When is a particle speeding up, slowing down or turning?
You will be able to: Use velocity and acceleration signs and verify direction changes across zero velocity.
When is a particle speeding up, slowing down or turning?
A car moving left at −4 m/s and accelerating left gets faster. A car moving left with rightward acceleration gets slower until its velocity reaches zero.
A useful starting point: How do position, velocity and acceleration describe one motion? →
Words and symbols before equations
- Speeding up
- Increasing magnitude of velocity.
- Slowing down
- Decreasing magnitude of velocity.
- Turning time
- An instant when velocity changes sign across it.
- Stationary instant
- An instant with v=0; it need not be a turn.
What this picture assumes
Original mathematical model; readouts are rounded. s=t²−4t+5 m on 0≤t≤4 s. Track-positive is right; the position graph is not a physical path. Speed derivative is not assigned at v=0.
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: position 2 m, velocity -2 m/s, speed 2 m/s, acceleration 2 m/s². leftward; slowing down. On [0,4], displacement=0 m and distance=8 m.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
Where v≠0, differentiating abs(v) gives speed′=v a/abs(v). Thus equal signs of v and a mean speeding up; opposite signs mean slowing down.
For s=t²−4t+5, v=2t−4 and a=2. On 0<t<2 the signs oppose and speed decreases. On 2<t<4 both are positive and speed increases.
At t=2, v=0. Check the signs on each side to establish a turn from leftward to rightward. The speed graph abs(2t−4) has a corner there, so the formula dividing by abs(v) is not used at zero.
A zero velocity alone does not establish a turn: v=(t−2)² is nonnegative on both sides of t=2. The particle pauses without reversing direction.
A worked example, step by step
A particle has v(t)=(t−1)(t−3) m/s. Is it speeding up at t=2.5?
- Evaluate velocity: v(2.5)=1.5(−0.5)=−0.75 m/s.
- Differentiate velocity: a(t)=2t−4.
- At 2.5, a=1 m/s².
- Opposite signs mean slowing down; the particle is moving left while its velocity becomes less negative.
Do not use the sign of acceleration alone to decide whether speed increases, or label every zero of velocity a turn.
Does v=0 force a=0?
Compare with an explanation
No. In this model v(2)=0 while a(2)=2 m/s².
Predict. Change one thing. Explain.
Compare t=1, 2 and 3. State direction and speed trend separately; explain why the zero-velocity instant needs a side-sign check.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
t=1 s: position 2 m, velocity -2 m/s, speed 2 m/s, acceleration 2 m/s². leftward; slowing down. On [0,4], displacement=0 m and distance=8 m.
Original mathematical model; readouts are rounded. s=t²−4t+5 m on 0≤t≤4 s. Track-positive is right; the position graph is not a physical path. Speed derivative is not assigned at v=0.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the relevant rate relationship, tangent estimate, or limit argument and its conditions. 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 questionFor v(t)=t²−4 m/s on t≥0, determine motion direction and speed trend on (0,2) and (2,4), and decide whether t=2 is a turn.
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Compare with the answer and four-point rubric
- 1 point: a=2t>0 on both open intervals.
- 1 point: On (0,2), v<0: negative direction and slowing down.
- 1 point: On (2,4), v>0: positive direction and speeding up.
- 1 point: Velocity changes from negative to positive at 2, so the particle turns there.
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 1Which signs imply speeding up?
Velocity and acceleration have the same sign, with v≠0.
RECALL 2What proves a direction change?
Opposite velocity signs on the two sides of the instant.
RECALL 3Can speed fail to be differentiable at a turn?
Yes; abs(v) can have a corner where v crosses zero.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
When is a particle speeding up, slowing down or turning?
- For v≠0: speed′=v a/abs(v).
- A turn requires a velocity sign change.
Remember: Do not use the sign of acceleration alone to decide whether speed increases, or label every zero of velocity a turn.
Conditions: Original mathematical model; readouts are rounded. s=t²−4t+5 m on 0≤t≤4 s. Track-positive is right; the position graph is not a physical path. Speed derivative is not assigned at v=0.
Refresh Kid · AP Calculus AB Unit 4 · Objectives CHA-3.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 4.2, CHA-3.B. CED effective Fall 2020, with Fall 2026 clarifications, checked September 17, 2026. Unit 4 has seven official topics. Topic 4.7 assesses 0/0 and ∞/∞ quotient forms; other indeterminate forms are excluded from the core lessons. Focused lesson titles and questions are original Refresh Kid teaching material.
Derivative units and signs are interpreted in context. Speed is the magnitude of velocity; turning requires a sign change. Related-rate equations hold at nearby times and are differentiated before snapshot values are inserted. Cone and ladder models have explicit physical domains. Tangent approximations remain estimates; error direction requires behavior on the relevant interval. L’Hôpital’s rule requires an eligible quotient form, nearby differentiability, nonzero denominator derivative and an existing finite or infinite derivative-ratio limit. A failed derivative-ratio limit is inconclusive about the original quotient.
The Organic Chemistry Tutor video creators and relevant descriptions were checked; full videos were not reviewed. Khan Academy’s unit destination was checked; its JavaScript lesson content was not fully readable by the research tool. OpenStax Sections 3.4, 4.1, 4.2 and 4.8 were consulted for conceptual cross-checking. No creator scripts, questions, diagrams or artwork were copied. Refresh Kid is not endorsed by these providers.
GitHub’s 3D website collection and its camera-control example informed optional spatial inspection. Graph geometry is generated from the stated original equations; self-hosted Three.js retains its MIT license. The optional 3D model shows the circular water surface and axial cross-section of a tip-down conical tank. The water radius and height obey the same similar-triangle ratio in 2D and 3D. Camera rotation only changes the view; signed flow and height controls describe an instantaneous state. Complete labeled 2D geometry, rates and equations remain available without WebGL.
Independent teacher review and observation of students remain pending. Checks do not certify scientific accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus AB questions and scoring guides. The archive spans multiple units; no entire exam question is assigned to this lesson.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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