Derive the constant-acceleration equations
You will be able to: Derive constant-acceleration relations by integration and select a useful equation.
Where do the familiar motion equations come from?
A scooter starts at 2 m/s and gains 2 m/s each second for 3 s. Its final speed is 8 m/s. Its displacement is 15 m because its speed is not 8 m/s for the whole trip.
A useful starting point: Read position, velocity and acceleration graphs →
Words and symbols before equations
- Constant acceleration a
- The same acceleration at every time in the interval.
- Initial values x₀ and v₀
- Position and velocity at the chosen t = 0.
- Eliminate time
- Combine equations to obtain a relation that does not contain t.
What this picture assumes
x₀ = 0 m and acceleration is constant throughout the interval. Graph scales may change as parameters change; always read the labeled axes.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- t = 3 s: x = 15 m, v = 8 m/s, a = 2 m/s². Displacement 15 m; distance 15 m; velocity change 6 m/s.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Integrating constant a gives v = v₀ + at. Integrate that velocity to obtain x = x₀ + v₀t + (1/2)at². The factor one-half comes from the integral of t, or from the triangular area added under the velocity graph.
With constant acceleration, velocity changes linearly, so displacement also equals [(v₀ + v)/2]t. Combine this with v − v₀ = at to obtain v² = v₀² + 2aΔx. Squaring removes the sign of v; interpret direction from the situation.
Choose an equation whose known quantities fit the problem. These relations can be applied to x or y components separately, but the acceleration must be constant for that component throughout the chosen interval. For varying acceleration, return to derivatives or integrals.
A worked example, step by step
A cart starts at 3 m/s and accelerates at 2 m/s² over a displacement of 8 m. Find its final velocity if it keeps moving right.
- Take right as positive: v₀ = +3 m/s, a = +2 m/s², Δx = +8 m.
- Time is not given, so use v² = v₀² + 2aΔx.
- v² = 9 + 32 = 41 m²/s², giving |v| ≈ 6.40 m/s.
- Choose v = +6.40 m/s because it continues right; the equation by itself only fixes the magnitude.
A constant acceleration equation with an average a inserted is not generally valid when acceleration varies.
For a = 0, what do the equations become?
Compare with an explanation
v stays v₀ and displacement is v₀t.
Predict. Change one thing. Explain.
Keep v₀ fixed, then change a. Compare displacement with the rectangle v₀t plus the triangular contribution ½at².
On narrow screens, swipe or scroll diagrams sideways to read all labels.
t = 3 s: x = 15 m, v = 8 m/s, a = 2 m/s². Displacement 15 m; distance 15 m; velocity change 6 m/s.
x₀ = 0 m and acceleration is constant throughout the interval. Graph scales may change as parameters change; always read the labeled axes.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant position, velocity, acceleration or integral relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA cart has x₀ = 5 m, v₀ = 1 m/s and constant a = 2 m/s². Derive x(t), then find v, displacement and position at 3 s.
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Compare with the answer and four-point rubric
- 1 point: Integration gives x(t) = 5 + t + t² in SI units.
- 1 point: v(3) = 1 + 2(3) = 7 m/s.
- 1 point: Displacement = 3 + 9 = 12 m.
- 1 point: Position = 5 + 12 = 17 m.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Why is there a ½ in ½at²?
Integrating the linear velocity contribution at gives ½at².
RECALL 2Which equation avoids time?
v² = v₀² + 2aΔx.
RECALL 3When do these equations fail?
When the chosen acceleration component is not constant over the interval.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Derive the constant-acceleration equations
- v = v₀ + at; Δx = v₀t + ½at².
- v² = v₀² + 2aΔx.
- Conditions: fixed coordinate direction and constant acceleration over the interval.
Remember: A constant acceleration equation with an average a inserted is not generally valid when acceleration varies.
Conditions: x₀ = 0 m and acceleration is constant throughout the interval. Graph scales may change as parameters change; always read the labeled axes.
Refresh Kid · AP Physics C: Mechanics Unit 1 (official Unit 1) · Objectives 1.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 1.3, objectives 1.3.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. This is Mechanics Unit 1: Kinematics. Quantitative motion problems stay in one or two dimensions; the optional spatial view clarifies vector notation. Students should study calculus alongside this course. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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