Travel graphs, rates and area under speed
Read a graph's axes first, then connect distance gradients, speed gradients and speed–time area to a journey.
Core: practical graphs, conversion graphs and gradients as rates, including distance–time journeys. Extended adds acceleration, deceleration, speed–time area and tangent gradients.
Before you begin
- Calculate speed from distance and time with compatible units.
- Find areas of triangles and rectangles.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Convert 3 minutes to seconds.
What you will learn
- Draw and interpret distance–time and conversion graphs.
- Use gradients as rates of change.
- Calculate speed–time area and estimate tangent gradients in Extended work.
The axes decide what a slope means
On a distance–time graph, the gradient is change in distance divided by change in time, so it describes speed for the stated distance measure. A line rising from 0 km at 0 h to 60 km at 1.5 h has gradient 40 km/h. A horizontal section means that recorded distance is not changing. If the vertical axis is distance from a starting point, a falling section can represent returning towards it.
On a conversion graph, plot pairs in the stated units and interpolate between reliable points. For a linear model 1 mile ≈ 1.6 km, a graph of kilometres against miles passes through (0, 0) and (5, 8), with gradient about 1.6 km per mile. Extrapolating a real data relationship beyond its measured range may be unreliable.
Extended: a speed–time gradient is acceleration
The gradient of a speed–time line is change in speed divided by elapsed time. Rising from 0 to 12 m/s in 6 s gives acceleration 2 m/s². A horizontal speed section means constant speed, not rest unless its value is zero. A falling speed section describes deceleration; its gradient is negative.
For a curve, estimate a local gradient by drawing a tangent at the specified point. Choose two well-separated points on that tangent, which need not lie on the curve, and divide their vertical change by horizontal change. The gradient depends on axis units and scale. It estimates the rate at that point rather than the average between distant curve points.
Extended: distance is speed–time area
Over a constant-speed interval, distance equals speed multiplied by time, the area of its rectangle. A linear increase from zero produces a triangle. A linear change between two non-zero speeds produces a trapezium with area half the sum of the speeds multiplied by the duration. Add areas of linear sections to get total distance travelled.
A speed measured in m/s and time in seconds gives area in metres. Convert minutes to seconds before combining them with m/s. Do not use area under a distance–time graph as distance: its units would be distance multiplied by time. The graph shown here reaches 12 m/s over six seconds, stays there for four seconds and stops over four seconds, giving 36 + 48 + 24 = 108 m.
Pause and explain
What does a horizontal non-zero speed–time segment mean?
Worked example
Extended: speed rises uniformly from 0 to 12 m/s in 6 s, remains 12 m/s for 4 s, then falls uniformly to zero in 4 s. Find acceleration in the first stage and total distance.
Show the worked solution
- Initial acceleration is (12 − 0)/6 = 2 m/s².
- The three areas are 1/2 × 6 × 12 = 36 m; 4 × 12 = 48 m; and 1/2 × 4 × 12 = 24 m.
- Total distance is 108 m. The last stage has gradient −3 m/s², consistent with stopping in four seconds.
Answer 2 m/s² initially; total distance 108 m
From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
On a distance–time graph, distance from home rises from 10 km at 0.5 h to 40 km at 1.25 h. Find the average outward speed in that segment.
Give me a hint
Use changes, rather than the final distance divided by the final time.
Compare my reasoning
- Distance change is 40 − 10 = 30 km.
- Time change is 1.25 − 0.5 = 0.75 h.
- The gradient is 30/0.75 = 40 km/h.
40 km/h
Look for these in your work
- Used differences from both endpoints.
- Reported compatible units.
Extended: speed decreases linearly from 15 to 5 m/s over 8 s. Find its acceleration and distance travelled.
Give me a hint
Use the gradient for acceleration and the trapezium area for distance.
Compare my reasoning
- Acceleration is (5 − 15)/8 = −1.25 m/s².
- Mean speed for this straight segment is (15 + 5)/2 = 10 m/s.
- Distance is 10 × 8 = 80 m; speed stays positive throughout.
−1.25 m/s²; 80 m
Look for these in your work
- Kept the negative acceleration sign.
- Used area rather than gradient for distance.
Extended: a tangent to a distance–time curve at 20 s passes through (10 s, 20 m) and (30 s, 100 m). Estimate instantaneous speed at 20 s.
Give me a hint
Calculate the gradient of the tangent, not a chord of the curve.
Compare my reasoning
- The tangent's vertical change is 100 − 20 = 80 m.
- Its horizontal change is 30 − 10 = 20 s.
- Estimated local speed is 80/20 = 4 m/s; the result depends on the accuracy of the drawn tangent.
Approximately 4 m/s
Look for these in your work
- Used points on the tangent.
- Described the result as an estimate.
Common mistakes
- Calling a horizontal positive speed segment rest.
- Using speed–time gradient to calculate distance.
- Ignoring different units or reading a curve's chord as its tangent.
What can you explain now?
A conversion graph uses K = 1.6M, where M is miles and K is kilometres. Estimate kilometres for 7.5 miles.
Compare with the explanation
12 km
Read or calculate 1.6 × 7.5 = 12; the conversion coefficient is approximate.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Draw a short journey as a distance–time graph. Extended: draw a speed–time graph and explain separately what its gradients and areas represent.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.
Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.
Number contains 19 sequenced lessons; Algebra and graphs contains 21; Coordinate geometry contains 10 teaching lessons and a mixed checkpoint. Nine earlier overviews support selected topics across the syllabus. The other six syllabus areas, full cumulative assessment and human teacher review are not yet complete. Written lesson practice is self-checked, not automatically graded. The checkpoint samples skills and does not save an exam grade or certify mastery.
Check the official syllabus 2025–2027 Open related practice and resources