Kinematic Equations for Constant Acceleration

AP Physics 1· AP Physics 1 CED — Kinematics· 14 min read

1. Core Concepts and Standard Notation★★☆☆☆⏱ 3 min

Kinematic equations for constant acceleration are a set of algebraic relationships describing 1D motion of an object with unchanging acceleration (fixed magnitude and direction), the most common introductory motion scenario tested on AP Physics 1. This topic accounts for 5-7% of total AP Physics 1 exam score, and acts as a foundation for larger problems involving forces, energy, and momentum.

📘 Definition

Constant Acceleration Kinematics

A set of algebraic relationships that solve for unknown motion quantities without integration, valid only when acceleration is constant in magnitude and direction.

Example:

Near-Earth free fall (air resistance negligible), uniformly accelerating vehicles

Standard notation used consistently across AP Physics 1 materials is:

$x_{0}$ = initial position, $x$ = final position
$v_{0}$ = initial velocity, $v$ = final velocity
$a$ = constant acceleration
$t$ = time elapsed (with initial time $t_{0} = 0$ for simplicity)
$Δ x = x - x_{0}$ = net displacement

2. Derivation of the Four Core Equations★★☆☆☆⏱ 4 min

🔬 Derivation

Goal:

Derive all four core kinematic equations from first principles for constant acceleration

Starting from:

Definition of constant acceleration and average velocity

1
Start from the definition of constant acceleration, with $Δ t = t - 0 = t$ :
2
$a = \frac{v - v_0}{t}$
3
Rearranging gives the first equation, relating velocity to time:
4
$v = v_0 + at \tag{1}$
5
For constant acceleration, velocity changes linearly with time, so average velocity equals the average of initial and final velocity. By definition, average velocity also equals displacement over time:
6
$v_{\text{avg}} = \frac{v_0 + v}{2} = \frac{\Delta x}{t}$
7
Rearranging gives the second equation, relating displacement to average velocity and time:
8
$\Delta x = \frac{1}{2}(v_0 + v)t \tag{2}$
9
Substitute Equation (1) into Equation (2) to eliminate final velocity $v$ , and simplify to get the third equation, relating displacement to time:
10
$\Delta x = v_0 t + \frac{1}{2} a t^2 \tag{3}$
11
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 time-independent equation:
12
$v^2 = v_0^2 + 2 a \Delta x \tag{4}$

Result:

Each equation omits exactly one of the five kinematic quantities ( $v, v_{0}, a, Δ x, t$ ). To solve problems, always select the equation that includes all your known quantities and your single unknown.

📐 Worked Example

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

1
List all known quantities and the unknown we need to find:
2
$v_{0} = 0 m/s$ (starting from rest), $a = 2.0 m/s^{2}$ , $t = 5.0 s$ , unknown $Δ x = ?$
3
Identify the correct equation: Equation (3) $Δ x = v_{0} t + \frac{1}{2} a t^{2}$ includes all three knowns and solves directly for $Δ x$ .
4
Substitute values into the equation:
5
$\Delta x = (0)(5.0) + \frac{1}{2}(2.0)(5.0)^2$
6
Calculate the final result:
7
$\Delta x = 25 \text{ m}$

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. Vector Sign Conventions and Free Fall★★★☆☆⏱ 3 min

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. Consistency is far 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.

For near-Earth free fall, acceleration due to gravity always points downward, regardless of the object's direction of motion: if upward is positive, $a = - g = - 9.8 m/s^{2}$ ; if downward is positive, $a = + g = + 9.8 m/s^{2}$ .

📐 Worked Example

A ball is thrown straight upward from ground level with an initial speed of $19.6 m/s$ . 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 $a = - 9.8 m/s^{2}$ .
2
List known quantities and unknown: $v_{0} = + 19.6 m/s$ , at maximum height the ball stops before falling back, so final velocity $v = 0 m/s$ , unknown $t = ?$ .
3
Choose the correct equation that includes all knowns and solves for $t$ : $v = v_{0} + a t$ .
4
Rearrange and solve for $t$ :
5
$t = \frac{v - v_0}{a} = \frac{0 - 19.6}{-9.8} = 2.0 \text{ s}$

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★★★☆☆⏱ 3 min

AP Physics 1 heavily tests the connection between kinematic equations and motion graphs. For constant acceleration, each graph type has a predictable shape that aligns directly with the core equations:

Acceleration vs. time (a vs t): A horizontal line at $y = a$ , since acceleration is constant. The area under the a vs t graph equals $Δ v$ , matching $Δ v = a t$ from Equation (1).
Velocity vs. time (v vs t): A straight line, with slope equal to acceleration $a$ and y-intercept equal to $v_{0}$ , matching Equation (1). The area under the v vs t graph equals displacement $Δ x$ , matching Equation (2).
Position vs. time (x vs t): A parabola, since $x = x_{0} + v_{0} t + \frac{1}{2} a t^{2}$ from Equation (3). It opens upward for positive $a$ , downward for negative $a$ , and the slope of the tangent at any time equals instantaneous velocity.

📐 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 $(0, 10 m/s)$ to $(4 s, 2 m/s)$ . 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: $v_{0} = 10 m/s$ , $v = 2 m/s$ , $t = 4 s$ , unknown $Δ x$ .
3
Substitute into Equation (2):
4
$\Delta x = \frac{1}{2}(10 + 2)(4) = 24 \text{ m}$
5
This matches the trapezoid area formula for graphs, confirming the 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. AP-Style Worked Problem Examples★★★★☆⏱ 4 min

📐 Worked Example

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, $g = 9.8 m/s^{2}$ . Options: A) 5 m/s, B) 10 m/s, C) 15 m/s, D) 20 m/s

1
List knowns and unknown: $v_{0} = 0 m/s$ , $Δ x = 5.0 m$ , $a = 9.8 m/s^{2}$ , unknown final speed $v$ . Time is not given or required, so select the time-independent equation.
2
Substitute into $v^{2} = v_{0}^{2} + 2 a Δ x$ :
3
$v^2 = 0 + 2(9.8)(5.0) = 98$
4
Take the square root: $v = 98 \approx 10 m/s$ . The other options come from common errors: A from wrong time substitution, C/D from arithmetic errors. Correct answer is B.

📐 Worked Example

A bicyclist is moving at a constant speed of $8.0 m/s$ when she passes a runner who is just starting to accelerate from rest at a constant $2.0 m/s^{2}$ 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? (c) Explain why the runner is moving faster than the bicyclist when they meet.

1
Part (a): Let $t$ = time after the bicyclist passes the runner. The bicyclist has zero acceleration, so their displacement is $Δ x_{b} = 8.0 t$ . The runner's displacement is $Δ x_{r} = \frac{1}{2} (2.0) t^{2} = t^{2}$ . When the runner catches up, displacements are equal:
2
$t^2 = 8.0 t \implies t(t - 8.0) = 0$
3
The non-zero solution is $t = 8.0 s$ .
4
Part (b): Use $v = v_{0} + a t$ for the runner:
5
$v = 0 + (2.0)(8.0) = 16 \text{ m/s}$
6
Part (c): For the first 4 seconds, the bicyclist is faster, so the distance between them increases. The runner's speed equals the bicyclist's at $t = 4 s$ , after which the runner is faster and the distance decreases until catch at $t = 8 s$ . By the time of catch, the runner has been faster for 4 seconds, so their speed is higher at meeting.

📐 Worked Example

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 20 m/s. The police car accelerates from rest at a constant $4 m/s^{2}$ to catch the truck. How far from the traffic light does the police car catch up to the truck?

1
Let $t$ = time after the light turns green. The truck's displacement is $Δ x_{T} = 20 t$ , and the police car's displacement is $Δ x_{P} = \frac{1}{2} (4) t^{2} = 2 t^{2}$ . When the police car catches the truck, displacements are equal:
2
$2 t^2 = 20 t \implies t = 10 \text{ s} (non-zero solution)$
3
Substitute back to find displacement from the traffic light:
4
$\Delta x_P = 2(10)^2 = 200 \text{ m}$

6. Common Pitfalls

Wrong move:

Using $v_{avg} = \frac{v _{0} + v}{2}$ for motion with non-constant acceleration.

Why:

This relationship only holds when acceleration is constant, since velocity is only linear in time for unchanging acceleration.

Correct move:

Always confirm the problem states acceleration is constant before using any kinematic equations; for non-constant acceleration, use only $v_{avg} = \frac{total displacement}{total time}$ .

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 net displacement and total distance; 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 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 $Δ x = 0$ in the displacement equation.

Wrong move:

Rearranging $v^{2} = v_{0}^{2} + 2 a Δ x$ 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: $v^{2} - v_{0}^{2} = 2 a Δ x$ , so $v^{2} = v_{0}^{2} + 2 a Δ x$ , before taking the square root.

7. Quick Reference Cheatsheet

Formula/Relationship	Omits Quantity	Key Notes
$v = v_{0} + a t$	$Δ x$	Valid only for constant acceleration
$Δ x = \frac{1}{2} (v_{0} + v) t$	$a$	Valid only for constant acceleration
$Δ x = v_{0} t + \frac{1}{2} a t^{2}$	$v$	Valid only for constant acceleration
$v^{2} = v_{0}^{2} + 2 a Δ x$	$t$	Valid only for constant acceleration
$a = \pm 9.8 m/s^{2}$	N/A	Free fall: sign depends on coordinate system, negative if upward is positive
a vs t graph (constant $a$ )	N/A	Horizontal line; area under graph = change in velocity $Δ v$
v vs t graph (constant $a$ )	N/A	Straight line with slope $a$ ; area under graph = displacement $Δ x$
x vs t graph (constant $a$ )	N/A	Parabola; slope of tangent at time $t$ = instantaneous velocity

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Free fall speed calculation
2022 · FRQ
Catch-up motion problem

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 sub-topic, 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 directly 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.