How do initial values connect acceleration, velocity and position?
You will be able to: Recover velocity and position with separate initial conditions.
How do initial values connect acceleration, velocity and position?
A cart accelerates steadily, but it may already be moving when the timer starts. Knowing its acceleration alone does not tell its starting speed or location.
A useful starting point: Why can a trip have small displacement but large distance? →
Words and symbols before equations
- Acceleration a(t)
- Velocity rate in m/s².
- Initial velocity v(0)
- Velocity at the reference time.
- Initial position s(0)
- Location at the reference time.
- Net velocity change
- Integral of acceleration.
What this picture assumes
Original model; numerical labels are rounded. a=2 m/s², v(0)=−1 m/s, s(0)=4 m. v=−1+2t and s=4−t+t². The turn is at 0.5 seconds.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- t=1 s; a=2 m/s²; v=1 m/s; s=4 m. Position derivative equals velocity; velocity derivative equals acceleration.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
Integrating acceleration gives a change in velocity: v(t)=v(0)+∫₀ᵗ a(u)du. Add the velocity initial value only once.
Then integrate the resulting velocity: s(t)=s(0)+∫₀ᵗ v(u)du. This second step requires its own position initial value.
For a=2 m/s², v(0)=−1 m/s and s(0)=4 m, v=−1+2t and s=4−t+t². The cart initially moves backward even though acceleration is positive.
Check both derivative relationships s′=v and v′=a. Positive acceleration does not always mean speeding up; speed decreases when velocity and acceleration have opposite signs.
A worked example, step by step
Given a(t)=6t m/s², v(0)=2 m/s and s(0)=1 m, find v(2) and s(2).
- Integrate acceleration: v(t)=2+3t².
- Thus v(2)=14 m/s.
- Integrate velocity: s(t)=1+2t+t³.
- Thus s(2)=13 m; the two integrations have different units and initial values.
The integral of acceleration is velocity change, not displacement. Position requires integrating velocity, including its initial value.
Does a positive acceleration require a positive velocity?
Compare with an explanation
No. It tells how velocity changes, not its current sign.
Predict. Change one thing. Explain.
For a=2, v(0)=−1 and s(0)=4, move time through 0.5 s. Explain the initial slowdown and later reversal using the velocity and position graphs.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
t=1 s; a=2 m/s²; v=1 m/s; s=4 m. Position derivative equals velocity; velocity derivative equals acceleration.
Original model; numerical labels are rounded. a=2 m/s², v(0)=−1 m/s, s(0)=4 m. v=−1+2t and s=4−t+t². The turn is at 0.5 seconds.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the units, signed change, strip direction, radius distances or cross-sectional area. Identify what the representation cannot tell you.
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=−2, v(0)=6 and s(0)=10 in SI units. Find velocity and position at 2 s and the time of a turn.
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Compare with the answer and four-point rubric
- 1 point: v(t)=6−2t.
- 1 point: s(t)=10+6t−t².
- 1 point: v(2)=2 m/s and s(2)=18 m.
- 1 point: v crosses zero at t=3, so the particle reverses then.
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 does ∫a give?
Velocity change.
RECALL 2Why two initial conditions?
Acceleration leaves both starting velocity and starting position unspecified.
RECALL 3When does positive acceleration slow a particle?
When its velocity is negative.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do initial values connect acceleration, velocity and position?
- v(t)=v(a)+∫ₐᵗ a(u)du.
- s(t)=s(a)+∫ₐᵗ v(u)du.
- Check both derivatives and both initial values.
Remember: The integral of acceleration is velocity change, not displacement. Position requires integrating velocity, including its initial value.
Conditions: Original model; numerical labels are rounded. a=2 m/s², v(0)=−1 m/s, s(0)=4 m. v=−1+2t and s=4−t+t². The turn is at 0.5 seconds.
Refresh Kid · AP Calculus BC Unit 8 · Objectives CHA-4.C · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 8.2, CHA-4.C. CED effective Fall 2020 and Fall 2026 clarifications checked September 17, 2026. Unit 8 covers BC topics 8.1–8.13, including smooth planar graph arc length and distance traveled. Focused lesson titles, examples, questions and illustrations are original Refresh Kid material. Unit 7 differential equations is available separately. Parametric and polar curves belong to later units.
Average value is distinguished from average rate. Motion uses velocity for displacement and speed for distance, with initial values stated separately. Area bounds and ordering are checked, including multiple crossings. Volumes derive the slice area before integration, distinguish diameter from radius, and use distances from the specified axis. Disks and washers use perpendicular slices and a consistent integration variable. A region crossing an axis requires checking the actual swept disk rather than inventing a hole.
The Organic Chemistry Tutor Disk & Washer Method and Arc Length Calculus Problems video titles and creator were checked; the full video was not reviewed. Use the free video as an optional companion; no paid material is required. Khan Academy’s unit destination was checked, but its JavaScript lesson contents were not fully readable by the research tool. OpenStax Sections 6.1 and 6.2 and the arc-length subsection of Volume 2 Section 2.4 were consulted for conceptual cross-checking. No provider scripts, questions, artwork or diagrams were copied. Refresh Kid is not affiliated with or endorsed by these providers.
GitHub’s 3D website collection informed optional spatial inspection and camera controls. All cross-section and revolution geometry is original and uses existing self-hosted Three.js with its MIT license. Spatial coordinates preserve the mathematical lengths; sampled mesh surfaces illustrate exact formulas. Teal shows the solid and orange a selected zero-thickness section. The volume is for the entire solid, not the highlighted plane. No autoplay is used; camera rotation changes only the view. Labeled 2D regions, cross-section diagrams, readouts and calculations remain available without WebGL.
Independent teacher review and observation of students remain pending. Technical checks do not certify mathematical accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus BC questions and scoring guides. The archive spans multiple units; no entire exam question is assigned here.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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