton's Laws, relate work and kinetic energy via integration of force over position, calculate impulse as the integral of force over time, and extend all these concepts to rotational motion, where angular position, velocity, and acceleration follow the exact same calculus relations as their linear counterparts. Mastering this unit is non-negotiable for scoring a 5 on the exam, as every later unit builds directly on these skills.

If you have questions about any of the concepts, worked examples, or practice problems in this guide, you can ask Ollie our AI tutor anytime on the homepage. You can also move on to our study guide for Newton's Laws of Motion for AP Physics C: Mechanics next, to apply the kinematics skills you just learned to dynamics problems.