How can you tell whether planar motion is speeding up?
You will be able to: Use velocity magnitude and the component dot product to analyze speed.
How can you tell whether planar motion is speeding up?
A car can accelerate by turning even if its speed stays constant. In two dimensions, an acceleration arrow does not automatically mean the car speeds up.
A useful starting point: Why does acceleration need two initial vectors? →
Words and symbols before equations
- Speed v
- √(vₓ²+vᵧ²), a scalar.
- Dot product V·A
- vₓaₓ+vᵧaᵧ.
- Speeding up
- Positive derivative of speed.
- Stationary instant
- Both velocity components are zero.
What this picture assumes
Original model. Coordinates have equal visual scales; angles are radians. Curves are sampled for display, while formulas determine the readouts. R=⟨2t,1+4t−t²⟩ m, V=⟨2,4−2t⟩ m/s, A=⟨0,−2⟩ m/s². This mathematical trajectory is restricted to 0≤t≤4, above y=0. The tangent shows direction, not a velocity scale.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- t=1 s; R=⟨2, 4⟩ m; V=⟨2,2⟩ m/s. Speed 2.82843 m/s; V·A=-4 m²/s³; speed derivative -1.41421 m/s².
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
Differentiate v²=V·V to get 2v v′=2V·A. Where v>0, v′=(V·A)/v.
A positive dot product means speeding up; a negative one means slowing down. Zero dot product gives zero instantaneous speed derivative when v>0.
For R=⟨2t,1+4t−t²⟩, V=⟨2,4−2t⟩ and A=⟨0,−2⟩. V·A=4t−8, so speed decreases before t=2 and increases afterward.
At zero speed the quotient is not valid. Use the definition or nearby speed values; do not divide by zero.
A worked example, step by step
At an instant V=⟨3,4⟩ m/s and A=⟨2,−1⟩ m/s². Find speed and its rate of change.
- Speed=√(9+16)=5 m/s.
- V·A=3(2)+4(−1)=2.
- Since speed is nonzero, v′=2/5 m/s².
- The positive derivative means speeding up, even though one acceleration component is negative.
Do not decide speeding up from one coordinate’s signs or from acceleration magnitude alone.
Can nonzero acceleration coexist with constant speed?
Compare with an explanation
Yes. Acceleration can change direction while leaving speed unchanged, as in uniform circular motion.
Predict. Change one thing. Explain.
Move across t=2 in the projectile model. Compare the dot product with the displayed speed. Explain why acceleration is unchanged while its effect on speed changes.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
t=1 s; R=⟨2, 4⟩ m; V=⟨2,2⟩ m/s. Speed 2.82843 m/s; V·A=-4 m²/s³; speed derivative -1.41421 m/s².
Read the representation: Teal shows a curve; orange marks the selected point, tangent, ray or swept region described above. Dashed segments are auxiliary comparisons. Read the model conditions and units before comparing lengths.
Original model. Coordinates have equal visual scales; angles are radians. Curves are sampled for display, while formulas determine the readouts. R=⟨2t,1+4t−t²⟩ m, V=⟨2,4−2t⟩ m/s, A=⟨0,−2⟩ m/s². This mathematical trajectory is restricted to 0≤t≤4, above y=0. The tangent shows direction, not a velocity scale.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the units, coordinate rates, parameter direction, tracing count or radial boundaries. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionAt an instant V=⟨−3,4⟩ and A=⟨1,−2⟩, find speed, V·A, speed derivative and whether the particle is slowing.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Speed=5.
- 1 point: V·A=−3−8=−11.
- 1 point: Speed derivative is −11/5.
- 1 point: It is slowing because that derivative is negative.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What is V·A in components?
vₓaₓ+vᵧaᵧ.
RECALL 2When can you divide by speed?
Only when speed is positive.
RECALL 3Can a turning particle have constant speed?
Yes; acceleration may be perpendicular to velocity.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How can you tell whether planar motion is speeding up?
- Speed=√(vₓ²+vᵧ²).
- v′=(vₓaₓ+vᵧaᵧ)/speed for speed>0.
- At rest, investigate separately.
Remember: Do not decide speeding up from one coordinate’s signs or from acceleration magnitude alone.
Conditions: Original model. Coordinates have equal visual scales; angles are radians. Curves are sampled for display, while formulas determine the readouts. R=⟨2t,1+4t−t²⟩ m, V=⟨2,4−2t⟩ m/s, A=⟨0,−2⟩ m/s². This mathematical trajectory is restricted to 0≤t≤4, above y=0. The tangent shows direction, not a velocity scale.
Refresh Kid · AP Calculus BC Unit 9 · Objectives FUN-8.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 9.6, FUN-8.B. CED effective Fall 2020 and Fall 2026 clarifications checked September 17, 2026. This unit covers BC topics 9.1–9.9. Parametric and vector motion is planar. Polar arc length, surface area and three-dimensional vector calculus are not assigned as required Unit 9 material.
Parametric slope and second derivatives retain their nonzero-denominator conditions. Speed is the magnitude of velocity; displacement and distance use different integrals. Polar coordinates allow signed radii, with distance abs(r). Polar tangents use Cartesian angular rates. Areas use squared radial boundaries with correct angular intervals, tracing counts and boundary switches; the pole is checked separately.
All focused explanations, examples, practice and models are original Refresh Kid work. OpenStax was consulted for mathematical cross-checking. Khan Academy’s destination was checked, but JavaScript lesson content was not fully readable by the research tool. Organic Chemistry Tutor video titles, creator and destinations were checked; full videos were not reviewed. No questions, diagrams or provider scripts were copied. Resources are optional; Refresh Kid is not affiliated with or endorsed by these providers.
GitHub’s 3D website collection informed optional camera and spatial inspection. The original parameter-lift model uses self-hosted Three.js with its MIT license. The teal projection is the physical circle. The vertical axis of the orange lift is time, not spatial height, and its scale is explicitly stated. No spatial arc-length calculation is made from that lift. Keyboard controls and labeled 2D alternatives remain available, without autoplay or required WebGL.
Independent teacher review and observation of students remain pending. Technical checks do not certify mathematical accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus BC questions and scoring guides. The archive spans multiple units; no entire exam question is assigned here.
Learn → Explore → Practice → Review is informed by the IES learning guide; this implementation has not been evaluated with learners.
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