When acceleration changes with time
You will be able to: Integrate a nonconstant acceleration and use initial conditions.
How do you handle a motor that accelerates more strongly each second?
A cart’s motor produces a(t) = (2 m/s³)t while v(0) = 0. Acceleration is 2 m/s² at 1 s and 4 m/s² at 2 s. The final acceleration is not the acceleration throughout the trip.
A useful starting point: Free fall and the top of a throw →
Words and symbols before equations
- Time-dependent acceleration
- Acceleration specified as a function of time.
- Coefficient units
- Units attached to a constant so the whole expression has the correct units.
- Definite integral bounds
- The starting and ending times of the accumulation.
What this picture assumes
a(t) = kt from t = 0. This idealized prescribed acceleration varies with time; constant-a formulas do not apply when k is nonzero.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- At 2 s: a = 4 m/s², v = 4 m/s, x = 2.667 m. k = 2 m/s³; initial values are retained.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Integrate a(t) = kt from 0 to t: Δv = kt²/2. With initial velocity v₀, v = v₀ + kt²/2. Here k has units m/s³, so kt² has units m/s.
Integrate again: x = x₀ + v₀t + kt³/6. One integration accumulates velocity change; a second accumulates displacement. Each requires an initial value.
Using a(t_final) in Δx = ½at² overestimates displacement for this rising-acceleration example from rest. Even an interval-average acceleration only determines endpoint velocity change; it does not generally determine position with the constant-a equation. Different time histories can yield the same Δv but different displacements.
A worked example, step by step
For a(t) = (6 m/s³)t, v(0) = 2 m/s and x(0) = 1 m, find v and x at 2 s.
- Integrate acceleration: v(t) = 2 + 3t² in SI units.
- Integrate velocity: x(t) = 1 + 2t + t³.
- At 2 s, v = 2 + 3(4) = 14 m/s.
- At 2 s, x = 1 + 4 + 8 = 13 m. Differentiating twice returns a = 6t.
The constant-acceleration equations cannot use an endpoint acceleration to represent a changing acceleration.
Does doubling k double final position when x₀ and v₀ are nonzero?
Compare with an explanation
Not generally. Only the kt³/6 contribution doubles; the initial-position and initial-velocity contributions stay fixed.
Predict. Change one thing. Explain.
Double k while keeping time, x₀ and v₀ fixed. Predict how the acceleration-dependent contributions to v and x change.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At 2 s: a = 4 m/s², v = 4 m/s, x = 2.667 m. k = 2 m/s³; initial values are retained.
a(t) = kt from t = 0. This idealized prescribed acceleration varies with time; constant-a formulas do not apply when k is nonzero.
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 starts from rest at x = 0 with a(t) = (3 m/s³)t. Derive v(t), derive x(t), find x at 2 s and explain why ½a(2)(2 s)² is wrong.
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Compare with the answer and four-point rubric
- 1 point: v(t) = (1.5 m/s³)t².
- 1 point: x(t) = (0.5 m/s³)t³.
- 1 point: x(2 s) = 4 m.
- 1 point: Using the final 6 m/s² as constant predicts 12 m, incorrectly applying that larger acceleration for the whole interval.
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 1How do you find v from a(t)?
Integrate over time and add the initial velocity.
RECALL 2Why keep coefficient units?
They make each term dimensionally consistent.
RECALL 3How can you verify a derived x(t)?
Check initial values and differentiate twice to recover a(t).
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
When acceleration changes with time
- If a = kt: v = v₀ + kt²/2.
- x = x₀ + v₀t + kt³/6.
- Use integration for the specified a(t); do not assume a constant.
Remember: The constant-acceleration equations cannot use an endpoint acceleration to represent a changing acceleration.
Conditions: a(t) = kt from t = 0. This idealized prescribed acceleration varies with time; constant-a formulas do not apply when k is nonzero.
Refresh Kid · AP Physics C: Mechanics Unit 1 (official Unit 1) · Objectives 1.3.A; 1.2.C · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 1.3, objectives 1.3.A; 1.2.C. 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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