Overview
What you’ll learn
- Use the kinematics relations v = ds/dt and a = dv/dt = d²s/dt².
- Find velocity and displacement by integrating acceleration and velocity, using initial conditions for the constants.
- Distinguish displacement from distance, and velocity from speed.
- Find when a particle is at rest or changes direction by solving v = 0.
- Find total distance travelled by splitting the motion at rest instants.
- Find maximum velocity, maximum displacement and other optimisation facts about a motion.
In 30 seconds
- 1The chain of calculus: v = ds/dt, a = dv/dt = d²s/dt² — differentiate down the chain, integrate up it.
- 2Going up needs initial conditions: each integration adds a constant fixed by a given value of v or s.
- 3Displacement is position relative to the start, with direction; distance is the total path length and is never negative.
- 4Velocity carries a sign; speed = |velocity|.
- 5The particle is at rest when v = 0; a direction change needs v = 0 and a sign change of v.
- 6Total distance: split the time interval at every rest instant and add the absolute displacement of each leg.
- 7Displacement on [0, T] is simply s(T) − s(0) — no splitting.
- 8Standard units: metres and seconds, velocity m/s, acceleration m/s².
Notes · 1 of 5
Displacement, velocity, acceleration
The vocabulary and the sign conventions that decide half the marks.
A particle moves along a straight line with a chosen positive direction. Its displacement s is position measured from the start: s = −3 means 3 units on the negative side. Velocity v is the rate of change of displacement, so it inherits the sign; speed is |v|. Acceleration a is the rate of change of velocity. Distances and speeds are never negative — displacements and velocities can be.
| Relation | Differentiate (↓) | Integrate (↑) |
|---|---|---|
| s = f(t) | v = ds/dt | s = ∫v dt + C₁ |
| v = g(t) | a = dv/dt | v = ∫a dt + C₂ |
| a = h(t) | — | a = d²s/dt², and s = ∫∫a with two constants |
Differentiating goes down the chain without extra information. Integrating goes up, but every integral introduces a constant that only the given initial conditions can pin down: "the particle starts from rest" means v = 0 at t = 0; "starts 5 m from O" means s = 5 at t = 0.
Notes · 2 of 5
Differentiating displacement and velocity
Two derivatives turn s into v into a — with rest instants falling out of v = 0.
Given s(t), differentiate once for velocity and twice for acceleration. Questions then ask for instants of rest (solve v = 0), direction changes (v = 0 with a sign change of v), or maximum displacement (a stationary-value problem: solve v = 0, then check the nature). This is Chapter 4 in motion clothing — the classification skills transfer unchanged.
Notes · 3 of 5
Integrating acceleration and velocity
Integrate up the chain, then let the initial conditions name the constants.
Given acceleration: v = ∫a dt + C₂, and then s = ∫v dt + C₁. Given velocity directly, skip straight to s = ∫v dt. Each integration constant comes from one initial condition — a stated velocity or position at a known time. Write the condition as an equation ("v = 3 when t = 0") and solve for the constant immediately; carrying two unknown constants at the end is where errors breed.
Notes · 4 of 5
Distance versus displacement
One subtraction, or a split at every rest instant — know which the question wants.
Displacement over [0, T] is s(T) − s(0), one subtraction, sign included. Total distance is the length of the actual path: find every rest instant in the interval, evaluate s at 0, at each rest instant and at T, then add the absolute changes between consecutive times. If the particle never rests, distance equals |displacement|.
Velocity–time descriptions work the same way: area between the v-graph and the t-axis is displacement, and total distance splits the area at the zero crossings of v, flipping below-axis pieces — exactly the area-below-the-axis technique from Chapter 5.
Notes · 5 of 5
Full motion studies
Assembling the pieces — rest instants, direction, maxima — into one coherent answer.
Long kinematics questions reward a fixed order of operations. (1) Write down v = ds/dt and solve v = 0 for the rest instants. (2) Check the sign of v either side of each rest instant to locate direction changes. (3) Evaluate s at t = 0, at every rest instant, and at the time asked about. (4) Answer precisely: displacement is one subtraction; distance is the sum of the legs. The same protocol finds a maximum velocity (optimise v(t) itself) or a maximum displacement (optimise s via v = 0).
Revise
Key terms
The words examiners expect you to use precisely.
- Displacement
- Position measured from the starting point in a stated positive direction; it can be negative.
- Distance
- The total length of path travelled, regardless of direction; never decreases.
- Velocity
- Rate of change of displacement: v = ds/dt; negative velocity means motion in the negative direction.
- Speed
- The magnitude of velocity, |v|; always non-negative.
- Acceleration
- Rate of change of velocity: a = dv/dt = d²s/dt².
- At rest
- An instant when v = 0; the particle is momentarily stationary.
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- Initial conditions
- Given values of v and s at a particular time (often t = 0) used to fix the constants of integration.
- Motion study
- A full description of a motion: rest instants, direction changes, displacement, distance, maximum velocity or displacement.
Revise
Common mistakes
Where students most often lose marks in this chapter — and what to do instead.
Mistake: Treating distance and displacement as the same quantity.
Displacement is position relative to the start with direction; distance is the whole path length.
Instead: Displacement = s(T) − s(0); distance = sum of absolute changes between rest instants.
Mistake: Forgetting the constants of integration when finding v from a or s from v.
An uncounted family of motions all share the same acceleration; only the initial condition picks out the real one.
Instead: Add + C, then substitute the given initial condition immediately.
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Mistake: Applying the initial condition before integrating.
Substituting numbers freezes the formula: ∫a(t₀) dt is a constant, not a function.
Instead: Integrate symbolically first, then fix the constant with the condition.
Mistake: Solving v = 0 and calling every solution a direction change.
v = 0 is only an instant of rest; the direction changes only if v switches sign there.
Instead: Check the sign of v on both sides of the root (or quote the sign change) before claiming a reversal.
Mistake: Adding displacements instead of distances for total distance.
Negative displacement legs cancel positive ones, undercounting the path length.
Instead: Take the absolute value of each leg between consecutive rest instants, then add.
Mistake: Splitting the motion at the wrong times.
Total distance splits where v = 0 — the rest instants — not at arbitrary or endpoints-only times.
Instead: Solve v = 0, keep the roots inside the interval, evaluate s at 0, each root, and T.
Mistake: Mixing up velocity and acceleration when the question gives v in terms of t.
If v(t) is given, acceleration is one derivative away — but displacement needs an integral, not another derivative.
Instead: Identify which quantity you hold and which is asked; move one step at a time along the s–v–a chain.
Mistake: Dropping units, or answering a distance in metres per second.
Kinematics answers carry physical units: m for s and distance, m/s for v, m/s² for a.
Instead: State units with every final answer.
Mistake: Using t = 0 conditions as "the particle starts from rest at the origin" without checking.
Some motions begin at a non-zero position or velocity; assuming s(0) = 0 invents data.
Instead: Use only the conditions actually given; s(0) may be 12 and v(0) may be 5.
Mistake: Reporting the time of maximum displacement without the displacement value (or vice versa).
The question asks for a quantity, not just an instant — and the maximum needs its nature justified.
Instead: Solve v = 0, justify the maximum (sign change or second derivative), then state both the time and the value.
Check yourself
Quick quiz
15 questions with instant explanations. About 5 minutes.
Question 1 of 3
- Question 1: current question
- Question 2: not answered
- Question 3: not answered
Recall the calculus chain
In kinematics notation, v equals
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Practise
Topical questions
Exam-style practice in three tiers, from recall to data questions.
- 4Foundation
- 3Standard
- 2Challenging
9 questions worth 47 marks, each with a model answer, mark scheme and examiner tips.
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Practice paper
A 30-mark paper set out like a school paper, with a separate mark scheme. Give yourself 40 min.
Question paper
Chapter 6 Practice Paper
30 marks · 40 min · 5 pages
5 multiple-choice questions and structured questions with answer lines. Chapter 6 (C1 Kinematics: calculus for straight-line motion).
Mark scheme
PROChapter 6 Practice Paper — mark scheme
Answers · Marking points · 3 pages
Model answers with a mark for each creditworthy point — for marking at home or in class.
Where this chapter is tested
Each school sets its own scope — check your assessment outline.
Questions
Frequently asked questions
What does Sec 4 A-Math Chapter 6 cover?
The kinematics part of syllabus sub-topic C1 of O-Level 4049 / SEC G3 K341: using differentiation and integration for straight-line motion — v = ds/dt, a = dv/dt, v = ∫a dt, s = ∫v dt — including initial conditions, rest instants and total distance. Differentiation is Chapter 4 and integration is Chapter 5.
How do I find total distance travelled?
Solve v = 0 to find every rest instant inside the time interval, evaluate the displacement at the start, at each rest instant and at the end, then add the absolute sizes of the legs. Do not add signed displacements.
Where do the constants of integration come from?
From the initial conditions stated in the question — a given velocity or position at a known time, often t = 0. Each integration introduces one constant, so a velocity-from-acceleration question needs one condition and a displacement-from-acceleration question needs two.
Does v = 0 always mean the particle turned around?
No. It means the particle is momentarily at rest. It turned around only if v changes sign through that instant — check the sign of v on both sides.