Kinematics in Cambridge A Level Mathematics (9709) is about a particle moving in a straight line, and almost every question is one of two kinds. Either the acceleration is constant, and you work from a velocity-time graph or the constant acceleration equations, or the acceleration changes with time, and you differentiate or integrate. Most marks are lost by using the wrong tool for the kind, or by misreading what a phrase such as "returns to O" asks for. This post shows what the Mechanics mark schemes credit, walks through one question in full, and gives you three real questions to try.
The two kinds of question
Constant acceleration. A velocity-time graph made of straight line segments is the usual setting. The gradient of a segment is the acceleration, and the area under the graph is the displacement. You can also use v = u + at and the other constant acceleration equations, one segment at a time.
Variable acceleration. The question gives v or a in terms of t. The connections are calculus:
$$a = \frac{dv}{dt}, \qquad v = \frac{ds}{dt}$$Going down the chain from displacement to acceleration, you differentiate. Going up, you integrate, and you need a constant of integration or limits.
The test is simple: if the acceleration depends on t, it is not constant, and the constant acceleration equations do not apply. A velocity such as 44t - 6t^2 - 36 has a t^2 term, so its acceleration changes with time.
What the mark schemes reward
9709 schemes use M marks for method, A marks for accuracy that depend on the method, and B marks that stand alone. In kinematics this has a clear practical effect.
The method mark is often just an attempt. Paper 42, May/June 2024 gave M1 for "Attempt to differentiate given v" and later M1 for "Attempt to integrate given v". Paper 43, May/June 2024 gave M1 "For attempt at integration". Showing the attempt clearly, even if you slip later, is worth a mark.
Intermediate values are B marks. In Paper 43, May/June 2025, a velocity-time graph question gave B1 for the velocity at t = 20 being 20 m s⁻¹ and B1 for the speed at t = 50 being 10 m s⁻¹, before any areas. Write such values down as separate lines.
Follow-through on "their" values. The same scheme gave a method mark for the displacement written with "their 20". If your velocity is wrong but your area method is right, the method mark can still be earned. This only works if the examiner can see the method, so write the area expression, not just a number.
Correct answers only, where stated. Some accuracy marks say "ONLY", as in "t = 2, 9 (and 0) ONLY". Extra wrong values cost the mark.
For a fuller explanation of the codes, see how Cambridge mark schemes work.
Phrases that tell you what to do
| The question says | Which means |
|---|---|
| at instantaneous rest | v = 0 |
| returns to O | displacement s = 0 |
| acceleration is positive | \frac{dv}{dt} > 0 |
| minimum or maximum velocity | \frac{dv}{dt} = 0 |
| distance travelled | add sizes of displacements, splitting where v = 0 |
| no change in velocity at t = 1 | the two velocity expressions are equal at t = 1, which finds a constant |
The last row comes from Paper 42, Oct/Nov 2024, where acceleration was given in two pieces. The mark for the second constant was for using the velocity of 5 m s⁻¹ at t = 1, found in the first part.
Setting out a velocity-time graph question
For a graph made of straight segments, work in the same order every time. First, write the velocity at every corner of the graph, using v = u + at or the gradient. Second, write the area of each segment as its own expression, using a triangle, a rectangle or a trapezium. Third, form the equation the question needs, such as total distance equal to a given value.
Paper 42, Oct/Nov 2024, Question 1 shows why. A particle accelerates from rest to V in 4 s, then decelerates at \frac{5}{3} m s⁻² to rest in 6 s. The first M1 was for using v = u + at, or the gradient, to get V = 10. The method mark in the next part was for an area equation written with "their |V|", so a wrong V there did not cost the method mark. Corners first, then areas, then the equation.
Worked walk-through: Paper 42, May/June 2024, Question 2
A particle P moves in a straight line. At time t s after leaving O, its velocity is
$$v = 44t - 6t^2 - 36$$The question is worth 5 marks.
(a) Find the set of values of t for which the acceleration is positive, 2 marks.
Differentiate to get the acceleration. That attempt is the M1.
$$a = \frac{dv}{dt} = 44 - 12t$$Set 44 - 12t > 0, which gives t < \frac{11}{3}. That is the A1. A common slip is to solve v > 0 instead, which answers a different question.
(b) Find the two values of t at which P returns to O, 3 marks.
"Returns to O" means displacement zero, so integrate v. The attempt is M1, and the correct expression is A1:
$$s = 22t^2 - 2t^3 - 36t + c$$At t = 0 the particle is at O, so c = 0. Now set s = 0:
$$-2t(t^2 - 11t + 18) = 0 \quad\Rightarrow\quad -2t(t - 2)(t - 9) = 0$$So t = 0, 2 or 9. The scheme's final A1 is for "t = 2, 9 (and 0) ONLY". The particle starts at O at t = 0, so the two returns are at t = 2 and t = 9.
Notice the order of thinking: read the phrase, turn it into a condition, then choose differentiate or integrate. Try it yourself and have each mark checked against the scheme.
Common ways marks are lost
- Using suvat with variable acceleration. If a depends on t, constant acceleration equations give the wrong answer and earn nothing.
- Dropping the constant of integration. It is often zero, but say why: the particle is at O when t = 0.
- Confusing distance with displacement. If the velocity changes sign, split the motion where v = 0 and add the sizes.
- Answering for velocity when the question asks about acceleration. In part (a) above, the condition is on \frac{dv}{dt}, not on v.
- Keeping t = 0 as a "return". The question asked for two values. Extra values can cost an A mark marked ONLY.
- Reading a graph area as one shape when it is two. Split trapeziums into a rectangle and a triangle if that is safer, and show each area.
Two more real questions to try
Paper 43, May/June 2025, Question 3 is a velocity-time graph with three straight segments. The particle accelerates, decelerates through zero and travels back to O. You find the time T when it gets back, which needs the area above the axis to equal the area below it. The scheme's answer is T = 130. Then you find the acceleration on the last segment, given as \frac{1}{8} m s⁻². Write down the B mark values first.
Paper 43, May/June 2024, Question 4 gives v = kt^2 - 4t + 3 and the distance travelled in the first 2 s. You integrate, substitute the limits and solve for k, then find the minimum velocity by setting \frac{dv}{dt} = 0. The question tells you that you do not need to show that it is a minimum, so do not spend time on it.
You do not need an account to try these: as a guest, each question part can be marked twice for free. Write your working by hand or type it, and the marking shows which M, A and B marks you earned.
Practise more kinematics questions
The free 9709 kinematics topical page lists 58 structured kinematics questions from 21 papers, from 2023 to 2026, each linked to its paper. Working through them in order of year is a good way to see the same phrases come back. For full papers, open the A Level Mathematics past papers and choose a Mechanics paper.
Frequently asked questions
- When do I use suvat and when do I use calculus in 9709 kinematics?
- Use the constant acceleration equations, or areas and gradients on a velocity-time graph, only when the acceleration is constant. If the acceleration depends on t, for example when the velocity has a t squared term, the acceleration is changing, so differentiate or integrate instead.
- What does returns to O mean in a kinematics question?
- The displacement from O is zero again. Integrate the velocity to get the displacement, set it equal to zero and solve. The solution t = 0 is the start of the motion, so the question usually wants the other values.
- How do I find distance rather than displacement?
- If the particle changes direction, the displacement can be smaller than the distance travelled. Find where the velocity is zero, split the motion there, and add the sizes of the separate displacements. On a velocity-time graph, add the areas above and below the axis as positive amounts.
- Do I need to include the constant of integration?
- Yes, whenever you integrate without limits. Find it from a known value, such as the displacement being zero at O when t = 0, or a velocity given at a particular time.