Kinematic Equations for Constant Acceleration — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: This chapter covers the four core kinematic equations for constant acceleration, their derivation, graphical interpretation, 1D problem-solving, vector sign conventions, and application to near-Earth free fall motion for all AP Physics 1 question types.

You should already know: Displacement, velocity, and acceleration definitions and vector properties; average velocity definition; coordinate system setup for 1-dimensional motion.

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 1 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 Kinematic Equations for Constant Acceleration?

Kinematic equations for constant acceleration are a set of algebraic relationships that describe the motion of an object moving with unchanging acceleration (fixed magnitude and direction) in one dimension, the most common introductory motion scenario tested in AP Physics 1. Per the College Board AP Physics 1 Course and Exam Description (CED), this topic makes up roughly 5-7% of the total exam score, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a foundational building block for larger problems involving forces, energy, or momentum.

Standard notation for this topic is consistent across all AP materials: $x_0$ = initial position, $x$ = final position, $v_0$ = initial velocity, $v$ = final velocity, $a$ = constant acceleration, and $t$ = time elapsed (we set initial time $t_0=0$ for simplicity). Displacement is defined as $\Delta x=x-x_0$, which simplifies the form of all core equations. This topic is sometimes called "uniform acceleration kinematics" or "constant acceleration motion equations." Unlike general kinematics that relies on calculus, these equations let you solve for unknown motion quantities without integration, as long as acceleration is truly constant.

2. Derivation and Core Kinematic Equations

We derive all four core kinematic equations from first principles, starting with the definition of constant acceleration: $a=\frac{\Delta v}{\Delta t}=\frac{v-v_0}{t}$ (since $\Delta t=t-0=t$). Rearranging this gives the first core equation, which describes velocity as a function of time: $$v=v_0+at\text{\hspace{1em}(1)}$$

Next, for constant acceleration, velocity changes linearly with time, so average velocity is the simple average of initial and final velocity: $v_{\text{avg}}=\frac{v_0+v}{2}$. We also know average velocity equals displacement over time: $v_{\text{avg}}=\frac{\Delta x}{t}$. Setting these equal gives the second core equation, which relates displacement to initial and final velocity and time: $$\Delta x=\frac{1}{2}(v_0+v)t\text{\hspace{1em}(2)}$$

Substitute equation (1) into equation (2) to eliminate final velocity $v$, and simplify to get the third equation, which describes displacement as a function of time: $$\Delta x=v_{0}t+\frac{1}{2}a{t}^2\text{\hspace{1em}(3)}$$

Finally, rearrange equation (1) to solve for $t=\frac{v-v_0}{a}$, substitute into equation (2) to eliminate time, and simplify to get the fourth core equation, independent of time: $$v^2=v_0^2+2a\Delta x\text{\hspace{1em}(4)}$$

The key pattern here: each equation omits exactly one of the five kinematic quantities ($v,v_0,a,\Delta x,t$), so you always pick the equation that includes all your known quantities and the single unknown you need to find.

Worked Example

A car starting from rest accelerates at a constant $2.0{\text{ m/s}}^2$ for 5.0 seconds. What is the total displacement of the car over this time interval?

1. List all known quantities and the unknown: $v_0=0\text{ m/s}$ (starting from rest), $a=2.0{\text{ m/s}}^2$, $t=5.0\text{ s}$, unknown $\Delta x=?$.
2. Identify the correct equation: Equation (3) $\Delta x=v_{0}t+\frac{1}{2}a{t}^2$ includes all three knowns and solves directly for displacement, so it is the right choice.
3. Substitute values into the equation: $\Delta x=(0)(5.0)+\frac{1}{2}(2.0)(5.0{)}^2$.
4. Calculate the result: $\Delta x=(1.0)(25)=25\text{ m}$.

Exam tip: Always list your knowns and unknown before picking an equation. This simple step eliminates 90% of common mistakes from choosing an equation that leaves you with two unknowns.

3. Sign Conventions for Vector Quantities

All kinematic quantities (displacement, velocity, acceleration) are vectors, so their sign depends entirely on the coordinate system you choose at the start of the problem. The standard convention is: set the origin at the object's initial position ($x_0=0$), assign a positive sign to any vector pointing in the positive x-direction, and a negative sign to any vector pointing opposite to the positive x-direction. Consistency is more important than which direction you choose as positive: as long as every vector's sign matches your coordinate system, you will get the correct result.

This is especially critical for free fall motion near Earth's surface, where acceleration due to gravity always points downward, regardless of the object's direction of motion. If you choose upward as the positive direction, acceleration $a=-g=-9.8{\text{ m/s}}^2$. If you choose downward as positive, $a=+g=+9.8{\text{ m/s}}^2$. Many student errors come from memorizing "gravity is negative" instead of matching the sign to their coordinate system.

Worked Example

A ball is thrown straight upward from ground level with an initial speed of $19.6\text{ m/s}$. Take upward as positive, and find the time it takes for the ball to reach its maximum height.

1. Define the coordinate system: origin at ground level, upward = positive, so acceleration from gravity is $a=-9.8{\text{ m/s}}^2$.
2. List knowns and unknown: $v_0=+19.6\text{ m/s}$, at maximum height the ball stops before falling back, so final velocity $v=0\text{ m/s}$, unknown $t=?$.
3. Choose the correct equation: $v=v_0+at$ includes all knowns and solves for $t$.
4. Rearrange and solve: $t=\frac{v-v_0}{a}=\frac{0-19.6}{-9.8}=2.0\text{ s}$.

Exam tip: If you get a negative time or an unreasonably large final speed, the first thing to check is the sign of your acceleration relative to your chosen coordinate system.

4. Graphical Interpretation of Constant Acceleration

AP Physics 1 heavily tests the connection between kinematic equations and motion graphs, so understanding this relationship is critical for exam success. For constant acceleration, each type of motion graph has a predictable shape that directly matches the kinematic equations:

1. Acceleration vs. time (a vs t): Since acceleration is constant, this is a horizontal line at $y=a$. The area under the a vs t graph between $t=0$ and $t$ equals the change in velocity $\Delta v$, which matches $\Delta v=at$ from equation (1).
2. Velocity vs. time (v vs t): From equation (1), $v$ is linear in $t$, so v vs t is a straight line with slope equal to acceleration $a$ and y-intercept equal to $v_0$. The area under the v vs t graph between $t=0$ and $t$ equals displacement $\Delta x$, which matches equation (2) (the area of a trapezoid is exactly $\frac{1}{2}(v_0+v)t$).
3. Position vs. time (x vs t): From equation (3), $x=x_0+v_{0}t+\frac{1}{2}a{t}^2$, so this is a parabola. It opens upward if $a$ is positive, and opens downward if $a$ is negative. The slope of the tangent line to the parabola at any time $t$ equals the instantaneous velocity at that time.

Worked Example

The velocity vs time graph for a car slowing to a stop along a straight road is a straight line that goes from $(0,10\text{ m/s})$ to $(4\text{ s},2\text{ m/s})$. What is the total displacement of the car over the 4-second interval?

1. For v vs t graphs, displacement equals the area under the line, which can be confirmed with kinematic equation (2).
2. List knowns: $v_0=10\text{ m/s}$, $v=2\text{ m/s}$, $t=4\text{ s}$, unknown $\Delta x$.
3. Substitute into equation (2): $\Delta x=\frac{1}{2}(10+2)(4)=0.5\times 12\times 4=24\text{ m}$.
4. Confirm with the trapezoid area formula for graphs: $\frac{(v_0+v)}{2}\times \Delta t=24\text{ m}$, matching the algebraic result.

Exam tip: For AP Physics 1 MCQ questions about motion graphs, you can almost always solve displacement problems faster using the area rule for v-t graphs than full algebraic substitution.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using $v_{\text{avg}}=\frac{v_0+v}{2}$ for motion with non-constant acceleration. Why: Students memorize this relationship and forget it only holds when acceleration is constant, because velocity is only linear in time when acceleration is unchanging. Correct move: Always confirm the problem states acceleration is constant before using any of the four kinematic equations; for non-constant acceleration, use $v_{\text{avg}}=\frac{\text{total displacement}}{\text{total time}}$ only.
• Wrong move: Using $-g$ for acceleration when you chose downward as the positive direction in free fall problems. Why: Students memorize "gravity is negative" without connecting the sign to their own coordinate system choice. Correct move: After choosing your coordinate system, explicitly write the sign of acceleration based on direction before substituting into any equation.
• Wrong move: Reporting displacement as total distance traveled when an object reverses direction. Why: Students confuse the two quantities, and kinematic equations only give net displacement, not the sum of distances traveled in each direction. Correct move: If asked for total distance, split the motion into segments before and after the direction reversal (where $v=0$), calculate the magnitude of displacement for each segment, then add them.
• Wrong move: Using the time to maximum height as the total time of flight for a ball thrown up from ground level that lands back at ground level. Why: Students solve for time to max height and stop, forgetting the ball has to fall back down to the starting height. Correct move: If asked for total time of flight when launch and landing height are equal, double the time to maximum height, or solve directly using $\Delta x=0$ in the displacement equation.
• Wrong move: Rearranging $v^2=v_0^2+2a\Delta x$ incorrectly, leading to a negative value under the square root. Why: Students rush algebra and move $v_0^2$ to the wrong side of the equation. Correct move: Always rearrange step by step: $v^2-v_0^2=2a\Delta x$, so $v^2=v_0^2+2a\Delta x$, before taking the square root.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

A tennis ball is dropped from rest from a height of 5.0 m above the ground. Air resistance is negligible. What is the approximate speed of the ball just before it hits the ground? Take downward as positive, $g=9.8{\text{ m/s}}^2$. A) 5 m/s B) 10 m/s C) 15 m/s D) 20 m/s

Worked Solution: We are given $v_0=0\text{ m/s}$, $\Delta x=5.0\text{ m}$, $a=9.8{\text{ m/s}}^2$, and we need to find final speed $v$. We do not have any information about time, so we select the time-independent kinematic equation $v^2=v_0^2+2a\Delta x$. Substitute the given values: $v^2=0+2(9.8)(5.0)=98$, so $v=\sqrt{98}\approx 10\text{ m/s}$. The other options come from common student errors: option A comes from using the wrong time in $v=at$, while options C and D come from basic arithmetic errors. The correct answer is B.

Question 2 (Free Response)

A bicyclist is moving at a constant speed of $8.0\text{ m/s}$ when she passes a runner who is just starting to accelerate from rest at a constant $2.0{\text{ m/s}}^2$ in the same direction as the bicyclist. (a) How long does it take the runner to catch up to the bicyclist? (b) What is the speed of the runner when they meet, in m/s? (c) Explain why the runner is moving faster than the bicyclist when they meet, even though the runner started from rest.

Worked Solution: (a) Let $t$ = time after the bicyclist passes the runner. The bicyclist has zero acceleration, so their displacement is $\Delta x_b=v_{b}t=8.0t$. The runner starts from rest, so their displacement is $\Delta x_r=v_{r0}t+\frac{1}{2}{a}_r{t}^2=0+\frac{1}{2}(2.0)t^2=t^2$. When the runner catches up, their displacements are equal: $t^2=8.0t$. Solving gives $t(t-8.0)=0$, so the non-zero solution is $t=\boxed{8.0\text{ s}}$. (b) Use $v=v_0+at$ for the runner: $v=0+(2.0)(8.0)=\boxed{16\text{ m/s}}$. (c) For the first 4 seconds after passing, the bicyclist is moving faster than the runner, so the distance between them increases. The runner's speed equals the bicyclist's speed at $t=4\text{ s}$, after which the runner is faster and the distance between them decreases until the catch at $t=8\text{ s}$. By the time of the catch, the runner has already been moving faster than the bicyclist for 4 seconds, so their speed is higher when they meet.

Question 3 (Application / Real-World Style)

A police car is stopped at a traffic light. When the light turns green, a truck traveling at constant speed passes the police car going 72 km/h (20 m/s). The police car accelerates from rest at a constant $4{\text{ m/s}}^2$ to catch the truck. How far from the traffic light does the police car catch up to the truck?

Worked Solution: Let $t$ = time after the light turns green. The truck has constant speed, so its displacement is $\Delta x_T=v_{T}t=20t$. The police car starts from rest, so its displacement is $\Delta x_P=v_{0}t+\frac{1}{2}a{t}^2=0+\frac{1}{2}(4)t^2=2t^2$. When the police car catches the truck, their displacements are equal: $2t^2=20t$, so the non-zero solution is $t=10\text{ s}$. Substitute back to find displacement: $\Delta x_P=2(10{)}^2=200\text{ m}$. This means the police car catches the truck 200 meters past the traffic light, a realistic distance for a standard police car accelerating from rest.

7. Quick Reference Cheatsheet

8. What's Next

Kinematic equations for constant acceleration are the foundation for almost all motion topics in AP Physics 1, starting with projectile motion, where we split motion into independent horizontal and vertical components, each with constant acceleration (zero for horizontal, $g$ for vertical). Without mastering the sign conventions and equation selection in this chapter, you will not be able to correctly solve projectile motion problems, which make up a large portion of the kinematics unit on the AP exam. This topic also feeds into dynamics (Newton's laws), where you will use Newton's second law to calculate acceleration, then apply the kinematic equations here to find the resulting motion of objects. This core connection means constant acceleration kinematics is used in nearly all units of AP Physics 1, from momentum to circular motion.

• Projectile Motion
• Velocity and Acceleration Graphs
• Newton's Second Law
• Free Fall Motion