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SCIENCE · INVESTIGATE AND EXPLAIN

Reconstruct displacement and distance from velocity

A velocity–time graph answers two different questions. Its slope gives acceleration, with units (m/s)/s. Its signed area gives displacement, with units (m/s)×s. A point on the vertical axis is velocity itself, not displacement or acceleration. Before calculating, write the quantity and the units you need.

Learning outcome: Compare (v₀,a)=(3,−1), (4,−1), and (−3,1). Calculate each area before checking the table. Locate any interior reversal and explain the boundary case where velocity reaches zero exactly at 4 s. Record both displacement and distance.

Starting recap

Use signed quantities, algebraic substitution, graph slopes and areas. State the system, coordinate axes and units before calculating.

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Try

For v₀=3 m/s and a=−1 m/s², the graph crosses zero at 3 s. From 0 to 3 s the area is +4.5 m. From 3 to 4 s it is −0.5 m. Displacement is 4 m, distance is 5 m, average velocity is 1 m/s and average speed is 1.25 m/s.

Open the experiment

Predict

For a graph with equal positive and negative triangular areas, can displacement vanish while distance remains positive?

Experiment

Compare (v₀,a)=(3,−1), (4,−1), and (−3,1). Calculate each area before checking the table. Locate any interior reversal and explain the boundary case where velocity reaches zero exactly at 4 s. Record both displacement and distance.

Reconstruct signed area and total distance

Worked teaching model · synthetic values · Synthetic linear velocity v=v₀+at over 0–4 s. Position starts at zero. Acceleration continues through v=0; no stop controller. Exact area uses any interior zero crossing.

Reset keeps your prediction, captures and practice history.

043-1time (s)velocity (m/s)

Axes rescale with settings; compare numerical values, not screen angles or apparent lengths. Solid/points show the selected model relationship or observations; the dashed comparison is explained below.

Calculated evidence ledger — synthetic teaching model
QuantityValueUnit
Signed displacement4m
Total distance5m
Final velocity-1m/s
Average velocity1m/s
Interior reversal time3s
Accessible plot coordinates
Primary plotted points
time (s)velocity (m/s)
03
0.252.75
0.52.5
0.752.25
12
1.251.75
1.51.5
1.751.25
21
2.250.75
2.50.5
2.750.25
30
3.25-0.25
3.5-0.5
3.75-0.75
4-1

The negative-velocity region contributes negative displacement but positive distance.

Area under velocity is signed. Average speed uses total distance/time, not the absolute value of average velocity when the motion reverses. A zero velocity at an endpoint is not an interior reversal.

The negative-velocity region contributes negative displacement but positive distance.

Area under velocity is signed. Average speed uses total distance/time, not the absolute value of average velocity when the motion reverses. A zero velocity at an endpoint is not an interior reversal.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

1. Orient: Reconstruct displacement and distance from velocity

A velocity–time graph answers two different questions. Its slope gives acceleration, with units (m/s)/s. Its signed area gives displacement, with units (m/s)×s. A point on the vertical axis is velocity itself, not displacement or acceleration. Before calculating, write the quantity and the units you need.

2. Explain the model

For a straight velocity graph, the signed area can be found from trapezoids or triangles. If the graph crosses zero, split at the crossing. Area below zero subtracts from displacement because the object moves along the negative axis. The magnitude of that same area adds to distance because distance never cancels a return journey.

If v=v₀+at, a possible zero occurs at t=−v₀/a when a is nonzero. Count a reversal only if that time lies strictly inside the observed interval and the sign changes. A zero at the final instant is not evidence that the object has already returned; a zero velocity with zero acceleration can persist indefinitely.

Average velocity divides signed displacement by elapsed time. Average speed divides distance by elapsed time. They agree in magnitude only when the object does not reverse direction over the interval. Negative acceleration can increase speed if the object is already moving negatively.

3. Work an example

For v₀=3 m/s and a=−1 m/s², the graph crosses zero at 3 s. From 0 to 3 s the area is +4.5 m. From 3 to 4 s it is −0.5 m. Displacement is 4 m, distance is 5 m, average velocity is 1 m/s and average speed is 1.25 m/s.

4. Predict and investigate

Predict first: For a graph with equal positive and negative triangular areas, can displacement vanish while distance remains positive?

Compare (v₀,a)=(3,−1), (4,−1), and (−3,1). Calculate each area before checking the table. Locate any interior reversal and explain the boundary case where velocity reaches zero exactly at 4 s. Record both displacement and distance. Use the lesson experiment controls and retain the settings with your evidence.

5. Transfer and explain

Try a new practice case

An object has v=4−2t m/s from t=0 to t=3 s. Find total distance traveled.

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Reflect and retain evidence

Why is |average velocity| insufficient to reconstruct distance? Include a counterexample with an actual direction change.

Record a claim, numerical or graphical evidence, assumptions, a limitation and a changed-case prediction. A supplied synthetic investigation is not evidence that you performed a physical laboratory.

Framework and source notes

Original teaching prose, cases and synthetic datasets. Framework mapping: 1.2.B; 1.3.A. College Board AP Physics 1 CED, Unit 1, checked 30 October 2026. This mapping is a content cross-reference, not AP authorization or a claim of full course completion.

Apply

A velocity graph runs linearly from +2 m/s at 0 s to −2 m/s at 4 s. What are displacement and distance?

Public practice checks a specific claim. It does not establish independent mastery.

Why is |average velocity| insufficient to reconstruct distance? Include a counterexample with an actual direction change.

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

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