Speed-time graph questions in IGCSE Physics (0625) come down to two rules: the gradient of the line is the acceleration, and the area under the line is the distance travelled. Most of the marks then go to applying those rules carefully, with the right numbers read off the graph, the right units, and a description that uses the words the mark scheme expects. This post shows what recent Paper 4 mark schemes credit, walks through one full question, and gives you three real questions to try and have marked.
What the mark schemes reward
The definition of acceleration. Paper 42, Oct/Nov 2023 asked for it "in words". The scheme accepted "change / increase in velocity per unit time OR rate of change of velocity". Note the "per unit time". "How fast something speeds up" is not a definition.
Gradient for acceleration. For a straight line, acceleration equals the change in speed divided by the time taken. The scheme for a braking car in Paper 43, May/June 2025 gave a mark for "deceleration = gradient" or \Delta v \div t written out, before any number.
Area for distance. This is the most frequently credited idea. Paper 41, Oct/Nov 2024 asked directly which feature of the graph represents distance, and the mark was for the "area under the (graph) line". In calculation parts, a mark is given for writing "area under the speed-time graph" or "average speed × time taken", even before the arithmetic.
Describing the graph. In Paper 41, May/June 2025 the motion of a train had to be described for two marks. One mark needed the full sequence, "accelerates AND constant speed AND decelerates". The other was for saying the acceleration or deceleration was constant. For a curved line, Paper 41, Oct/Nov 2024 credited "gradient is changing" or simply that the line is curved, as evidence that the deceleration is not constant.
Vectors and scalars. Velocity and acceleration are vectors, and speed is a scalar. A vector "has direction (as well as magnitude)". These come up as one-mark starters.
How the calculation marks work
0625 schemes label the final answer with A and give C marks for working. In Paper 42, Oct/Nov 2023, the acceleration calculation was marked A2 for the answer 4.1 m/s², with C1 for "(a =) (∆)v / (∆)t OR 13(.0) / 3.2". So a correct answer earns both marks. A wrong answer with the correct equation or substitution can still earn one. That is the practical reason to write the equation down every time: it costs a few seconds and protects a mark.
The same pattern holds for distance. In Paper 41, May/June 2025, the distance between two stations was worth three marks. The correct answer earned all three (A3). A wrong answer could still collect one C1 for "area under the speed-time graph" or "average speed × time taken", and another for the correct areas written out with numbers.
Areas made of more than one shape
Most journeys on a speed-time graph have three sections: speeding up, constant speed and slowing down. The area under the whole line is then a triangle, a rectangle and another triangle. The scheme for the train in Paper 41, May/June 2025 shows both ways of setting it out. You can add the three areas,
$$\tfrac{1}{2} \times 80 \times 56 \;+\; 56 \times 400 \;+\; \tfrac{1}{2} \times 50 \times 56$$or treat the whole shape as a trapezium, with the two parallel sides being the total time and the time at constant speed:
$$\tfrac{1}{2} \times (400 + 530) \times 56$$Both give 26 000 m, which the scheme also accepts as 26 km. Pick whichever you are less likely to get wrong, and write the areas with their numbers before you add them, because that line is a mark on its own.
Worked walk-through: Paper 42, Oct/Nov 2023, Question 1
A car accelerates uniformly from rest at t = 0. At t = 3.2 s its speed is 13.0 m/s. The question is worth 13 marks.
(a)(i) Calculate the acceleration, 2 marks. Write the equation, then substitute:
$$a = \frac{\Delta v}{\Delta t} = \frac{13.0}{3.2} = 4.06 \text{ m/s}^2$$The scheme's answer is 4.1 m/s². The equation alone is the C1.
(a)(ii) Explain acceleration in words, 1 mark. Change in velocity per unit time.
(b)(i) Plot the speed-time graph, 2 marks. The car keeps 13.0 m/s from 3.2 s to 12.0 s. The marks are for a straight line joining (0, 0) to (3.2, 13.0), and a horizontal line from 3.2 s to 12.0 s. Use a ruler and plot the corner point accurately.
(b)(ii) Distance from 0 to 3.2 s, 2 marks. This is the area of a triangle:
$$\text{distance} = \tfrac{1}{2} \times 3.2 \times 13.0 = 20.8 \text{ m}$$The scheme gives 21 m. Writing "area under the graph" or "average velocity × time" earns the C1 if the arithmetic slips.
(c) Show that the work done while braking is about 1.1 \times 10^5 J, 4 marks. The car of mass 1350 kg stops from 13.0 m/s in 2.0 s, travelling 13 m. Each step is a separate mark: the equation W = Fd, the equation F = ma, the force 1350 \times 6.5 = 8775 N (the deceleration is 13.0 \div 2.0 = 6.5 m/s²), and then 8775 \times 13.0 giving about 1.1 \times 10^5 J. In a "show that" question the final value is given to you, so every mark is for the working. Skip a line and you lose that mark.
(d) Suggest and explain why the stopping distance increases, 2 marks. One mark for a sensible suggestion that increases the stopping distance, and one for an explanation that matches it.
Try the whole question yourself, including the graph, and have it marked against the official scheme.
Common ways marks are lost
- Using the gradient when the question wants distance. Distance is area. Gradient is acceleration. Check which one the question asks for before you start.
- Reading the scale wrongly. Count what one small square is worth on each axis before you read any value.
- Forgetting the triangle is half. The area of the acceleration section is \frac{1}{2} \times base \times height, not base \times height.
- Half a description. "It speeds up then slows down" misses the constant speed section and the word "constant" for the acceleration, which were separate marks.
- No unit, or the wrong one. Acceleration is m/s², distance is m.
- Skipping lines in "show that" questions. The answer is given, so the marks are all in the steps.
Two more real questions to try
Paper 41, Oct/Nov 2024, Question 2 is about a drag car with a rocket engine and a parachute. It asks for the maximum momentum, the feature of the graph that represents distance, and the distance in the first 6.5 s, which is the area of a triangle: \frac{1}{2} \times 130 \times 6.5, given as 420 m in the scheme. It then asks how the graph shows that the car decelerates and that the deceleration is not constant, and finishes with the energy transfer as the car slows, from the kinetic store to the internal (thermal) store.
Paper 41, May/June 2025, Question 1 gives a train's speed-time graph between two stations. You read off the maximum speed, describe the motion for two marks, then calculate the distance between the stations from three areas added together. The last part asks what change to the motion would make the journey take longer, and the scheme accepts a lower acceleration, a lower deceleration or a lower maximum speed.
You do not need an account to try these: as a guest, each question part can be marked twice for free. If the A and C letters in your marked answer are new to you, how Cambridge mark schemes work explains them.
Practise more motion questions
The free 0625 motion topical page lists 119 motion questions from 79 papers, from 2023 to 2026, each linked to its paper. Most are multiple choice, and the 43 structured ones are the best practice for the graph calculations above. For full papers, open the IGCSE Physics past papers and choose a Paper 4.
Frequently asked questions
- How do I find distance from a speed-time graph?
- Find the area under the line. Split the shape into triangles and rectangles, or use a trapezium, and add the areas. Recent 0625 mark schemes give a compensation mark for writing that distance is the area under the speed-time graph, or average speed times time, even if the final number is wrong.
- How do I find acceleration from a speed-time graph?
- Acceleration is the gradient: the change in speed divided by the time taken for that change. A negative gradient means the object is decelerating. If the line is curved, the acceleration is changing.
- What definition of acceleration do the mark schemes accept?
- A recent 0625 scheme accepted change in velocity per unit time, or rate of change of velocity. Saying acceleration is speeding up is not a definition and does not earn the mark.
- Do I lose marks if I do not show working in 0625 calculations?
- In the calculation parts of recent 0625 schemes, the correct final answer carries the A marks for the whole part. Working matters when the answer is wrong, because the scheme then awards C marks for a correct equation or substitution. Always write the equation first.