SCIENCE · INVESTIGATE AND EXPLAIN

Match piecewise motion without inventing jumps

Carry boundary values into each integration interval and stop a model when its constraint changes.

Learning outcome: Derive switch values for a=2; compare braking 1,2 and 4. Check continuity at the switch and first rest, then show why extrapolation through rest violates the stated hold.

Starting recap

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

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Try

With initial acceleration 2 and braking 2, the switch at 2 s has x=6,v=5; first rest is 4.5 s at 12.25 m.

Open the experiment

Predict

Does a sudden acceleration change require a sudden velocity change?

Your prediction and reason

Experiment

Derive switch values for a=2; compare braking 1,2 and 4. Check continuity at the switch and first rest, then show why extrapolation through rest violates the stated hold.

Match integration constants across an acceleration switch

Worked teaching model · synthetic values · Original x(0)=0,v(0)=1; initial acceleration a until t=2 s, then acceleration −b until first rest, then held at rest. Instantaneous changes in acceleration idealized; velocity and position remain continuous. Hold is a new constraint, not extrapolation of braking through reversal.

Initial acceleration: 2 m/s² Braking magnitude after switch: 2 m/s²

Reset keeps your prediction, captures and practice history.

Matched position

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.

Matched position: numerical coordinates

time (s)	position (m)
0	0
0.1375	0.156406
0.275	0.350625
0.4125	0.582656
0.55	0.8525
0.6875	1.16016
0.825	1.50562
0.9625	1.88891
1.1	2.31
1.2375	2.76891
1.375	3.26563
1.5125	3.80016
1.65	4.3725
1.7875	4.98266
1.925	5.63063
2	6
2.0625	6.30859
2.2	6.96
2.3375	7.57359
2.475	8.14937
2.6125	8.68734
2.75	9.1875
2.8875	9.64984
3.025	10.0744
3.1625	10.4611
3.3	10.81
3.4375	11.1211
3.575	11.3944
3.7125	11.6298
3.85	11.8275
3.9875	11.9873
4.125	12.1094
4.2625	12.1936
4.4	12.24
4.5	12.25
4.5375	12.25
4.675	12.25
4.8125	12.25
4.95	12.25
5.0875	12.25
5.225	12.25
5.3625	12.25
5.5	12.25

Matched 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.

Matched velocity: numerical coordinates

time (s)	velocity (m/s)
0	1
0.1375	1.275
0.275	1.55
0.4125	1.825
0.55	2.1
0.6875	2.375
0.825	2.65
0.9625	2.925
1.1	3.2
1.2375	3.475
1.375	3.75
1.5125	4.025
1.65	4.3
1.7875	4.575
1.925	4.85
2	5
2.0625	4.875
2.2	4.6
2.3375	4.325
2.475	4.05
2.6125	3.775
2.75	3.5
2.8875	3.225
3.025	2.95
3.1625	2.675
3.3	2.4
3.4375	2.125
3.575	1.85
3.7125	1.575
3.85	1.3
3.9875	1.025
4.125	0.75
4.2625	0.475
4.4	0.2
4.5	0
4.5375	0
4.675	0
4.8125	0
4.95	0
5.0875	0
5.225	0
5.3625	0
5.5	0

Calculated values under the declared model
Quantity	Value	Unit
Switch velocity	5	m/s
Switch position	6	m
First-rest time	4.5	s
Rest position	12.25	m
Velocity jump at switch	0	m/s

Matching initial values at each boundary preserves position and velocity while acceleration may change.

After t=2 use elapsed time z=t−2, x=xs+vs z−bz²/2 and v=vs−bz. Continuing this braking expression past first rest would reverse the object, contradicting the stipulated hold. At switching instants acceleration has one-sided values; the velocity derivative is not a unique ordinary value there.

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

Matching initial values at each boundary preserves position and velocity while acceleration may change.

Observation to attach to a capture

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

Understand

Give each interval its own local clock

The object starts at x=0,v=1. During 0≤t≤2, constant acceleration A gives v=1+At and x=t+At²/2. At the switch, vs=1+2A and xs=2+2A. These are the initial conditions for the next interval, not values to discard when acceleration changes.

Match the braking segment

Use z=t−2 after the switch. With acceleration−B, integrate to v=vs−Bz and x=xs+vs z−Bz²/2. At z=0 these return vs and xs, ensuring continuity. Simply using x=−Bt²/2 would silently invent a different position and velocity at the boundary.

Locate a second boundary

First rest occurs when z=vs/B, at total time 2+vs/B. Its position is xs+vs²/(2B), obtained by substituting the stop time into the integrated position. Thereafter this model explicitly holds the object at rest. Continuing the braking polynomial would make v negative and reverse the object, contradicting that constraint.

Work the reference history

For A=2,B=2, vs=5 m/s and xs=6 m. Braking lasts 2.5 s and adds 6.25 m, giving first rest at 4.5 s and 12.25 m. Halving B doubles the braking time and distance but leaves the first segment unchanged. A graph may therefore share an initial segment and then diverge.

Interpret corners and transfer

Position and velocity are continuous, but acceleration switches abruptly. At such ideal instants the ordinary two-sided derivative of velocity is not unique; one-sided acceleration values describe the model. Real actuators smooth the transition. For an incoming speed 6 and braking magnitude 3, integrate 6−3z to its zero at 2 s to obtain 6 m. No physical stopping-distance safety guidance is inferred from this fictional trace.

Public numerical transfer practice

Try a new practice case

An object enters a braking stage at 6 m/s with constant braking magnitude 3 m/s². Find additional stopping distance.

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.

Apply

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

Which derivatives exist at ideal switching instants, and which have only one-sided values?

Your explanation and limitations

Continue

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

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Previous: Integrate distance across a velocity reversal Next: Differentiate a moving-origin transformation

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