Mathematics 0580 · syllabus C2.9, E2.9
Cambridge IGCSE Mathematics 0580 — Distance–time and speed–time graphs topical past papers
Reading and drawing travel and conversion graphs, speed as the gradient of a distance–time graph, and acceleration and distance from speed–time graphs.
- Questions
- 122
- Years
- 2009–2026
- With mark scheme
- 88%
- Questions 2022–2026
- 31
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What Distance–time and speed–time graphs covers
Graphs in practical situations show how one quantity changes with another. On a distance–time graph the gradient is the speed: a steeper line means faster travel, a horizontal line means the object has stopped, and a line sloping down means it is heading back towards the start. You read times and distances from travel graphs, draw them from a description of a journey, and use conversion graphs, such as one between kilometres and miles or between two currencies, in both directions.
Extended candidates also use speed–time graphs. Their gradient is the acceleration, negative for a deceleration, and the area under the graph is the distance travelled; for sections made of straight lines, that area splits into rectangles, triangles and trapeziums. Questions ask for an acceleration, a total distance, an average speed, or an unknown time or speed that makes the distance a given value. On a curved graph you estimate the rate of change at a point by drawing a tangent and finding its gradient.
Travel and conversion graphs are Core content, so they can be set on all four papers; speed–time graphs, acceleration and the area under the graph are Extended, examined only on Papers 2 and 4. The topic uses gradient from coordinate geometry and the formula for average speed, and the idea of a gradient as a rate of change leads into differentiation.
In the syllabus
The headings this topic has in the Cambridge International 2025–2027 syllabus, as the board prints them.
- C2.9 Graphs in practical situations
- E2.9 Graphs in practical situations
What you should be able to do
- Read and interpret distance–time graphs, including stops and return journeys
- Find a speed as the gradient of a distance–time graph, with units
- Draw a travel graph from a description, and use conversion graphs both ways
- Find accelerations and decelerations from a speed–time graph (Extended)
- Work out a distance from the area under a speed–time graph (Extended)
- Estimate a rate of change from the gradient of a tangent to a curve (Extended)
Key ideas
- Gradient of a distance–time graph = speed
- Gradient of a speed–time graph = acceleration, in m/s² when the speed is in m/s and the time in seconds
- Area under a speed–time graph = distance travelled
- Average speed = total distance ÷ total time, read from the whole graph
How Distance–time and speed–time graphs has been examined
From 2022 to 2026, Distance–time and speed–time graphs was on Paper 3 · Calculator (Core) in 8 of 14 exam series, worth about 3% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 1 · Non-calculator (Core) | 2 | 1% | 2 of 14 |
| Paper 2 · Non-calculator (Extended) | 11 | 2% | 8 of 14 |
| Paper 3 · Calculator (Core) | 11 | 3% | 8 of 14 |
| Paper 4 · Calculator (Extended) | 7 | 1% | 5 of 14 |
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: 4.
Filed in Momentum under: Distance-time Graphs, Speed-time Graphs.
A worked example
A cyclist accelerates uniformly from rest to 8 m/s in 10 seconds, travels at 8 m/s for 30 seconds, then decelerates uniformly to rest in 6 seconds. Find the acceleration, the deceleration and the total distance travelled.
- Sketch the speed–time graph: up from 0 to 8 m/s over 10 s, level at 8 m/s for 30 s, then down to 0 over 6 s.
- Acceleration = gradient of the first section = 8 ÷ 10 = 0.8 m/s².
- Deceleration = 8 ÷ 6 = 1.33 m/s² to 3 significant figures.
- Distance = area under the graph: ½ × 10 × 8 + 30 × 8 + ½ × 6 × 8 = 40 + 240 + 24 = 304 m.
- Check: the whole shape is a trapezium with parallel sides 46 s and 30 s and height 8 m/s, so its area is ½ × (46 + 30) × 8 = 304 m.
Answer: Acceleration 0.8 m/s²; deceleration 1.33 m/s² (3 s.f.); distance 304 m
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Reading the height of a speed–time graph as the distance: on a speed–time graph the distance is the area underneath, not the value on the vertical axis.
- Mixing units, such as a time axis in minutes with a speed in km/h: convert before calculating a gradient or an area.
- Counting grid squares for a gradient without using the scales on the axes, which gives a number with no meaning.
- Giving a deceleration as a negative number: when a question asks for the deceleration, give its size, such as 2.5 m/s²; a negative value is right only when it asks for the acceleration.
- Using the area of a single triangle for a trapezium-shaped section, which misses the rectangle underneath.
Exam technique
- Mark the two points you use for a gradient on the graph and show the division, so the method mark is visible.
- Split the area under a speed–time graph into simple shapes, find each area, and add them.
- For a tangent, rule a long line that touches the curve at the point, then use two well-separated points on it.
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.