SCIENCE · INVESTIGATE AND EXPLAIN

Split a motion interval at its actual reversal

Negative acceleration does not always mean decreasing speed. With positive initial velocity, a negative acceleration first slows the object. If the same acceleration continues after velocity reaches zero, the object moves increasingly fast in the negative direction. The model must say whether a controller stops the process at rest; this one does not.

Learning outcome: Compare v0=3,4,5 m/s at b=1, then compare b=1 and 2 at v0=3. Retain the event time, interval classification and separately calculated distance/displacement. Explain why extrapolating after 4 s requires another model claim.

Starting recap

Use SI units, signed components, algebraic ratios and graph coordinates. Identify the measurement rule and distinguish a record from an interpolated model.

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Try

At v0=4 m/s and b=1 m/s², zero velocity occurs exactly at 4 s. The object travels 8 m forward, ends at x=10 m and has no backward travel within the modeled interval.

Open the experiment

Predict

Does reaching zero velocity at the final instant require adding a backward distance?

Your prediction and reason

Experiment

Compare v0=3,4,5 m/s at b=1, then compare b=1 and 2 at v0=3. Retain the event time, interval classification and separately calculated distance/displacement. Explain why extrapolating after 4 s requires another model claim.

Separate distance and displacement across a reversal

Worked teaching model · synthetic values · Starts at x=2 m and v0>0, constant acceleration−b for a full4 s; no controller stops the motion at reversal. Uniform acceleration is assumed only in that declared interval. Mathematical point motion, not a stopping-distance safety rule.

Initial positive velocity: 3 m/s Magnitude of negative acceleration: 2 m/s²

Reset keeps your prediction, captures and practice history.

Position and exact reversal boundary

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: constant-acceleration position (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Position and exact reversal boundary

constant-acceleration position

time (s)	position (m)
0	2
0.05	2.1475
0.1	2.29
0.15	2.4275
0.2	2.56
0.25	2.6875
0.3	2.81
0.35	2.9275
0.4	3.04
0.45	3.1475
0.5	3.25
0.55	3.3475
0.6	3.44
0.65	3.5275
0.7	3.61
0.75	3.6875
0.8	3.76
0.85	3.8275
0.9	3.89
0.95	3.9475
1	4
1.05	4.0475
1.1	4.09
1.15	4.1275
1.2	4.16
1.25	4.1875
1.3	4.21
1.35	4.2275
1.4	4.24
1.45	4.2475
1.5	4.25
1.55	4.2475
1.6	4.24
1.65	4.2275
1.7	4.21
1.75	4.1875
1.8	4.16
1.85	4.1275
1.9	4.09
1.95	4.0475
2	4
2.05	3.9475
2.1	3.89
2.15	3.8275
2.2	3.76
2.25	3.6875
2.3	3.61
2.35	3.5275
2.4	3.44
2.45	3.3475
2.5	3.25
2.55	3.1475
2.6	3.04
2.65	2.9275
2.7	2.81
2.75	2.6875
2.8	2.56
2.85	2.4275
2.9	2.29
2.95	2.1475
3	2
3.05	1.8475
3.1	1.69
3.15	1.5275
3.2	1.36
3.25	1.1875
3.3	1.01
3.35	0.8275
3.4	0.64
3.45	0.4475
3.5	0.25
3.55	0.0475
3.6	-0.16
3.65	-0.3725
3.7	-0.59
3.75	-0.8125
3.8	-1.04
3.85	-1.2725
3.9	-1.51
3.95	-1.7525
4	-2

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.

Signed velocity

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

Equivalent numerical coordinates for Signed velocity

velocity

time (s)	velocity (m/s)
0	3
4	-5

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 position	-2	m
Signed displacement	-4	m
Path distance	8.5	m
Final signed velocity	-5	m/s
Average velocity	-1	m/s
Average speed	2.125	m/s
Time of zero velocity in continued model	1.5	s
Reversal strictly inside interval	1	flag

Distance requires splitting at a reversal; displacement is just the final minus initial position.

The zero-velocity event is t=v0/b. If it lies after4 s, no reversal occurs in the observed interval; if exactly at4 s, arrival is a zero-speed endpoint, not an extra backward segment. Negative acceleration can first reduce speed and later increase it in the negative direction.

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

Distance requires splitting at a reversal; displacement is just the final minus initial position.

Observation to attach to a capture

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

Understand

Begin with a measurement question

Build the algebraic model

The supplied object begins at x=2 m, velocity v0 and constant acceleration−b for 4 s. Position is x=2+v0t−bt²/2 and velocity is v0−bt. The zero-velocity event occurs at t=v0/b. Only a zero inside the declared interval creates a forward segment followed by a backward segment in the measured distance.

Check a concrete implication

Displacement is xfinal−xinitial whether or not reversal occurs. Distance adds the nonnegative lengths of the separate segments. If tturn lies strictly between 0 and4 s, compute xmax first, then distance=(xmax−xinitial)+(xmax−xfinal). If tturn is later than 4 s, the whole observed interval remains forward. At exactly 4 s, the endpoint has zero velocity but no subsequent backward segment has yet occurred.

Challenge the evidence boundary

For v0=3 m/s and b=2 m/s², reversal occurs at 1.5 s and x=4.25 m. Final position is −2 m. Displacement is −4 m, while distance is 2.25+6.25=8.5 m; average velocity −1 m/s and average speed 2.125 m/s answer different questions. These virtual parameters do not supply a safe braking or equipment stopping rule.

Worked original case

At v0=4 m/s and b=1 m/s², zero velocity occurs exactly at 4 s. The object travels 8 m forward, ends at x=10 m and has no backward travel within the modeled interval.

Predict and investigate

Does reaching zero velocity at the final instant require adding a backward distance? Compare v0=3,4,5 m/s at b=1, then compare b=1 and 2 at v0=3. Retain the event time, interval classification and separately calculated distance/displacement. Explain why extrapolating after 4 s requires another model claim.

Changed numerical practice

Try the changed case before feedback

An object begins at x=0 with v0=4 m/s and a=−2 m/s² for3 s. Find its traveled distance in metres.

Numerical result in the requested units

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Reflect and source-check

How can an object have negative average velocity but positive average speed?

Original explanations, synthetic records and models; checked 30 October 2026 against OpenStax College Physics 2e. The guided calculation is algebra-based; no calculus prerequisite is assumed. All activity is virtual. No physical launch, drop, water crossing, machinery operation or medical decision is requested. A public answer or saved capture does not verify independent mastery or actual measurement technique.

Apply

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

How can an object have negative average velocity but positive average speed?

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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