IGCSE Maths · Algebra and Graphs · Core & Extended

Graphs in Practical Situations: Conversion, Travel and Speed-Time Graphs

A graph is just a story told in pictures — a rising line means increasing, a flat line means unchanging, a falling line means decreasing, and how steep it is tells you how fast that change happens. Conversion graphs, distance-time graphs, and drawing graphs from data are shared by both tiers. Speed-time graphs and everything built from them are Extended-only.

1. Conversion Graphs

A conversion graph is a straight line connecting two different units — read a value on one axis, trace across/up to the line, then across/down to read the equivalent on the other axis.

Reading a conversion graph — miles to kilometres Miles Kilometres 30 48 50 80 To convert 30 miles to kilometres: find 30 on the horizontal axis, go straight up to the line, then straight across to the vertical axis — reading off 48 km.
Try it — Conversion Graphs (using the graph above)
  1. Using the ratio shown (30 miles = 48 km), how many km is 10 miles?
  2. Using the same ratio, how many miles is approximately 64 km?
  3. What is the gradient of this conversion line, and what does it represent?
Show answers
  1. 48 ÷ 3 = 16 km
  2. 64 ÷ 1.6 = 40 miles
  3. Gradient = 48/30 = 1.6 — this represents the number of km in one mile (the conversion rate).

2. Distance-Time Graphs

On a distance-time graph, the gradient represents speed. A steep line means fast, a shallow line means slow, a flat (horizontal) section means stationary, and a line returning toward the time axis means travelling back toward the starting point.

Reading a journey story Time (min) Distance (km) 0 20 35 50 6 This graph tells a story: from 0–20 minutes, the person walks 6 km to a friend’s house (rising line). From 20–35 minutes, they stay there (flat line — stationary). From 35–50 minutes, they walk home again — and since this section is steeper, covering the same 6 km in only 15 minutes instead of 20, they walked home faster than they walked there.
Worked example — calculating speeds from the graph Find the speed for each stage of the journey above.
Outward journey: 6 km in 20 minutes = 6 km in 1/3 hour → speed = 6 ÷ (1/3) = 18 km/h
At the friend’s house: distance isn’t changing → speed = 0 km/h
Return journey: 6 km in 15 minutes = 6 km in 1/4 hour → speed = 6 ÷ (1/4) = 24 km/h
Try it — Distance-Time Graphs
  1. A cyclist’s distance-time graph shows they travel 15 km in the first 30 minutes at a constant speed. Find their speed in km/h.
  2. A distance-time graph is flat between 10 and 25 minutes. What does this tell you?
  3. A car travels 40 km in 24 minutes on one section of a journey, and 40 km in 30 minutes on another. Which section was faster, and by how much (in km/h)?
Show answers
  1. 15 km in 0.5 h → 30 km/h
  2. The object was stationary (not moving) for that 15-minute period.
  3. Section 1: 40 ÷ 0.4 = 100 km/h. Section 2: 40 ÷ 0.5 = 80 km/h. Section 1 was faster, by 20 km/h.

3. Drawing Graphs from Given Data

Plot each data point accurately, use a ruler for straight sections, and label both axes clearly with units.

Worked example A bus’s distance from its depot is recorded every 10 minutes: 0 min → 0 km, 10 min → 8 km, 20 min → 16 km, 30 min → 16 km (a stop), 40 min → 10 km, 50 min → 0 km.
To draw this: plot the five points (0,0), (10,8), (20,16), (30,16), (40,10), (50,0), then join them with straight ruled segments — the flat section between 20 and 30 minutes shows the stop, and the falling sections show the bus heading back toward the depot.
Try it — Drawing from Data
  1. A car’s speed over time is: 0s→0, 5s→20, 10s→20, 15s→20, 20s→0 (in m/s). Describe what each stage of the journey represents.
  2. Given the bus data above, find the bus’s average speed for just the first 20 minutes (before the stop).
Show answers
  1. 0–5s: accelerates to 20 m/s. 5–15s: travels at constant 20 m/s. 15–20s: decelerates to a stop.
  2. 16 km in 20 minutes (1/3 h) → 16 ÷ (1/3) = 48 km/h

4. Speed-Time Graphs Extended only

On a speed-time graph, the gradient represents acceleration (or deceleration, if the line falls). A flat section means constant speed — not stationary, since the vertical axis is speed, not distance.

Reading a speed-time graph Time (s) Speed (m/s) 0 10 30 40 20 shaded area = distance 0–10s: the line rises — the object accelerates from 0 to 20 m/s. 10–30s: the line is flat — constant speed of 20 m/s. 30–40s: the line falls — the object decelerates back to 0 m/s (comes to a stop).
Worked example — calculating acceleration Find the acceleration during the first 10 seconds of the graph above (speed increases from 0 to 20 m/s).
Acceleration = gradient = change in speed ÷ change in time = (20 − 0) ÷ (10 − 0) = 2 m/s²
Worked example — calculating deceleration Find the deceleration during the last 10 seconds of the graph above (speed decreases from 20 to 0 m/s).
Gradient = (0 − 20) ÷ (40 − 30) = −2 m/s², so the deceleration is 2 m/s² (deceleration is often given as a positive value describing how fast the object is slowing down).
Try it — Speed-Time Graphs Extended
  1. A car accelerates from 0 to 15 m/s in 5 seconds. Find its acceleration.
  2. A train decelerates from 30 m/s to 0 m/s in 12 seconds. Find its deceleration.
  3. A speed-time graph is flat at 25 m/s between 5s and 20s. What does this tell you, and what is the acceleration during this stage?
Show answers
  1. 15 ÷ 5 = 3 m/s²
  2. 30 ÷ 12 = 2.5 m/s²
  3. The object is travelling at a constant 25 m/s — not accelerating or decelerating. The acceleration during this stage is 0 m/s².

5. Distance as Area Under a Speed-Time Graph Extended only

The area between a speed-time graph and the time axis equals the total distance travelled. For straight-line sections, break the shape into rectangles and triangles (or use the trapezium area formula) and add the areas together.

Worked example — using the graph from Section 4 Find the total distance travelled over the 40 seconds shown in the speed-time graph above.
Split the shape into three parts:
Triangle 1 (0–10s, accelerating): ½ × 10 × 20 = 100 m
Rectangle (10–30s, constant speed): 20 × 20 = 400 m
Triangle 2 (30–40s, decelerating): ½ × 10 × 20 = 100 m
Total distance: 100 + 400 + 100 = 600 m
Try it — Area Under the Graph Extended
  1. A speed-time graph shows constant acceleration from 0 to 12 m/s over 6 seconds, then constant speed at 12 m/s for a further 10 seconds. Find the total distance travelled.
  2. A car travels at a constant 18 m/s for 15 seconds, then decelerates to rest over the next 6 seconds. Find the total distance travelled.
Show answers
  1. Triangle: ½×6×12=36 m. Rectangle: 10×12=120 m. Total = 156 m
  2. Rectangle: 15×18=270 m. Triangle: ½×6×18=54 m. Total = 324 m

6. Estimating the Gradient of a Curve Using a Tangent Extended only

When a distance-time graph curves (speed changing continuously, not in straight sections), you can’t read a single gradient for the whole graph. Instead, draw a tangent — a straight line that just touches the curve at one point without crossing it — and find the gradient of that tangent to estimate the speed at that exact moment.

Reading a tangent on a curve Time (s) Distance (m) tangent at this point (Diagram illustrative, not to scale.) The curve shows an object speeding up over time — notice it gets steeper as time goes on. To estimate the speed at the marked point, draw a straight tangent line touching the curve only there, then treat that line like an ordinary straight-line gradient calculation.
Worked example A tangent drawn on a curved distance-time graph passes through the points (10, 20) and (25, 65), where time is in seconds and distance is in metres. Estimate the speed at the point of tangency.
Gradient = (65 − 20) ÷ (25 − 10) = 45 ÷ 15 = 3 m/s
Try it — Tangents to Curves Extended
  1. A tangent to a distance-time curve passes through (4, 10) and (12, 34). Estimate the speed at the point of tangency.
  2. A tangent to a distance-time curve passes through (0, 5) and (8, 45). Estimate the speed at the point of tangency.
Show answers
  1. (34−10)/(12−4) = 24/8 = 3 m/s
  2. (45−5)/(8−0) = 40/8 = 5 m/s
Common mistakes to watch for
  • Confusing distance-time and speed-time graphs — a flat line means “stationary” on one, but “constant speed” on the other.
  • Reading the gradient as rise/run the wrong way round, or misreading the scale on an axis.
  • Forgetting to convert time into hours (or the matching unit) before calculating a speed in km/h or m/s.
  • For area-under-the-graph questions, forgetting a triangular section entirely, or using the wrong base/height pairing.
  • Drawing a tangent that crosses the curve at two points instead of just touching it at one — a true tangent only touches.

7. Quick-Fire Challenge Round

True or False?
  1. A flat line on a distance-time graph means the object has stopped.
  2. A flat line on a speed-time graph means the object has stopped.
  3. The area under a speed-time graph gives the total distance travelled.
  4. A steeper distance-time graph always means a slower speed.
Show answers
  1. True.
  2. False — a flat speed-time line means constant (non-zero, unless it’s flat at zero) speed, not stopped.
  3. True. Ext.
  4. False — a steeper line means a faster speed, since gradient represents speed on this type of graph.
Spot the Error
  1. “A distance-time graph is flat between 10 and 20 minutes, so the object was travelling fastest during this time.”
  2. “To find the distance from a speed-time graph, I read the gradient of the line.”
  3. “A car travels 60 km in 45 minutes: speed = 60 ÷ 45 = 1.33 km/h.”
Show answers
  1. A flat section on a distance-time graph means the object wasn’t moving at all — the opposite of “fastest.”
  2. The gradient of a speed-time graph gives acceleration, not distance — distance comes from the area under the graph.
  3. 45 minutes must be converted to hours first: 45 min = 0.75 h, so speed = 60 ÷ 0.75 = 80 km/h, not 1.33 km/h (dividing by minutes without converting gives a meaningless unit).

8. Mixed Practice — Bringing It All Together

Core & Extended
  1. A distance-time graph shows a journey of 24 km completed in 40 minutes at a constant speed. Find the speed in km/h.
  2. A conversion graph shows that 5 kg = 11 lb. Find how many lb is equivalent to 20 kg.
  3. A distance-time graph rises for 15 minutes, is flat for 10 minutes, then falls back to zero over the next 20 minutes. Describe what’s happening at each stage.
Show answers
  1. 24 ÷ (2/3) = 36 km/h
  2. 11 × 4 = 44 lb
  3. Moving away from the start (0–15 min), stationary (15–25 min), returning to the start, slower than the outward journey since it takes longer for the same distance (25–45 min).
Extended-level mixed set Extended only
  1. A speed-time graph rises from 0 to 16 m/s over 8 seconds, then stays constant for 12 seconds. Find the total distance travelled.
  2. Find the acceleration for the first 8 seconds of the graph in question 1.
  3. A tangent to a curved distance-time graph passes through (2, 8) and (10, 40). Estimate the speed at that point.
Show answers
  1. Triangle: ½×8×16=64 m. Rectangle: 12×16=192 m. Total = 256 m
  2. 16 ÷ 8 = 2 m/s²
  3. (40−8)/(10−2) = 32/8 = 4 m/s

Practice at Home

A large set again — take it steadily.

Part A — Conversion graphs
  1. A conversion graph shows 8 km = 5 miles. How many miles is 24 km?
  2. Using the same conversion, how many km is 15 miles?
Part B — Distance-time graphs
  1. A runner covers 5 km in 25 minutes at a constant speed. Find their speed in km/h.
  2. A distance-time graph shows a journey out (steep rise), a stop (flat), then a return (less steep rise back to zero, i.e. falling). What does the difference in steepness between the outward and return journeys tell you?
  3. A cyclist travels 12 km in 30 minutes on one leg of a trip, and 18 km in 40 minutes on another. Which leg was faster, and by how much (km/h)?
Part C — Speed-time graphs and acceleration Extended
  1. A cyclist accelerates from 0 to 8 m/s in 4 seconds. Find their acceleration.
  2. A lift decelerates from 5 m/s to 0 m/s in 2.5 seconds. Find the deceleration.
  3. A speed-time graph is flat at 12 m/s for 20 seconds. State the acceleration during this time, and explain your reasoning.
Part D — Area under a speed-time graph Extended
  1. A speed-time graph rises from 0 to 10 m/s over 5 seconds, then stays constant for 15 seconds. Find the total distance.
  2. A car travels at a constant 22 m/s for 10 seconds, then decelerates to rest over 4 seconds. Find the total distance.
  3. A speed-time graph rises from 0 to 6 m/s over 3 seconds, stays constant for 7 seconds, then falls back to 0 over 5 seconds. Find the total distance.
Part E — Tangents to curves Extended
  1. A tangent to a curved distance-time graph passes through (3, 12) and (9, 42). Estimate the speed at that point.
  2. A tangent to a curved distance-time graph passes through (1, 4) and (6, 34). Estimate the speed at that point.
Part F — Applied / challenge
  1. Sketch (in words, describing each stage) a distance-time graph for this journey: a student walks to a bus stop (5 minutes), waits for the bus (8 minutes), then travels quickly to school by bus (12 minutes). Describe the shape of each of the three stages.
  2. A car accelerates uniformly from rest to 24 m/s in 6 seconds, travels at that speed for 20 seconds, then decelerates uniformly to rest in 8 seconds. Find the total distance travelled over the whole journey.
  3. Explain, using the ideas from this note, why the gradient of a distance-time graph and the area under a speed-time graph represent two genuinely different things, even though both graphs describe the same journey.

Key Vocabulary Recap

Gradient — the steepness of a line; represents rate of change
Conversion graph — a straight-line graph linking two different units
Acceleration — the rate at which speed increases (gradient of a speed-time graph)
Deceleration — the rate at which speed decreases
Tangent — a straight line touching a curve at exactly one point, used to estimate gradient there

Syllabus Reference

This note covers Cambridge IGCSE Mathematics (0580) section C2.9 (Core subject content) and section E2.9 (Extended subject content) — Graphs in practical situations. Conversion graphs, distance-time graphs, and drawing graphs from data are shared by both tiers. Speed-time graphs, acceleration and deceleration, calculating distance as the area under a speed-time graph, and estimating the gradient of a curve using a tangent are Extended-only, marked above. Areas under speed-time graphs are limited to linear (straight-line) sections at both tiers.

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