Accumulation and initial values
You will be able to: Integrate acceleration and velocity and use initial conditions to determine constants.
How do you reconstruct motion from acceleration?
A cart starts at x = 1 m with velocity 2 m/s. It gains 3 m/s of velocity every second. After 2 s, its velocity is 8 m/s and its position is 11 m.
A useful starting point: From a secant to a derivative →
Words and symbols before equations
- Integral ∫
- Accumulation of a rate over an interval; geometrically a signed area.
- Antiderivative
- A function whose derivative is the given function.
- Initial condition
- A known value at a specified starting time.
- Integration constant C
- The unknown starting offset left when reversing differentiation.
What this picture assumes
Constant acceleration; fixed x axis. Integrals give changes. Add the specified initial values. Teal shading is positive area; orange is negative.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- t = 2 s: x = 11 m, v = 8 m/s, a = 3 m/s². Displacement 10 m; distance 10 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
Acceleration tells how velocity changes, so ∫a dt gives Δv, not the entire velocity. Add the initial value: v(t) = v(0) + ∫₀ᵗa(τ)dτ. The dummy variable τ labels times being accumulated; t is the upper endpoint.
Integrate velocity to get displacement, then add the initial position: x(t) = x(0) + ∫₀ᵗv(τ)dτ. The reverse power rule is ∫tⁿdt = tⁿ⁺¹/(n+1) + C for n ≠ −1. Check an antiderivative by differentiating it.
For a = 3 m/s², velocity is 2 + 3t in SI units. Integrating gives x = 1 + 2t + 1.5t². Initial conditions determine the offsets. Two carts can share the same acceleration and still have different velocities and positions.
A worked example, step by step
With a(t) = (2 m/s³)t, v(0) = 3 m/s and x(0) = −1 m, find x at 2 s.
- Integrate acceleration: v(t) = (1 m/s³)t² + C.
- Use v(0) = 3 m/s: v(t) = 3 m/s + (1 m/s³)t².
- Integrate again: x(t) = (3 m/s)t + (1/3 m/s³)t³ + D; x(0) sets D = −1 m.
- At 2 s, x = −1 + 6 + 8/3 = 23/3 m ≈ 7.67 m. Differentiating x returns v.
An area under a(t) gives a velocity change. It does not supply an unknown initial velocity.
If acceleration is zero, must position stay constant?
Compare with an explanation
No. Velocity stays constant, but it may be nonzero.
Predict. Change one thing. Explain.
Keep acceleration and elapsed time fixed. Change only the initial velocity. Predict the change in final velocity and in displacement before using the control.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
t = 2 s: x = 11 m, v = 8 m/s, a = 3 m/s². Displacement 10 m; distance 10 m; velocity change 6 m/s.
Constant acceleration; fixed x axis. Integrals give changes. Add the specified initial values. Teal shading is positive area; orange is negative.
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 questionLet a = 4 m/s², v(0) = −2 m/s and x(0) = 3 m. Derive v(t) and x(t), then find both at 2 s.
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Compare with the answer and four-point rubric
- 1 point: v(t) = −2 m/s + (4 m/s²)t.
- 1 point: x(t) = 3 m − (2 m/s)t + (2 m/s²)t².
- 1 point: v(2 s) = +6 m/s.
- 1 point: x(2 s) = 7 m; differentiation recovers the prescribed velocity and acceleration.
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 1What fixes an integration constant?
A known value such as initial position or velocity.
RECALL 2How do you check an antiderivative?
Differentiate it and compare with the original rate.
RECALL 3Does ∫v dt give position or displacement?
A definite integral gives displacement; add initial position to obtain final position.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Accumulation and initial values
- v(t) = v₀ + ∫₀ᵗ a(τ)dτ.
- x(t) = x₀ + ∫₀ᵗ v(τ)dτ.
- Keep units and one integration constant per integration until initial values are used.
Remember: An area under a(t) gives a velocity change. It does not supply an unknown initial velocity.
Conditions: Constant acceleration; fixed x axis. Integrals give changes. Add the specified initial values. Teal shading is positive area; orange is negative.
Refresh Kid · AP Physics C: Mechanics Unit 1 (official Unit 1) · Objectives 1.2.C · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 1.2, objectives 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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