SCIENCE · INVESTIGATE AND EXPLAIN

Separate changes in speed from changes in direction

A vector derivative can be nonzero even when a scalar speed derivative is zero.

Learning outcome: Vary u at fixed b and then b at fixed u. Reconstruct dot and cross products, verify the component-square identity and explain the equal-scale path.

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

At u=3,b=1,t=2, speed=sqrt13, tangential acceleration2/sqrt13 and normal3/sqrt13. Their squared sum is1 m²/s⁴.

Open the experiment

Predict

For nonzero velocity, a·v divided by speed gives:

Your prediction and reason

Experiment

Vary u at fixed b and then b at fixed u. Reconstruct dot and cross products, verify the component-square identity and explain the equal-scale path.

Resolve acceleration along and across the trajectory

Worked teaching model · synthetic values · Parametric trajectory r=(u t,b t²/2), t from0 to4 s; evaluation at2 s. Positive u,b ensure nonzero speed and defined tangent/normal. Curvature is geometry, not proof of a particular force.

Constant horizontal velocity: 2 m/s Vertical acceleration: 1 m/s²

Reset keeps your prediction, captures and practice history.

Parametric path with equal spatial scales

Spatial axes use the same pixels per physical unit, preserving geometry. 1: r(t),0–4 s (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Parametric path with equal spatial scales

r(t),0–4 s

east position (m)	north position (m)
0	0
0.1	0.00125
0.2	0.005
0.3	0.01125
0.4	0.02
0.5	0.03125
0.6	0.045
0.7	0.06125
0.8	0.08
0.9	0.10125
1	0.125
1.1	0.15125
1.2	0.18
1.3	0.21125
1.4	0.245
1.5	0.28125
1.6	0.32
1.7	0.36125
1.8	0.405
1.9	0.45125
2	0.5
2.1	0.55125
2.2	0.605
2.3	0.66125
2.4	0.72
2.5	0.78125
2.6	0.845
2.7	0.91125
2.8	0.98
2.9	1.05125
3	1.125
3.1	1.20125
3.2	1.28
3.3	1.36125
3.4	1.445
3.5	1.53125
3.6	1.62
3.7	1.71125
3.8	1.805
3.9	1.90125
4	2
4.1	2.10125
4.2	2.205
4.3	2.31125
4.4	2.42
4.5	2.53125
4.6	2.645
4.7	2.76125
4.8	2.88
4.9	3.00125
5	3.125
5.1	3.25125
5.2	3.38
5.3	3.51125
5.4	3.645
5.5	3.78125
5.6	3.92
5.7	4.06125
5.8	4.205
5.9	4.35125
6	4.5
6.1	4.65125
6.2	4.805
6.3	4.96125
6.4	5.12
6.5	5.28125
6.6	5.445
6.7	5.61125
6.8	5.78
6.9	5.95125
7	6.125
7.1	6.30125
7.2	6.48
7.3	6.66125
7.4	6.845
7.5	7.03125
7.6	7.22
7.7	7.41125
7.8	7.605
7.9	7.80125
8	8

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
East velocity	2	m/s
North velocity	2	m/s
Speed	2.828427	m/s
Tangential acceleration	0.7071068	m/s²
Normal acceleration magnitude	0.7071068	m/s²
Curvature	0.08838835	1/m
Radius of curvature	11.31371	m

Acceleration can change speed and direction simultaneously.

v=(u,bt), acceleration=(0,b). The dot product a·v divided by speed gives tangential acceleration; the cross-product magnitude |v×a| divided by speed gives normal acceleration. Curvature=|v×a|/|v|³ and an=v² curvature. These relations require nonzero speed; the controls deliberately exclude the zero-speed singularity.

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

Acceleration can change speed and direction simultaneously.

Observation to attach to a capture

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

Understand

Separate changes in speed from changes in direction

A vector derivative can be nonzero even when a scalar speed derivative is zero.

Take r(t)=(u t,b t²/2) with constant positive u and b. Differentiating components gives v=(u,bt) and a=(0,b). Speed is sqrt(u²+b²t²). Its derivative is b²t/speed, which is the tangential component of acceleration, not generally its full magnitude.

The dot product sums matching components: a·v=b²t. Divide by speed to project acceleration onto the unit tangent. The magnitude of the planar cross product is |vx ay−vy ax|=ub. Divide this by speed to obtain the perpendicular acceleration magnitude. Their squared magnitudes sum to b², the square of the original acceleration magnitude.

Curvature measures direction change per path length. For this smooth nonzero-speed curve, curvature=|v×a|/|v|³ and radius of curvature is its reciprocal. Therefore normal acceleration is speed² times curvature. This is a local circle approximation, not a claim that the entire parabolic path is a circle.

At u=2,b=1,t=2, v=(2,2), speed=sqrt8 and both tangential and normal accelerations are1/sqrt2 m/s². Curvature is1/(8sqrt2) per metre and radius8sqrt2 m. A spatial plot must use equal axis scales to represent its geometric shape honestly. The controls exclude zero speed, where this tangent-based decomposition would need separate treatment.

Changed-case check

Predict before reading the comparison. At u=3,b=1,t=2, speed=sqrt13, tangential acceleration2/sqrt13 and normal3/sqrt13. Their squared sum is1 m²/s⁴.

Retained investigation

Vary u at fixed b and then b at fixed u. Reconstruct dot and cross products, verify the component-square identity and explain the equal-scale path. Why does a radius of curvature describe local geometry rather than the force law?

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.

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

Why does a radius of curvature describe local geometry rather than the force law?

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Previous: Integrate acceleration twice and locate a reversal Next: Differentiate the basis vectors of circular motion

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