SCIENCE · INVESTIGATE AND EXPLAIN

Integrate acceleration twice and locate a reversal

Acceleration alone does not specify one unique trajectory.

Learning outcome: Compare an interval ending before reversal, exactly at reversal where supported, and after it. Preserve both integration constants and piecewise distance calculations.

Starting recap

Read derivatives as local rates and integrals as signed accumulation. Review power and chain rules, vector components, units and initial conditions where used.

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Try

With α=2 and T=2, v=2 m/s, x=−1/3 m and displacement−4/3 m. Reversal is sqrt2 s; distance is(8sqrt2−4)/3≈2.43790 m.

Open the experiment

Predict

For α=1 from t0 to4 s, the traveled distance is:

Your prediction and reason

Experiment

Compare an interval ending before reversal, exactly at reversal where supported, and after it. Preserve both integration constants and piecewise distance calculations.

Recover two initial conditions and a reversal

Worked teaching model · synthetic values · One-dimensional prescribed a(t)=αt, initial velocity−2 m/s and position1 m. No forces or drag are inferred. Distance uses the absolute velocity integral split at its nonnegative zero, not absolute displacement.

Acceleration-ramp coefficient: 1 m/s³ Elapsed time: 4 s

Reset keeps your prediction, captures and practice history.

Integrated position with both initial conditions

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: x0+integral of velocity (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Integrated position with both initial conditions

x0+integral of velocity

time (s)	position (m)
0	1
0.05	0.9000208
0.1	0.8001667
0.15	0.7005625
0.2	0.6013333
0.25	0.5026042
0.3	0.4045
0.35	0.3071458
0.4	0.2106667
0.45	0.1151875
0.5	0.02083333
0.55	-0.07227083
0.6	-0.164
0.65	-0.2542292
0.7	-0.3428333
0.75	-0.4296875
0.8	-0.5146667
0.85	-0.5976458
0.9	-0.6785
0.95	-0.7571042
1	-0.8333333
1.05	-0.9070625
1.1	-0.9781667
1.15	-1.046521
1.2	-1.112
1.25	-1.174479
1.3	-1.233833
1.35	-1.289938
1.4	-1.342667
1.45	-1.391896
1.5	-1.4375
1.55	-1.479354
1.6	-1.517333
1.65	-1.551312
1.7	-1.581167
1.75	-1.606771
1.8	-1.628
1.85	-1.644729
1.9	-1.656833
1.95	-1.664187
2	-1.666667
2.05	-1.664146
2.1	-1.6565
2.15	-1.643604
2.2	-1.625333
2.25	-1.601563
2.3	-1.572167
2.35	-1.537021
2.4	-1.496
2.45	-1.448979
2.5	-1.395833
2.55	-1.336438
2.6	-1.270667
2.65	-1.198396
2.7	-1.1195
2.75	-1.033854
2.8	-0.9413333
2.85	-0.8418125
2.9	-0.7351667
2.95	-0.6212708
3	-0.5
3.05	-0.3712292
3.1	-0.2348333
3.15	-0.0906875
3.2	0.06133333
3.25	0.2213542
3.3	0.3895
3.35	0.5658958
3.4	0.7506667
3.45	0.9439375
3.5	1.145833
3.55	1.356479
3.6	1.576
3.65	1.804521
3.7	2.042167
3.75	2.289063
3.8	2.545333
3.85	2.811104
3.9	3.0865
3.95	3.371646
4	3.666667

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
Final velocity	6	m/s
Final position	3.666667	m
Signed displacement	2.666667	m
Traveled distance	8	m
First velocity reversal time	2	s
Reversal occurs in selected interval	1	flag

Integrating acceleration twice requires two independently specified initial conditions.

v=−2+αt²/2 and x=1−2t+αt³/6. Integrate absolute velocity in separate sign intervals to find distance. The positive zero sqrt(4/α) may lie beyond the selected interval; do not insert an unobserved reversal or confuse the integration constant with a force parameter.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Integrating acceleration twice requires two independently specified initial conditions.

Observation to attach to a capture

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

Understand

Integrate acceleration twice and locate a reversal

Acceleration alone does not specify one unique trajectory.

The supplied ramp a(t)=αt integrates to v(t)=C1+αt²/2. The independently supplied initial velocity−2 m/s fixes C1. Integrating again gives x(t)=C2−2t+αt³/6; initial position1 m fixes C2. Neither integration constant can be silently set to zero just because the acceleration formula has no constant term.

With α=1 m/s³, velocity is zero at t=2 s and changes from negative to positive. Position at this turning event is−5/3 m. At t=4 s position is11/3 m and signed displacement from1 m is8/3 m. Distance is instead the sum of8/3 m before reversal and16/3 m after it:8 m.

The difference is integral v dt versus integral absolute(v) dt. Find every relevant sign-change time, split the interval there and add positive path contributions. A reversal beyond the requested endpoint is not part of that interval; at t=1 s the object is still moving left.

Acceleration, displacement and distance are distinct records. Reconstructing a motion function from acceleration is a conditional mathematical model, not an experimental force measurement. A time-varying acceleration also cannot be replaced by one constant-acceleration shortcut unless its integral consequences have been justified.

Changed-case check

Predict before reading the comparison. With α=2 and T=2, v=2 m/s, x=−1/3 m and displacement−4/3 m. Reversal is sqrt2 s; distance is(8sqrt2−4)/3≈2.43790 m.

Retained investigation

Compare an interval ending before reversal, exactly at reversal where supported, and after it. Preserve both integration constants and piecewise distance calculations. What can and cannot be reconstructed if only acceleration, but no initial velocity, is supplied?

Source and evidence boundary

Conceptual reference checked30 October2026: OpenStax University Physics Volume1. All prose, examples and models are original. These are bounded supplied mathematical trajectories or synthetic observations, not observed physical experiments. No launch, impact or apparatus procedure is requested. Record assumptions, prior predictions, errors and independent checks; guided success is not independently demonstrated proficiency.

Apply

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

What can and cannot be reconstructed if only acceleration, but no initial velocity, is supplied?

Your explanation and limitations

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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