SCIENCE · INVESTIGATE AND EXPLAIN

Differentiate a moving-origin transformation

Distinguish a constant offset, a constant observer velocity and an accelerating observer.

Learning outcome: At observer acceleration 0 compare U=−2,0,2; then hold U=1 and compare accelerations −1,0,1. Differentiate the transformation and explicitly label which cases are noninertial.

Starting recap

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

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Try

At t=2,U=1,A=0, lab x=6,v=11,a=12 and relative x=2,v=10,a=12 in SI units.

Open the experiment

Predict

Which observer changes preserve acceleration, and why?

Your prediction and reason

Experiment

At observer acceleration 0 compare U=−2,0,2; then hold U=1 and compare accelerations −1,0,1. Differentiate the transformation and explicitly label which cases are noninertial.

Differentiate a moving-origin transformation

Worked teaching model · synthetic values · Laboratory motion x=t³−t, observer origin X=2+Ut+(A/2)t². Axes parallel and nonrotating. A=0 is inertial relative to stipulated inertial laboratory; A≠0 explicitly an accelerating diagnostic frame.

Observer initial velocity U: 1 m/s Observer acceleration A: 0 m/s²

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Laboratory 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.

Laboratory velocity: numerical coordinates

time (s)	velocity (m/s)
0	-1
0.1	-0.97
0.2	-0.88
0.3	-0.73
0.4	-0.52
0.5	-0.25
0.6	0.08
0.7	0.47
0.8	0.92
0.9	1.43
1	2
1.1	2.63
1.2	3.32
1.3	4.07
1.4	4.88
1.5	5.75
1.6	6.68
1.7	7.67
1.8	8.72
1.9	9.83
2	11
2.1	12.23
2.2	13.52
2.3	14.87
2.4	16.28
2.5	17.75
2.6	19.28
2.7	20.87
2.8	22.52
2.9	24.23
3	26

Observer-relative 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.

Observer-relative velocity: numerical coordinates

time (s)	velocity (m/s)
0	-2
0.1	-1.97
0.2	-1.88
0.3	-1.73
0.4	-1.52
0.5	-1.25
0.6	-0.92
0.7	-0.53
0.8	-0.08
0.9	0.43
1	1
1.1	1.63
1.2	2.32
1.3	3.07
1.4	3.88
1.5	4.75
1.6	5.68
1.7	6.67
1.8	7.72
1.9	8.83
2	10
2.1	11.23
2.2	12.52
2.3	13.87
2.4	15.28
2.5	16.75
2.6	18.28
2.7	19.87
2.8	21.52
2.9	23.23
3	25

Calculated values under the declared model
Quantity	Value	Unit
Laboratory position at 2 s	6	m
Observer origin at 2 s	4	m
Relative position at 2 s	2	m
Laboratory velocity at 2 s	11	m/s
Observer velocity at 2 s	1	m/s
Relative velocity at 2 s	10	m/s
Laboratory acceleration at 2 s	12	m/s²
Relative acceleration at 2 s	12	m/s²
Observer inertial relative to laboratory	1	flag

Constant-velocity translation changes velocity but preserves acceleration.

Differentiate x′=x−X once and twice. A constant origin offset disappears from velocity; constant observer velocity disappears from acceleration. This diagnostic does not authorize applying an inertial force law without accounting for an accelerating frame.

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

Constant-velocity translation changes velocity but preserves acceleration.

Observation to attach to a capture

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

Understand

Transform positions before memorizing a velocity rule

For parallel, nonrotating axes, relative position is x′=x−X, where X is the observer origin measured in the laboratory. The supplied laboratory is inertial and x=t³−t. The observer follows X=2+Ut+At²/2. The constant 2 is an origin offset in metres, U is initial observer velocity, and A is observer acceleration.

Differentiate once and twice

Laboratory velocity is 3t²−1 and observer velocity is U+At, so v′=3t²−1−U−At. Differentiating again gives a′=6t−A. Constant origin offsets disappear after one derivative; a constant observer velocity disappears after the second. This is why all such inertial translations agree on acceleration.

Compare an inertial case

At t=2,U=1,A=0, x=6 and X=4, so x′=2 m. Laboratory velocity is 11 and relative velocity 10 m/s. Both accelerations are 12 m/s². Different measured velocities do not imply a disagreement about the trajectory once both observers state their reference frame.

Challenge the inertial assumption

Set A=1 while U remains 1. At t=2, observer velocity is 3 and relative acceleration is 11 rather than 12 m/s². This is not a failure of the inertial transformation rule: the observer now accelerates. Applying an inertial force equation unchanged in that frame would require further analysis beyond this kinematic subtraction.

Extend and transfer

With fixed Cartesian bases, the same derivation holds componentwise: r′=r−R and v′=v−V. An observer moving only horizontally leaves the vertical component unchanged. A rotating basis would require additional derivative terms and is not represented. In a changed nonrotating case with laboratory acceleration 8 and observer acceleration 3, relative acceleration is 5 m/s², not 8. Label the frame before arguing that acceleration is invariant.

Public numerical transfer practice

Try a new practice case

At one instant lab acceleration is 8 m/s² and the nonrotating observer origin accelerates at 3 m/s² in the same direction. Find relative acceleration.

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 a constant position offset affect neither velocity nor acceleration?

Your explanation and limitations

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