Kinematics in one dimension — AP Physics C: Mechanics Study Guide

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

Covers: Position, velocity, acceleration definitions (average and instantaneous), kinematic equations for constant acceleration, graphical motion analysis, and calculus-based methods for non-constant acceleration in one dimension.

You should already know: Basic single-variable differentiation and integration; how to set up a coordinate system; quadratic equation solution methods.

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 / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Kinematics in one dimension?

Kinematics is the branch of mechanics that describes the motion of objects without reference to the forces that cause the motion. One-dimensional kinematics restricts all motion to a single straight line (almost always the x-axis by convention), so all motion quantities have only a magnitude and a sign (positive/negative, indicating direction along the line). Per the official AP Physics C: Mechanics CED, all kinematics topics (including this one) make up 10-15% of total exam weight, and one-dimensional kinematics appears in both multiple-choice (MCQ) and free-response (FRQ) sections, usually as the foundation for larger problems involving forces, energy, or rotation.

Standard notation conventions: $x(t)$ is position at time $t$, relative to an origin $x=0$; $\Delta x=x_f-x_i$ is displacement, which is distinct from distance traveled (the total length of the path taken). Average velocity is displacement divided by time, while average speed is total distance divided by time. AP exam problems regularly test this critical distinction, and one-dimensional kinematics is the prerequisite for every other topic in mechanics.

2. Calculus Definitions of Average and Instantaneous Motion

All kinematic quantities are defined by the relationships between position, time, velocity, and acceleration. Average quantities are defined over a finite time interval $\Delta t=t_f-t_i$: $$\bar{v}=\frac{\Delta x}{\Delta t}=\frac{x_f-x_i}{t_f-t_i},\bar{a}=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{t_f-t_i}$$ Instantaneous quantities are the limit of average quantities as $\Delta t$ approaches 0, which reduces to the derivative definitions that are core to AP Physics C: $$v(t)=\frac{dx}{dt},a(t)=\frac{dv}{dt}=\frac{d^{2}x}{d{t}^2}$$ To reverse this relationship (get velocity from acceleration, position from velocity), we use integration, with the constant of integration fixed by initial conditions (initial position $x_0=x(0)$ and initial velocity $v_0=v(0)$): v(t) = v_0 + \int_0^t a(\tau) d\tau, \quad x(t) = x_0 + \int_0^t v(\tau) d\tau} These definitions work for all motion, regardless of whether acceleration is constant or non-constant, so they are the most general tools for solving kinematics problems.

Worked Example

The position of a toy car moving along a straight track is given by $x(t)=2t^3-12t^2+10t+4$, where $x$ is in meters and $t$ in seconds for $t\ge 0$. Find (a) the instantaneous velocity at $t=2$ s, and (b) the average velocity between $t=1$ s and $t=3$ s.

1. For instantaneous velocity, use the derivative definition: $v(t)=\frac{dx}{dt}=6t^2-24t+10$.
2. Substitute $t=2$ s: $v(2)=6(2^2)-24(2)+10=24-48+10=-14\text{m/s}$. The negative sign indicates velocity points in the negative x-direction.
3. For average velocity, first calculate position at the interval endpoints: $x(1)=2(1)-12(1)+10(1)+4=4\text{m}$, $x(3)=2(27)-12(9)+10(3)+4=-20\text{m}$.
4. Calculate average velocity: $\bar{v}=\frac{x(3)-x(1)}{3-1}=\frac{-20-4}{2}=-12\text{m/s}$.

Exam tip: Always confirm the sign of your velocity/acceleration matches your coordinate system: AP problems accept the sign as a full answer for direction, so you do not need extra description if your coordinate system is clearly defined.

3. Kinematics with Constant Acceleration

When acceleration is constant ($a(t)=a$ for all $t$), the general integration rules simplify to a set of three widely used kinematic equations. Starting from the integral definition of velocity: $$v(t)=v_0+{\int }_0^{t}ad\tau =v_0+at$$ Integrate again to get position: $$x(t)=x_0+{\int }_0^t(v_0+a\tau )d\tau =x_0+v_{0}t+\frac{1}{2}a{t}^2$$ Eliminate time by rearranging the first equation to $t=\frac{v-v_0}{a}$ and substituting into the position equation to get the time-independent form: $$v^2=v_0^2+2a(x-x_0)$$ These equations are extremely useful for problems with constant acceleration (free fall near Earth's surface, constant braking, etc.), but they are only valid when acceleration is constant. AP problems frequently test whether students remember this restriction.

Worked Example

A ball is thrown straight upward from ground level with an initial speed of 24 m/s. Acceleration due to gravity is 9.8 m/s² downward. What is the maximum height the ball reaches before it begins falling back down?

1. Define the coordinate system: origin at ground level ($x_0=0$), positive x upward. Acceleration points downward, so $a=-9.8{\text{m/s}}^2$.
2. At maximum height, the ball stops moving upward before reversing direction, so its instantaneous velocity is $v=0$ at this point. We need to solve for maximum height $x=h$.
3. We know initial velocity, final velocity, and acceleration, so we use the time-independent kinematic equation to avoid solving for time first: $v^2=v_0^2+2a(x-x_0)$.
4. Substitute values: $0^2=(24{)}^2+2(-9.8)(h-0)⟹0=576-19.6h⟹h=\frac{576}{19.6}\approx 29.4\text{m}$.

Exam tip: For problems asking for maximum height, speed at a specific position, or stopping distance, the time-independent equation will almost always save you time compared to solving for time first.

4. Graphical Analysis of One-Dimensional Motion

AP Physics C heavily tests the ability to interpret the three common motion graphs: position vs. time ($x$ vs $t$), velocity vs. time ($v$ vs $t$), and acceleration vs. time ($a$ vs $t$). The relationships between these graphs follow directly from the derivative and integral definitions:

1. On an $x$ vs $t$ graph: The slope at any point equals the instantaneous velocity at that time; the average slope between two points equals average velocity.
2. On a $v$ vs $t$ graph: The slope at any point equals instantaneous acceleration, and the net area under the graph between two times equals displacement $\Delta x$ over that interval.
3. On an $a$ vs $t$ graph: The area under the graph between two times equals the change in velocity $\Delta v$ over that interval. These relationships are tested in nearly every exam, often in MCQ where you need to match one graph to another.

Worked Example

The velocity of a car accelerating from rest along a straight highway follows $v(t)=2t$ for $0\le t\le 10$ s, and $v(t)=40-2t$ for $10<t\le 20$ s, where $v$ is in m/s. What is the total displacement of the car between $t=0$ and $t=20$ s?

1. By definition, displacement equals the net area under the $v(t)$ curve between the two times.
2. The $v(t)$ curve forms a triangle with base equal to the total time interval (20 s) and height equal to the maximum velocity (20 m/s at $t=10$ s).
3. The area of a triangle is $\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 20\text{s}\times 20\text{m/s}=200\text{m}$.
4. Confirming with integration: $\Delta x={\int }_0^{10}2tdt+{\int }_{10}^{20}(40-2t)dt=100+100=200\text{m}$, matching the area calculation.

Exam tip: If the area under a graph is a simple geometric shape (triangle, rectangle, trapezoid), you do not need to integrate to find it—calculating the area directly saves valuable time on the exam.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using constant-acceleration kinematic equations when acceleration is given as a function of time or position. Why: Students memorize the simple constant-acceleration equations and default to them even when the problem explicitly states acceleration is non-constant. Correct move: Always check if acceleration is constant before using these equations; if not, use integration of $a(t)$ to find $v(t)$ and $x(t)$.
• Wrong move: Using displacement instead of total distance traveled to calculate average speed. For example, a ball thrown up and caught at the launch point has 0 displacement but non-zero distance traveled, so average speed is not zero. Why: Students confuse displacement (net change in position) with distance (total path length). Correct move: Always calculate average speed as total distance traveled divided by time, never displacement divided by time.
• Wrong move: Forgetting to add initial velocity/position when integrating acceleration/velocity. For example, integrating $a(t)=6t$ gives $v(t)=3t^2$ instead of $v(t)=v_0+3t^2$. Why: Students forget the constant of integration is fixed by initial conditions, not zero by default. Correct move: Always add the initial condition when writing the result of an integral for kinematics.
• Wrong move: Mixing up slope and area for motion graphs, e.g., taking the area under a position vs time graph as displacement. Why: Students confuse the order of differentiation between position, velocity, and acceleration. Correct move: Remember: slope of lower = higher (x→v→a, so slope of x is v, slope of v is a), area of higher = lower (area of a is Δv, area of v is Δx).
• Wrong move: Assuming gravity is always positive regardless of coordinate system direction. Why: Students memorize gravity as 9.8 m/s² and forget to change the sign if positive x is downward. Correct move: Always assign the sign of acceleration based on direction relative to your defined positive axis before plugging into equations.

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

Question 1 (Multiple Choice)

The position of an object moving along the x-axis is given by $x(t)=t^3-3t^2$, where $x$ is in meters and $t$ is in seconds. What is the average acceleration between $t=1$ s and $t=3$ s? (A) 0 m/s² (B) 3 m/s² (C) 6 m/s² (D) 12 m/s²

Worked Solution: Average acceleration is defined as $\bar{a}=\frac{\Delta v}{\Delta t}$, where $\Delta v=v(t_f)-v(t_i)$ and $\Delta t=t_f-t_i$. First, find velocity by differentiating the position function: $v(t)=\frac{dx}{dt}=3t^2-6t$. Evaluate at the endpoints: $v(1)=3(1{)}^2-6(1)=-3$ m/s, $v(3)=3(3{)}^2-6(3)=9$ m/s. Calculate $\Delta v=9-(-3)=12$ m/s, $\Delta t=3-1=2$ s, so $\bar{a}=12/2=6$ m/s². The correct answer is (C).

Question 2 (Free Response)

An object moving along the x-axis has an acceleration given by $a(t)=6t-14$ for $t\ge 0$, where $a$ is in m/s² and $t$ is in seconds. At $t=0$, the object is at $x_0=0$ m and has an initial velocity $v_0=15$ m/s. (a) Derive expressions for velocity $v(t)$ and position $x(t)$ as functions of time. (b) Find all times $t\ge 0$ when the object is momentarily at rest. (c) Calculate the total distance traveled by the object between $t=0$ and $t=3$ s.

Worked Solution: (a) We derive velocity by integrating acceleration, using initial velocity as the constant of integration: $$v(t)=v_0+{\int }_0^t(6\tau -14)d\tau =15+3t^2-14t=\boxed{3t^2-14t+15\text{m/s}}$$ We derive position by integrating velocity, using initial position as the constant of integration: $$x(t)=x_0+{\int }_0^t(3{\tau }^2-14\tau +15)d\tau =\boxed{t^3-7t^2+15t\text{m}}$$

(b) The object is at rest when $v(t)=0$. Solve the quadratic equation $3t^2-14t+15=0$ with the quadratic formula: $$t=\frac{14\pm \sqrt{196-180}}{6}=\frac{14\pm 4}{6}$$ This gives roots at $t=3$ s and $t=5/3\approx 1.67$ s. Both are non-negative, so the object is at rest at $\boxed{t=5/3\text{s}}$ and $\boxed{t=3\text{s}}$.

(c) Velocity is positive for $0\le t<5/3$ s and negative for $5/3<t<3$ s, so we split the interval to calculate total distance: $x(0)=0$, $x(5/3)=(5/3{)}^3-7(5/3{)}^2+15(5/3)=275/27\approx 10.19$ m, $x(3)=9$ m. Total distance $d=|x(5/3)-x(0)|+|x(3)-x(5/3)|=275/27+32/27=307/27\approx \boxed{11.4\text{m}}$.

Question 3 (Application / Real-World Style)

A high-speed passenger train is traveling at 80 m/s along a straight track when the driver spots a broken rail 1500 m ahead and triggers emergency braking. Emergency braking gives the train a constant deceleration of 2.5 m/s². Determine if the train stops before reaching the broken rail, and if so, how far the train is from the broken rail when it comes to rest.

Worked Solution: Define the origin at the train's position when braking is triggered, with positive x toward the broken rail. Initial conditions: $x_0=0$, $v_0=80$ m/s, $a=-2.5$ m/s² (deceleration opposite to motion). When the train stops, $v=0$. Use the time-independent constant acceleration equation: $$v^2=v_0^2+2a(x-x_0)⟹0=80^2+2(-2.5)x⟹x=\frac{6400}{5}=1280\text{m}$$ The broken rail is at 1500 m, so the distance from the broken rail when the train stops is $1500-1280=220$ m.

Interpretation: The emergency braking system stops the train 220 meters before the broken rail, avoiding a collision.

7. Quick Reference Cheatsheet

8. What's Next

One-dimensional kinematics is the fundamental foundation for all subsequent topics in AP Physics C: Mechanics. Immediately after this topic, you will extend the calculus-based definitions of position, velocity, and acceleration to two dimensions, to solve problems involving projectile motion and relative motion. Without mastering the sign conventions, derivative/integral relationships, and constant acceleration rules you learned here, multi-dimensional kinematics and all later topics involving motion will be significantly harder to master. This topic also feeds directly into Newton's second law of motion, where you will connect acceleration to net force, and later into energy and momentum, where you will integrate acceleration and velocity to find work and impulse.

Common follow-on topics to study next:

• Kinematics in two dimensions
• Newton's laws of motion
• Work and energy