What you'll learn
Distance–time graphs and speed–time graphs are fundamental tools for representing motion in CIE IGCSE Mathematics. You must interpret these graphs to extract information about journeys, calculate speeds and distances, and understand what different features represent. This topic appears regularly on both Paper 2 (Extended) and Paper 4 (Extended), typically worth 4-8 marks per question.
Key terms and definitions
Distance–time graph — a graph showing how distance travelled changes over time, with time on the horizontal axis and distance on the vertical axis.
Speed–time graph — a graph showing how speed changes over time, with time on the horizontal axis and speed on the vertical axis (sometimes called velocity–time graph when direction matters).
Gradient — the steepness of a line on a graph; on a distance–time graph, gradient represents speed; on a speed–time graph, gradient represents acceleration.
Speed — the rate at which distance changes with time, measured in units such as m/s, km/h or cm/s; calculated as distance ÷ time.
Acceleration — the rate at which speed changes with time, measured in units such as m/s² or km/h²; calculated as change in speed ÷ time taken.
Uniform speed — constant speed, represented by a straight line on a distance–time graph or a horizontal line on a speed–time graph.
Area under a graph — on a speed–time graph, the area between the line and the horizontal axis represents the total distance travelled.
Stationary — not moving; represented by a horizontal line on a distance–time graph (zero gradient) or a point on the time axis of a speed–time graph (speed = 0).
Core concepts
Reading distance–time graphs
Distance–time graphs plot cumulative distance travelled (vertical axis) against time elapsed (horizontal axis). The shape and gradient of the line reveal critical information about motion:
- Horizontal line: object is stationary (distance not changing)
- Straight sloping line: object moving at constant (uniform) speed
- Steep gradient: high speed
- Gentle gradient: low speed
- Curved line: speed is changing (accelerating or decelerating)
- Line curving upward: increasing speed (acceleration)
- Line curving downward: decreasing speed (deceleration)
When an object returns to its starting point, the graph comes back down to the original distance value. The distance shown is cumulative distance from the start, not displacement, so distance–time graphs for CIE IGCSE Mathematics never show negative values.
Calculating speed from distance–time graphs
Speed equals the gradient of a distance–time graph. To calculate speed between two points:
- Identify two points on the line segment
- Find the vertical change (change in distance)
- Find the horizontal change (change in time)
- Calculate: speed = change in distance ÷ change in time
For a straight-line section, the speed is constant throughout. For a curved section, you can find:
- Average speed over an interval: total distance ÷ total time
- Instantaneous speed at a point: gradient of the tangent to the curve at that point (rarely required at IGCSE level)
Units must match the axes. If distance is in kilometres and time in hours, speed will be in km/h. Always state your units in exam answers.
Reading speed–time graphs
Speed–time graphs plot instantaneous speed (vertical axis) against time (horizontal axis). Key features:
- Horizontal line: constant speed (zero acceleration)
- Sloping line upward: positive acceleration (speeding up)
- Sloping line downward: deceleration or negative acceleration (slowing down)
- Steeper slope: greater rate of acceleration or deceleration
- Line touching horizontal axis: object momentarily at rest
- Vertical line: theoretically instantaneous change in speed (unrealistic but sometimes used in idealised exam questions)
The speed shown can be zero (stationary) but is never negative in typical CIE IGCSE Mathematics questions, as we deal with speed not velocity.
Calculating acceleration from speed–time graphs
Acceleration equals the gradient of a speed–time graph. The process mirrors speed calculation from distance–time graphs:
- Identify two points on the line segment
- Find the vertical change (change in speed)
- Find the horizontal change (change in time)
- Calculate: acceleration = change in speed ÷ change in time
A negative result indicates deceleration (slowing down). For horizontal sections, acceleration is zero because speed is not changing.
Calculating distance from speed–time graphs
The area under the graph (between the line and the horizontal time axis) represents total distance travelled. This is a critical concept frequently tested in CIE IGCSE Mathematics papers.
For rectangles (constant speed):
- Distance = speed × time
- Area = base × height
For triangles (uniform acceleration or deceleration):
- Distance = ½ × base × height
- Distance = ½ × time × speed change
For trapeziums (combined sections):
- Distance = ½ × (sum of parallel sides) × height
- Distance = ½ × (initial speed + final speed) × time
For complex shapes: break the area into recognisable shapes (rectangles, triangles, trapeziums), calculate each area separately, then sum them. Count only areas above the time axis.
Units consistency matters: if speed is in m/s and time in seconds, distance will be in metres.
Converting between graph types
Understanding the relationship between distance–time and speed–time graphs strengthens your interpretation skills:
- A horizontal section on a distance–time graph (stationary) corresponds to speed = 0 on a speed–time graph
- A straight sloping section on a distance–time graph (constant speed) corresponds to a horizontal line on a speed–time graph
- A curved section on a distance–time graph (changing speed) corresponds to a sloping line on a speed–time graph
- The gradient of the distance–time graph at any moment equals the height (speed value) of the speed–time graph at that moment
CIE examiners may ask you to sketch one type of graph from information given in another, or to describe corresponding features.
Worked examples
Example 1: Distance–time graph interpretation and calculation
A cyclist travels from home to a shop and back. The distance–time graph shows her journey:
- From 0 to 20 minutes: straight line from (0, 0) to (20, 8)
- From 20 to 35 minutes: horizontal line at distance = 8 km
- From 35 to 50 minutes: straight line from (35, 8) to (50, 0)
(a) How far is the shop from her home? [1 mark]
Solution: At 20 minutes, the distance is 8 km. The shop is 8 km from home.
(b) For how long did she stay at the shop? [1 mark]
Solution: The horizontal section runs from 20 to 35 minutes = 15 minutes.
(c) Calculate her speed during the first 20 minutes. [2 marks]
Solution: Speed = distance ÷ time = 8 km ÷ 20 min = 8 km ÷ (20/60) h = 8 × 3 = 24 km/h
(Alternatively: 8 ÷ (1/3) = 24 km/h)
(d) Calculate her speed on the return journey. [2 marks]
Solution: Distance = 8 km, Time = 50 − 35 = 15 minutes Speed = 8 ÷ (15/60) = 8 ÷ 0.25 = 32 km/h
Example 2: Speed–time graph area calculation
A car accelerates uniformly from rest to 25 m/s in 10 seconds, travels at this constant speed for 30 seconds, then decelerates uniformly to rest in 20 seconds.
(a) Sketch the speed–time graph. [2 marks]
Solution: Graph shows three sections:
- Straight line from (0, 0) to (10, 25) — triangle
- Horizontal line from (10, 25) to (40, 25) — rectangle
- Straight line from (40, 25) to (60, 0) — triangle
(b) Calculate the total distance travelled. [4 marks]
Solution: Distance = sum of three areas
Area 1 (acceleration): ½ × 10 × 25 = 125 m Area 2 (constant speed): 30 × 25 = 750 m Area 3 (deceleration): ½ × 20 × 25 = 250 m
Total distance = 125 + 750 + 250 = 1125 m (or 1.125 km)
(c) Calculate the acceleration during the first 10 seconds. [2 marks]
Solution: Acceleration = change in speed ÷ time = (25 − 0) ÷ 10 = 2.5 m/s²
Example 3: Comparing two journeys
Two runners, A and B, run 100 m. Runner A runs at a constant speed of 8 m/s. Runner B accelerates uniformly from rest to 10 m/s, then maintains this speed.
(a) Calculate how long Runner A takes. [2 marks]
Solution: Time = distance ÷ speed = 100 ÷ 8 = 12.5 seconds
(b) Runner B reaches 10 m/s after 4 seconds. Calculate the distance covered during acceleration. [2 marks]
Solution: Distance = area of triangle = ½ × 4 × 10 = 20 m
(c) Calculate Runner B's total time for 100 m. [3 marks]
Solution: Remaining distance = 100 − 20 = 80 m Time at constant speed = 80 ÷ 10 = 8 s Total time = 4 + 8 = 12 seconds
Runner B is faster.
Common mistakes and how to avoid them
Confusing which axis is which: Students often swap distance and time on distance–time graphs, or speed and time on speed–time graphs. Always check axis labels carefully and remember time is conventionally the horizontal axis for both graph types.
Calculating gradient incorrectly: Dividing time by distance instead of distance by time when finding speed. Remember: gradient = rise ÷ run, so speed = vertical change ÷ horizontal change on a distance–time graph.
Forgetting to convert time units: Calculating speed when time is given in minutes but leaving the answer without converting to hours. If you use 20 minutes directly with distance in km, your speed will be in km/minute, not km/h. Convert: 20 min = 20/60 = 1/3 hour.
Not calculating area for distance on speed–time graphs: Students often read off the final speed value thinking it's the distance, or try to use speed = distance ÷ time. On speed–time graphs, you must calculate the area under the line to find distance travelled.
Misinterpreting horizontal lines: On distance–time graphs, horizontal means stationary (no movement). On speed–time graphs, horizontal means constant speed (no acceleration). These represent fundamentally different motions.
Adding areas incorrectly for trapeziums: Breaking complex shapes into triangles and rectangles but making arithmetic errors or missing sections. Draw vertical lines to separate sections clearly, label each area calculation, then sum carefully.
Exam technique for "Distance–time graphs and speed–time graphs: interpretation and calculation"
Command words matter: "Calculate" requires full working and numerical answers with units. "Describe" requires words explaining motion (e.g., "the cyclist is stationary" or "speed is increasing"). "Sketch" needs a correctly shaped graph with key values marked, but not graph paper precision.
Always show your working: Speed, distance and acceleration calculations typically award 1 method mark and 1 accuracy mark. Even if your final answer is wrong, correct method earns marks. Write out the formula you're using: speed = distance ÷ time.
State units every time: A numerical answer without units loses the final accuracy mark in most mark schemes. Check the units on the graph axes and ensure your answer matches appropriately (e.g., if using minutes, state whether speed is in km/min or convert to km/h).
For area calculations, annotate the graph: Draw lines to split composite shapes, label each section (A, B, C), and show each area calculation separately before summing. This structured approach reduces errors and shows clear method to the examiner.
Quick revision summary
Distance–time graphs show cumulative distance (vertical) against time (horizontal); gradient equals speed, horizontal sections mean stationary. Speed–time graphs show instantaneous speed against time; gradient equals acceleration, area under the graph equals distance travelled. Calculate speed from distance–time graphs using gradient (distance ÷ time). Calculate acceleration from speed–time graphs using gradient (change in speed ÷ time). Calculate distance from speed–time graphs by finding area under the line using shapes (rectangles, triangles, trapeziums). Always show working, include correct units, and check axis labels carefully in CIE IGCSE Mathematics examinations.