Mathematics 9709 · syllabus 4.2
Cambridge A Level Mathematics 9709 — Kinematics of motion in a straight line topical past papers
Displacement, velocity and acceleration in a straight line, from the constant-acceleration formulae to motion graphs and calculus with time.
- Questions
- 215
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 75
Anyone can browse these pages. Opening the questions and their mark schemes takes a free Momentum account.
What Kinematics of motion in a straight line covers
Kinematics (section 4.2 of Paper 4) describes motion in one dimension without asking what causes it. Distance and speed are scalars; displacement, velocity and acceleration are vectors, so their signs matter. When the acceleration is constant you use the constant-acceleration formulae, choosing the one that links the three quantities you know with the one you want.
Graphs carry much of the topic. The gradient of a displacement–time graph is velocity, the gradient of a velocity–time graph is acceleration, and the area under a velocity–time graph is displacement. When the acceleration varies with time, you differentiate displacement to get velocity and acceleration, and integrate the other way, using only the calculus of Paper 1. Questions often involve two particles, one setting off after the other, and ask when or where they meet.
Kinematics supplies the second half of most Newton's laws questions: the forces give the acceleration, and these formulae turn it into distances and times. The calculus is limited to Paper 1 techniques, so velocity and acceleration are given as polynomials or powers of t. Velocity–time graphs also help in energy and momentum questions with several stages. Projectiles are not part of the current 9709; all motion here is along a straight line.
In the syllabus
The headings this topic has in the Cambridge International 2026–2027 syllabus, as the board prints them.
- 4.2 Kinematics of motion in a straight line
What you should be able to do
- Choose and use the constant-acceleration formulae, including for vertical motion under gravity
- Sketch and interpret displacement–time and velocity–time graphs
- Find displacement as the area under a velocity–time graph
- Differentiate and integrate with respect to time when the acceleration is not constant
- Set up equations for two particles to find when and where they meet
- Tell distance from displacement when a particle changes direction
Key ideas
- v = u + at; s = ut + ½at²; s = ½(u + v)t; v² = u² + 2as; s = vt − ½at²
- v = ds/dt and a = dv/dt; s = ∫v dt and v = ∫a dt
- The area under a velocity–time graph is the displacement, and its gradient is the acceleration
- A particle is instantaneously at rest when v = 0, which is where it may change direction
How Kinematics of motion in a straight line has been examined
From 2022 to 2026, Kinematics of motion in a straight line was on Paper 4 · Mechanics in 11 of 11 exam series, worth about 37% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 4 · Mechanics | 75 | 37% | 11 of 11 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 8.
Filed in Momentum under: Kinematics of Motion in a Straight Line, General Motion in Straight Line.
A worked example
A particle moves in a straight line so that t seconds after leaving O its velocity is v = 6t − 3t² m s⁻¹. Find when the particle is next at rest, and the total distance it travels in the first 3 seconds.
- v = 0 when 3t(2 − t) = 0, so the particle is next at rest at t = 2.
- Integrate: s = 3t² − t³ + c, and s = 0 when t = 0, so c = 0.
- At t = 2, s = 12 − 8 = 4 m; at t = 3, s = 27 − 27 = 0 m.
- The particle moves 4 m away from O, turns round at t = 2, and returns 4 m to O.
- Total distance = 4 + 4 = 8 m, even though the displacement at t = 3 is zero.
Answer: At rest when t = 2 s; total distance 8 m
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Using the constant-acceleration formulae when the acceleration depends on t: they only apply when a is constant, so otherwise use calculus.
- Giving the final displacement as the total distance when the particle turns back: find when v = 0 and add the distances for each stage.
- Mixing signs in vertical motion: if upwards is positive, the acceleration due to gravity is −10 m s⁻².
- Leaving out the constant of integration when finding v or s, which loses the initial velocity or position.
Exam technique
- Start each stage by listing s, u, v, a and t with the known values and a chosen positive direction, then pick the formula.
- Sketch a velocity–time graph for a journey in several stages; areas of trapeziums and triangles often replace several formulae.
- Use the initial conditions to find constants of integration, and say so, for example 'when t = 0, s = 0, so c = 0'.
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.