Why can a trip have small displacement but large distance?
You will be able to: Integrate velocity for displacement and speed for total distance.
Why can a trip have small displacement but large distance?
A cart goes forward, turns, and rolls back. Its final position records net movement; its wheel counter records every meter traveled.
A useful starting point: What constant value would give the same total? →
Words and symbols before equations
- Position s(t)
- Signed location relative to an origin, in meters.
- Velocity v(t)
- Signed position rate in m/s.
- Speed
- abs(v(t)), a nonnegative rate.
- Displacement
- Final position minus initial position.
What this picture assumes
Original model; numerical labels are rounded. v=t−1 m/s, s(0)=4 m. Displacement=t²/2−t. Distance splits at the reversal t=1. Curves are time plots, not a physical path.
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=2 s; v=1 m/s; displacement=0 m; distance=1 m; position=4 m. Turn at t=1 s.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
For v(t)=t−1 on [0,3] seconds, the cart moves negatively before t=1 and positively afterward. Its displacement is ∫₀³(t−1)dt=3/2 m.
Total distance is ∫₀³ abs(t−1)dt. Split at the velocity zero 1: the first triangle has area 1/2 and the second area 2, giving 5/2 m.
If s(0)=4 m, final position is 4+3/2=5.5 m. Neither the integral of speed nor the initial position should be substituted for signed displacement.
Find velocity zeros and check side signs. A zero without a sign change is a stop without a reversal; splitting still does no harm when computing speed.
| Feature | Displacement | Distance |
|---|---|---|
| Integrand | Velocity | Speed = absolute velocity |
| Sign | Positive, negative or zero | Nonnegative |
| Opposite directions | Can cancel | Both add |
A worked example, step by step
A particle has v(t)=2t−4 m/s on [0,3], with s(0)=5 m. Find displacement, distance and final position.
- Velocity changes sign at t=2.
- Displacement=[t²−4t]₀³=−3 m.
- The negative triangle on [0,2] has magnitude 4 and the positive triangle on [2,3] area 1; distance=5 m.
- Final position is 5−3=2 m, and abs(displacement)≤distance as expected.
Taking the absolute value of the final signed integral is not the same as integrating speed when direction changes.
Can distance be zero if displacement is zero?
Compare with an explanation
It can, but need not be: an out-and-back trip has positive distance and zero displacement.
Predict. Change one thing. Explain.
Move time past t=1 for v=t−1 and s(0)=4. Compare displacement, total distance and position. Explain why distance never decreases.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
t=2 s; v=1 m/s; displacement=0 m; distance=1 m; position=4 m. Turn at t=1 s.
Original model; numerical labels are rounded. v=t−1 m/s, s(0)=4 m. Displacement=t²/2−t. Distance splits at the reversal t=1. Curves are time plots, not a physical path.
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 questionFor v=t−2 on [0,4] and s(0)=3, calculate displacement, distance and final position; identify the turn.
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Compare with the answer and four-point rubric
- 1 point: v=0 at t=2 and changes − to +.
- 1 point: The two signed triangles are −2 and +2, so displacement is 0.
- 1 point: Distance is 2+2=4.
- 1 point: Final position is 3; the cart reverses at 2 s.
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 1Which integral counts every meter?
The integral of speed.
RECALL 2What is needed for final position?
Initial position plus displacement.
RECALL 3What establishes a direction change?
A sign change in velocity.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why can a trip have small displacement but large distance?
- Displacement=∫v dt; distance=∫abs(v)dt.
- s(b)=s(a)+∫ₐᵇ v dt.
- Split at velocity sign changes.
Remember: Taking the absolute value of the final signed integral is not the same as integrating speed when direction changes.
Conditions: Original model; numerical labels are rounded. v=t−1 m/s, s(0)=4 m. Displacement=t²/2−t. Distance splits at the reversal t=1. Curves are time plots, not a physical path.
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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