SCIENCE · INVESTIGATE AND EXPLAIN

Integrate distance across a velocity reversal

Use zero crossings to distinguish signed displacement from accumulated distance.

Learning outcome: Compare T=1,2,3 and 5 at tau=2. Locate the velocity zero, integrate with signs and verify distance≥|displacement| including endpoint reversal.

Starting recap

Differentiate or integrate with stated units, initial values, event boundaries and evidence limits.

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At tau=2,T=3, displacement −3 m but distance 23/3 m; the path reverses at 2 s.

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Predict

Can a small net displacement hide a much longer route?

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Experiment

Compare T=1,2,3 and 5 at tau=2. Locate the velocity zero, integrate with signs and verify distance≥|displacement| including endpoint reversal.

Split an absolute-velocity integral at a reversal

Worked teaching model · synthetic values · Prescribed v(t)=k(t²−tau²) with k=1 m/s³, x(0)=0. Distance is integral |v|; displacement is integral v. Only nonnegative times considered; endpoint and no-reversal intervals included.

Reversal time tau: 2 s Observation duration: 3 s

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Signed velocity

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Signed velocity: numerical coordinates

time (s)	velocity (m/s)
0	-4
0.1	-3.99
0.2	-3.96
0.3	-3.91
0.4	-3.84
0.5	-3.75
0.6	-3.64
0.7	-3.51
0.8	-3.36
0.9	-3.19
1	-3
1.1	-2.79
1.2	-2.56
1.3	-2.31
1.4	-2.04
1.5	-1.75
1.6	-1.44
1.7	-1.11
1.8	-0.76
1.9	-0.39
2	0
2.1	0.41
2.2	0.84
2.3	1.29
2.4	1.76
2.5	2.25
2.6	2.76
2.7	3.29
2.8	3.84
2.9	4.41
3	5

Calculated values under the declared model
Quantity	Value	Unit
Signed displacement	-3	m
Total distance	7.66667	m
Final velocity	5	m/s
Final acceleration	6	m/s²
Interior reversal occurs	1	flag
Reversal time	2	s

Integrating velocity permits cancellation; integrating speed requires a sign-aware partition.

For T≤tau all velocity is nonpositive and distance = −displacement. For T>tau split at tau: distance = −F(tau)+F(T)−F(tau). A zero instantaneous velocity does not imply zero acceleration or zero distance traveled.

Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.

Integrating velocity permits cancellation; integrating speed requires a sign-aware partition.

Observation to attach to a capture

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Understand

Identify the sign before integrating

Displacement is ∫v dt; distance is ∫|v|dt. Oppositely directed displacements cancel in the first expression but never in the second. A speed integral is not generally the absolute value of one final velocity integral.

Find the relevant zero

The prescribed velocity is k(t²−tau²), with k=1 m/s³ and nonnegative time. It is negative before t=tau and positive afterward. The second algebraic root−tau is outside the observed interval. Acceleration is 2kt, so at the positive reversal the acceleration is nonzero even though velocity is momentarily zero.

Construct a piecewise integral

Let F(t)=k(t³/3−tau²t). If T≤tau, displacement is F(T) and distance is−F(T). If T>tau, distance is−F(tau)+[F(T)−F(tau)]. Splitting at the actual zero enforces the absolute value without changing the physical velocity function.

Audit the reference case

For tau=2,T=3, F(3)=−3 m and F(2)=−16/3 m. The first leg travels 16/3 m in the negative direction; the second travels 7/3 m in the positive direction. Distance is 23/3 m while displacement is −3 m. At T=2, the endpoint is a zero velocity but the accumulated distance is already 16/3 m.

Transfer and explain

For v=t²−1 over 0 to 2, F(1)=−2/3 and F(2)=2/3, so total distance is 2 m. Before reversal v and a have opposite signs, reducing speed; afterward they have the same sign, increasing speed. Explain this with signed rates rather than the phrase that positive acceleration always means speeding up.

Public numerical transfer practice

Try a new practice case

For v=t²−1 m/s over 0≤t≤2, find total distance in metres.

Numerical result in the requested units

This public, unsaved check is not a secure assessment. Explain the derivative, integral or boundary in the retained course evidence workspace.

Sources and evidence boundary

Original explanations and cases, checked 30 October 2026 against the official AP Physics C Mechanics framework, OpenStax calculus kinematics and vector kinematics. All numerical records are supplied or generated, not actual laboratory measurements. No physical launch, drop or moving apparatus is requested. Public feedback is practice, not independent certification.

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

Why does setting final velocity to zero not establish zero acceleration or zero journey length?

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Previous: Reconstruct motion with two integration constants Next: Match piecewise motion without inventing jumps

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