Step-by-step worked examples and graded practice questions on distance-time graphs — reading speed from the gradient, interpreting multi-stage journeys, and finding average speed.
📚 Foundation & Higher✅ 15 Practice Questions🔍 Full Worked Examples⚠️ Common Mistakes
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A distance-time graph shows the distance travelled from a starting point, plotted against time. The gradient of the line represents speed — the steeper the line, the faster the object is moving.
A straight, sloping line means travelling at a constant speed
A horizontal (flat) line means the object is stationary — not moving
A line sloping back down towards zero means the object is returning to its starting point
The steeper the line (ignoring direction), the faster the speed
A typical journey: steady speed away from home, a resting (flat) period, then a faster, steeper return.
Reading speed from the gradient
Speed is calculated exactly like gradient: speed = distance ÷ time. On a graph, this means dividing the change in distance (the rise) by the change in time (the run) between two points.
Worked Example 1 — Speed from a straight-line journey
A cyclist's distance-time graph is a straight line from (0, 0) to (2, 20), where x is in hours and y is in km. Find the cyclist's speed.
1
Speed = gradient = change in distance ÷ change in time
2
Speed = (20 − 0) / (2 − 0) = 10 km/h
Answer10 km/h
Multi-stage journeys
Most exam questions describe a journey with several stages — for example, travelling away from home, resting, then returning. Each straight section has its own gradient, so you must calculate the speed for each stage separately. A return journey has a negative gradient, since distance from the start is decreasing — but speed itself is always given as a positive value.
Worked Example 2 — A three-stage journey
A cyclist rides 8 km from home in 1 hour, rests for 30 minutes, then returns home in 1 hour. Find the speed for each stage of the journey.
1
Stage 1 (0 to 1 hour): speed = 8 / 1 = 8 km/h
2
Stage 2 (1 to 1.5 hours): distance stays at 8 km — the line is flat, so speed = 0 km/h (resting)
3
Stage 3 (1.5 to 2.5 hours): gradient = (0 − 8) / (2.5 − 1.5) = −8, so the speed is 8 km/h (returning)
AnswerStage 1: 8 km/h. Stage 2: 0 km/h (resting). Stage 3: 8 km/h (return).
The cyclist's journey: out at 8 km/h, resting (flat), then back at 8 km/h — shown by the negative gradient.
Average speed over a whole journey
Average speed uses the total distance travelled (not the net displacement) divided by the total time taken: average speed = total distance ÷ total time. For a there-and-back journey, the total distance includes both the outward and return legs, even though the person ends up back where they started.
Worked Example 3 — Average speed
Using the cyclist's journey from Worked Example 2, find the average speed for the whole 2.5-hour journey.
1
Total distance travelled = 8 km (out) + 8 km (back) = 16 km — not 0 km, even though the cyclist ends up at home
2
Total time = 2.5 hours
3
Average speed = 16 ÷ 2.5 = 6.4 km/h
Answer6.4 km/h
Practice questions
Work through each question before checking the answers. Difficulty is shown for each question.
Foundation (Grade 3–5)
Q1A cyclist travels at constant speed and covers 15 km in 1 hour. State the gradient of the distance-time graph representing this journey.Foundation
Q2A distance-time graph is a horizontal line for 20 minutes. What does this section of the graph show?Foundation
Q3A car travels 60 km in 1.5 hours at a constant speed. Find its speed.Foundation
Q4On a distance-time graph, one line has gradient 5 and another has gradient 12. Which represents the faster speed?Foundation
Q5A walker covers 3 km in the first hour, then rests for 30 minutes. What is their total distance travelled at the end of the 1.5 hours?Foundation
Higher (Grade 5–7)
Q6A train's distance-time graph passes through (0, 0) and (0.5, 45), where x is in hours and y is in km. Find the train's speed.Higher
Q7A cyclist rides 10 km to a shop in 40 minutes, rests for 20 minutes, then returns home in 30 minutes. Find the speed of the return journey, in km/h.Higher
Q8Using the journey in Q7, find the total time for the whole journey, in hours.Higher
Q9Using the journey in Q7, find the average speed for the whole journey.Higher
Q10A section of a distance-time graph has gradient −6 during a return journey. What does the negative sign represent, and what is the speed?Higher
Higher — Hard (Grade 8–9)
Questions 11–12 refer to the graph below, showing a cyclist's journey.
Graph for Questions 11–12.
Q11Using the graph above, find the cyclist's speed during each of the three stages of the journey.Grade 8–9
Q12Using the graph above, find the cyclist's average speed for the whole journey.Grade 8–9
Questions 13–14 refer to the graph below, showing a car's journey.
Graph for Questions 13–14.
Q13Using the graph above, find the car's speed during each of the two stages of the journey.Grade 8–9
Q14Using the graph above, find the car's average speed for the whole 3-hour journey.Grade 8–9
Q15A distance-time graph shows a walker already 5 km from home at t = 0. They stay at 5 km until t = 2 hours, then walk back home, arriving at t = 4 hours. Find their speed during the return stage.Grade 8–9
Answers
Foundation (Q1–Q5)
Q115 (km/h)
Q2The object is stationary — not moving
Q340 km/h(60 ÷ 1.5)
Q4Gradient 12 is steeper, so it represents the faster speed
Q53 km(distance stays the same while resting)
Higher (Q6–Q10)
Q690 km/h(45 ÷ 0.5)
Q720 km/h(10 km ÷ 0.5 h)
Q81.5 hours(40 + 20 + 30 = 90 minutes)
Q913⅓ km/h (13.3 km/h to 1 d.p.)(total distance 20 km ÷ total time 1.5 h)
Q10The negative gradient shows the object moving back towards the start; the speed is 6 km/h
Higher — Hard (Q11–Q15)
Q11Stage 1: 12 km/h. Stage 2: 0 km/h (resting). Stage 3: 24 km/h (return)(12/1; 0; (0−12)/(2−1.5) = −24)
Q1463⅓ km/h (63.3 km/h to 1 d.p.)(total distance 190 km ÷ total time 3 h)
Q152.5 km/h((0−5)/(4−2) = −2.5; speed = 2.5 km/h)
Common mistakes
These are the errors Alamin sees most frequently with distance-time graphs at GCSE. Recognising them now will save you marks in the exam.
Common Mistake 1
Reading a flat section as "not travelling far" instead of "stationary"
A horizontal line means distance is not changing at all — the object has stopped completely, not just slowed down.
Common Mistake 2
Giving a negative value as the speed
A return journey has a negative gradient, but speed itself is always positive. State the speed as a positive number, and describe the direction separately (e.g. "returning towards home").
Common Mistake 3
Using displacement instead of total distance for average speed
For a there-and-back journey, the net displacement is 0 — but average speed uses the total distance travelled (both legs added together), not the displacement.
Common Mistake 4
Applying one gradient to the whole journey
Multi-stage journeys need a separate gradient calculation for each straight section. Using the start and end points of the whole graph gives the average velocity, not the speed of an individual stage.
Common Mistake 5
Forgetting to convert time units
If the axis is in hours but the question gives minutes (or vice versa), convert before calculating. 30 minutes = 0.5 hours — mixing units gives a completely wrong speed.
Common Mistake 6
Not including units in the final answer
Speed always needs units (e.g. km/h or m/s) — a bare number is an incomplete answer and can lose marks.
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