Higher: differentiate displacement and velocity models
Derive signed velocity and acceleration, respect time domains and distinguish rest, direction, speed and total distance.
Exam scope and assessment notes
Higher only: Foundation has no section3.4 Calculus requirement. Higher also includes all shared Foundation content. These six original lessons teach variable/local versus interval rates, termwise integer-power differentiation, gradients/rates, stationary coordinates and graph-shape classification, one-dimensional kinematics and other simple practical models. H:3.4C requires both the gradient/rate and stationary-point lessons; H:3.4E requires both the motion and other-practical lessons with worked examples and all Guided, Independent and Transfer stages. The six-lesson split is StudyForA’s teaching design, not a syllabus-prescribed count. Small negative-integer powers with x≠0 conservatively retain integer-power breadth; fractional/trigonometric differentiation, integration and product/quotient/chain rules are not inferred requirements. H:3.4D classifies maxima and minima by general original-graph shape only; a second derivative is used for acceleration in motion, not as a substitute classification test. A derivative zero supplies an input, not the original stationary ordinate or an automatic turning point. Keep every original domain, physical restriction, admissible integer count and endpoint. Continuous relaxation supports differentiation for discrete counts; it does not make noninteger production feasible. Accepted boundary maxima need not be stationary. The separate H3.3D drawn-tangent construction is not replaced or newly credited by calculus. Fixed interactive examples support exact algebraic self-checks and approximate graph exploration; no general expression/root solver or written-work marker is supplied. Written work is not automatically marked. Attempt acknowledgements and revealed references do not award marks, prove a construction or establish mastery. All five Calculus source rows have structural teaching and staged-practice links; exhaustive subskill approval, specialist approval and cumulative formal assessment remain unestablished. The wider Maths A course and other Pearson qualifications remain incomplete. Publisher modular-map tags and uncorrected proof navigation are separate evidence: Book2 Graphs9 map pages386–415 differs from proof start385 and its later calculus subheadings. Full final-book and Teaching Hub body/pagination alignment remains unverified; chapter links confer no full-book or modular teaching credit.
Before you begin
- Differentiate polynomial sums twice.
- Factor quadratics and interpret signs on intervals.
- Read displacement and velocity graphs and distinguish coordinate position from total travel.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A particle has velocity −6 m/s. Which speed and direction statement is correct?
What you will learn
- Differentiate one-dimensional displacement to velocity and velocity to acceleration with correct units.
- Find permitted rest times and use signed velocity to distinguish reversal from momentary rest.
- Separate displacement from distance and speed from velocity, including zero-acceleration and endpoint cases.
s(t)=−t³ + 9t². Original domain: t in [0, 8]. Original values use m; first rates use m/s and acceleration uses m/s².
The panels share the same numbered horizontal scale and have independent vertical scales. Original height, derivative value and derivative zero have different meanings. No input, derivative solution or reference point is preselected.
No input is preselected. Choose a point on your own drawing or a rate input you want to check.
Blue: original; purple: first derivative; ochre: acceleration. Vertical scales differ. Filled boundary markers are included; open markers show excluded finite limits or holes. Traces split at original exclusions and are clipped to each panel; offscreen points are omitted rather than moved to a border. A wider view does not change a domain.
Position, velocity and acceleration have different meanings
In straight-line motion, displacement s records a signed coordinate from a fixed origin O. It can be negative when the particle lies on the chosen negative side. Velocity v=ds/dt is the signed rate of change of that coordinate, and acceleration a=dv/dt is the signed rate of change of velocity. With s in metres and t in seconds, their units are m, m/s and m/s² respectively.
Differentiate the displacement polynomial once to get velocity and again to get acceleration. Keep their graphs separately labelled: a height on the displacement graph is a position, not a speed or an acceleration. A practical time interval remains part of the model even if the polynomial expressions make algebraic sense outside it. At a boundary use the permitted side when interpreting a physical rate.
Rest and direction depend on velocity
The particle is at rest at a permitted time when v=0. Solve that equation, then apply the time domain. A root at the initial endpoint is initial rest, not evidence of an earlier movement or reversal. At an interior rest time inspect velocity on both sides: a sign change reverses direction, while the same sign on both sides gives a momentary stop without reversal.
Speed is |v|, so it cannot be negative. Zero acceleration means the instantaneous velocity rate is zero, not necessarily that velocity or speed is zero. Negative acceleration does not always mean slowing down: when velocity and acceleration are both negative, the particle moves in the negative direction with increasing speed. Relate these statements to the displacement and velocity graphs.
Distance is not the final displacement
The change in displacement from time a to b is s(b)−s(a), a signed net result. Total distance counts all travel positively. When direction reverses, split the interval at the known rest/reversal times and add the absolute changes of displacement on the separate one-direction pieces. This uses positions from the model and does not require an integration method.
A graph of displacement can rise, fall or pass through a negative coordinate; cumulative distance cannot decrease. Keep the original time interval and any model assumptions in all interpretations. A computed stationary input outside it is not a physical rest time, and a tangent or polynomial extension outside the model is not an added motion prediction.
Sources for this lesson
Pause and explain
On both sides of a permitted rest time, v(t) is positive. What follows?
Worked example
For displacement s(t)=9t²−t³ metres from a fixed origin on 0≤t≤8 seconds, derive velocity and acceleration and evaluate both at t=2 s. Find every permitted rest time, describe the later change of direction and mark the relevant displacement and velocity points. Compare the net displacement change and total distance over 0≤t≤8 using the positions at reversal times.
Show the worked solution
- Differentiate with respect to seconds: v(t)=18t−3t² m/s and a(t)=18−6t m/s². At t=2, v=36−12=24 m/s, so speed is 24 m/s, and a=18−12=6 m/s². Keep displacement, velocity and acceleration on separately labelled graphs over the stated time interval.
- Solve v=3t(6−t)=0: permitted rest times are t=0 and t=6 s. On 0<t<6 velocity is positive; on 6<t≤8 it is negative. The particle reverses at 6 s. The t=0 root is initial rest at an endpoint, with no modeled time before it from which to claim a reversal.
- The positions are s(0)=0, s(6)=108 and s(8)=64 m. Mark the original stationary point (6,108) separately from the velocity zero (6,0). Net displacement change is 64−0=64 m, while distance is |108−0|+|64−108|=108+44=152 m. The returning part adds distance even though it reduces displacement.
Answer v=18t−3t² m/s; a=18−6t m/s². At 2 s: velocity 24 m/s, speed 24 m/s, acceleration 6 m/s². Rest: 0 and 6 s; reversal at 6 s. Net displacement change over the model is 64 m; total distance is 152 m.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A particle has displacement s(t)=4t³−18t²+24t+1 metres for 0≤t≤3 seconds. Derive v and a, find every rest time and the direction on each intervening interval. At t=1.5 s give velocity, speed and acceleration and explain why zero acceleration is not rest here. Show the original stationary heights separately from velocity zeros.
Give me a hint
Differentiate twice, factor v and test its sign between its roots. Speed is the absolute value of velocity; acceleration is a different quantity.
Compare my reasoning
- Differentiate: v=12t²−36t+24=12(t−1)(t−2) m/s and a=24t−36 m/s². Both rest times, 1 s and 2 s, are inside the stated domain.
- Velocity is positive on [0,1), negative on (1,2) and positive on (2,3]. Thus the particle reverses at both rest times. The original stationary positions are s(1)=11 m and s(2)=9 m, so mark (1,11),(2,9) on displacement and (1,0),(2,0) on velocity.
- At t=3/2, v=12·9/4−36·3/2+24=27−54+24=−3 m/s. Speed is 3 m/s and acceleration is 24·3/2−36=0 m/s². It is still moving in the negative direction; zero acceleration does not imply zero velocity.
v=12(t−1)(t−2) m/s, a=24t−36 m/s². Rest at 1 and 2 s, with directions positive, negative, positive. At 1.5 s: velocity −3 m/s, speed 3 m/s, acceleration 0 m/s².
Look for these in your work
- Derive both rates and retain every permitted rest time.
- Use velocity signs for direction and correctly label the original stationary heights.
- Distinguish negative velocity, positive speed and zero acceleration with units.
For displacement s(t)=t³−6t²+12t−2 metres on 0≤t≤4 seconds, derive v and a and find every permitted rest time. Show the original stationary position and velocity zero separately, and explain whether the particle reverses there. State what the initial negative displacement means, without calling it negative distance.
Give me a hint
Factor the velocity before deciding what rest means. Check its sign on both sides of a root and return to s for the original stationary position.
Compare my reasoning
- Differentiate: v=3t²−12t+12=3(t−2)² m/s and a=6t−12 m/s². The only distinct rest time is t=2 s, which belongs to the model interval.
- At that time, s(2)=8−24+24−2=6 m, v(2)=0 and a(2)=0. Mark (2,6) on the displacement curve and (2,0) on the velocity graph. The displacement is s=(t−2)³+6, an increasing cubic that flattens at this point without turning.
- For permitted times on either side of 2, the squared velocity is positive, so there is no reversal. The particle stops momentarily and continues in the positive direction. The initial value s(0)=−2 m is a coordinate 2 m on the negative side of O, not a negative amount of travel.
v=3(t−2)² m/s, a=6t−12 m/s². Rest only at 2 s, at position 6 m, with speed and acceleration both zero there but no reversal. Initial displacement −2 m describes position, not negative distance.
Look for these in your work
- Find the single distinct permitted rest time and its original position.
- Use velocity signs and the original rising-through shape to explain no reversal.
- Keep displacement, speed and acceleration meanings and units separate.
For s(t)=15t−t² metres on 0≤t≤10 seconds, derive velocity and acceleration, locate rest and the maximum displacement, and give velocity and speed at t=8 s. Use the original displacement shape to justify the maximum. Explain why the negative acceleration does not mean decreasing speed throughout the model.
Give me a hint
The original displacement is a downward-opening parabola. Solve v=0 within the time interval, then compare the signs of velocity and acceleration before and after that time.
Compare my reasoning
- Differentiate: v=15−2t m/s and a=−2 m/s². Rest occurs at 15−2t=0, so t=15/2=7.5 s, inside the domain. Velocity is positive before that time and negative after it.
- The original downward parabola rises to then falls from this stationary point. Its maximum displacement is s(15/2)=225/2−225/4=225/4 m, at 7.5 s. Mark this original point separately from the velocity zero. The time domain stays 0≤t≤10.
- At t=8, v=15−16=−1 m/s, so speed is 1 m/s and acceleration is still −2 m/s². Before rest the positive velocity decreases toward zero, so speed falls; after rest the increasingly negative velocity has increasing magnitude, so speed rises. Negative acceleration alone does not determine whether speed is increasing or decreasing.
v=15−2t m/s, a=−2 m/s². Rest and maximum displacement at 7.5 s, with s=225/4 m. At 8 s velocity is −1 m/s and speed 1 m/s; speed is increasing then.
Look for these in your work
- Derive velocity and acceleration with correct units and time domain.
- Use the original shape and permitted rest time to find maximum displacement.
- Distinguish velocity from speed and explain the effect of negative acceleration before and after reversal.
Common mistakes
- Calling signed displacement a cumulative distance, or giving a negative speed.
- Treating acceleration zero as rest, or assuming negative acceleration always reduces speed.
- Accepting a rest time outside the model or assuming an endpoint rest or repeated velocity root must be a reversal.
What can you explain now?
At an instant v=−4 m/s and a=0 m/s². Is the particle at rest, and what is its speed?
Compare with the explanation
It is moving, with speed 4 m/s.
Rest requires velocity zero. Zero acceleration says the velocity is instantaneously not changing; it does not make its current value −4 vanish.
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Make it stick
Explain s→v→a with units, then distinguish rest from zero acceleration and reversal from a repeated rest root. Show how total distance can exceed the net displacement change.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.