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IGCSE / O Level Mathematics · Topic 2 Algebra and graphs · Chapter 17 · pp. 172–192

Graphs in practical situations

Graphs turn real situations into pictures you can read. A straight line converts one unit to another. A travel graph shows a journey at a glance: its gradient tells you how fast something is going, and on a speed–time graph the area underneath tells you how far it went.

Learning objectives

  1. Draw and use conversion graphs (currency, distance, temperature, charges).
  2. Use distance = speed × time and average speed = total distance ÷ total time.
  3. Draw and read distance–time graphs, and use the gradient as the speed.
  4. Read speed–time graphs and find acceleration and deceleration from the gradient. (Extended)
  5. Find the distance travelled from the area under a speed–time graph. (Extended)
  6. Estimate speed or acceleration on a curved graph with a tangent. (Extended)

Book exercises: 17.1 to 17.8 and Student assessments 1 to 3 (pp. 172–192).

Conversion graphs

🔑 Key ideas

  • A conversion graph is a straight line that changes one unit into another.
  • To draw it you need only two points: often (0, 0) and one known rate, such as €10 = 88 rand.
  • Read it like a map: go up from one axis to the line, then across to the other axis (or the reverse).
  • Off the edge of the graph? Read a smaller value and multiply: 200 rand = 4 × 50 rand.
  • Not every line starts at 0. A fixed charge (a call-out fee, the “add 30” in a temperature rule) lifts the start of the line.
012345678910 0102030405060708090 €5 → 44 rand 25 rand → about €2.80 Euros (€) Rand
The book's rand–euro graph (€1 = 8.80 rand). Dashed lines show the two ways to read it: up then across, or across then down.

📘 Example 1 · Rand and euros (€1 = 8.80 rand)

  1. €5 to rand: go up from €5 to the line, then across to the rand axis. It meets the axis at about 44 rand. Check: 5 × 8.80 = 44. ✔
  2. A 25 rand drink: go across from 25 rand to the line, then down. It meets the euro axis at about €2.80.
  3. A 200 rand meal is off the graph. Read a factor of 200 instead: 50 rand ≈ €5.70.
  4. Multiply back up: 200 rand = 4 × 50 rand ≈ 4 × €5.70 = €22.80.

📘 Example 2 · Given 80 km = 50 miles, convert 30 miles to km

  1. Plot the two points (0 km, 0 miles) and (80 km, 50 miles), and join them with a ruler.
  2. Go across from 30 miles to the line, then down to the km axis.
  3. It reads about 48 km. Check: 30 ÷ 50 × 80 = 48. ✔

📘 Example 3 · Rough rule: °F ≈ 2 × °C + 30

  1. Make two points: 0 °C → 2 × 0 + 30 = 30 °F, and 50 °C → 2 × 50 + 30 = 130 °F.
  2. Plot (0, 30) and (50, 130) and join them. The line starts at 30, not at 0.
  3. Read 20 °C: up to the line, across: about 70 °F (2 × 20 + 30 = 70).

✏️ Try it

  1. Using €1 = 8.80 rand, how many rand is €8?
  2. Using 80 km = 50 miles, how many miles is 120 km?
  3. A taxi charges 40 baht plus 6 baht per km. What does a 10 km ride cost? Does its graph pass through (0, 0)?
Answers

1. 8 × 8.80 = 70.4 rand · 2. 120 ÷ 80 × 50 = 75 miles · 3. 40 + 60 = 100 baht. No: at 0 km it already costs 40 baht.

Next: speed, distance and time →

Speed, distance and time

D S T Speed = distance ÷ time Distance = speed × time Time = distance ÷ speed
Cover the quantity you want. D over S means divide; S beside T means multiply.

🔑 Key ideas

  • Average speed = . Rest stops count in the total time.
  • Make the units match before you divide or multiply: km with hours, m with seconds.
  • Minutes to hours: divide by 60. 12 min = 0.2 h, so 2 h 12 min = 2.2 h (not 2.12 h).
  • 1 m/s = 3.6 km/h. So km/h ÷ 3.6 gives m/s, and m/s × 3.6 gives km/h.
TimeIn hoursSpeedIn the other unit
15 min0.25 h10 m/s36 km/h
40 min⅔ h ≈ 0.667 h20 m/s72 km/h
2 h 30 min2.5 h90 km/h25 m/s

📘 Example 4 · Average speed for 165 km in 2 h 12 min

  1. Change the minutes to hours: 12 ÷ 60 = 0.2 h, so the time is 2.2 h.
  2. Speed = distance ÷ time = 165 ÷ 2.2.
  3. = 75 km/h. Check: 75 × 2.2 = 165. ✔

📘 Example 5 · How far in 1 h 15 min at 48 km/h?

  1. 1 h 15 min = 1.25 h.
  2. Distance = speed × time = 48 × 1.25 = 60 km.

📘 Example 6 · How long to travel 3 km at 15 m/s?

  1. The speed is in metres, so change 3 km to 3000 m.
  2. Time = distance ÷ speed = 3000 ÷ 15 = 200 s.
  3. 200 s = 3 min 20 s.

📘 Example 7 · 60 km at 40 km/h, then 60 km at 60 km/h. Average speed?

  1. Time for the first part: 60 ÷ 40 = 1.5 h.
  2. Time for the second part: 60 ÷ 60 = 1 h.
  3. Total distance 120 km, total time 2.5 h.
  4. Average speed = 120 ÷ 2.5 = 48 km/h. It is not 50 km/h: you spend longer at the slow speed.

✏️ Try it

  1. Find the average speed for 45 km in 45 min.
  2. How far is 2 h 30 min at 64 km/h?
  3. How long does 500 m take at 4 m/s?
Answers

1. 45 ÷ 0.75 = 60 km/h · 2. 64 × 2.5 = 160 km · 3. 500 ÷ 4 = 125 s = 2 min 5 s

Next: distance–time graphs →

Distance–time (travel) graphs

🔑 Key ideas

  • A straight line means a constant speed.
  • A horizontal line means no movement: the object is stopped.
  • The gradient is the speed: . Steeper means faster.
  • A line sloping down means moving back towards the start.
  • Two journeys on the same axes meet where the lines cross.
01020304050 051015202530 1.5 m/s stopped (0 m/s) 0.5 m/s Time (s) Distance (m)
The book's girl walking: 15 m in 10 s (1.5 m/s), stopped for 20 s, then 10 m in 20 s (0.5 m/s).
steep: fast flat part: stopped shallow: slow down: going back
Read the shape before you read the numbers.

📘 Example 8 · A straight line through (0, 0) and (8 s, 80 m)

  1. Time to cover 30 m: go across from 30 m to the line, then down: 3 s.
  2. Pick two points on the line, such as (0, 0) and (4, 40).
  3. Gradient = 40 ÷ 4 = 10.
  4. The gradient of a distance–time graph is the speed: 10 m/s.

📘 Example 9 · Cars X and Y from A to B (100 km)

  1. Car X, 07 00 to 08 00: it goes 60 km in 1 h, so speed = 60 ÷ 1 = 60 km/h.
  2. Car X, 08 00 to 09 00: the line is flat, so the distance does not change. Car X is stationary.
  3. Car Y, 09 00 to 11 00: it goes 100 km in 2 h, so speed = 100 ÷ 2 = 50 km/h.

📘 Example 10 · Walking towards each other

Two houses are 6 km apart. At 09 00 Ann walks from one at 4 km/h and Ben from the other at 5 km/h, towards each other. When and where do they meet?

  1. On a graph, Ann's line rises from 0 km and Ben's falls from 6 km. They meet where the lines cross.
  2. To check by calculation: together they close the gap at 4 + 5 = 9 km/h.
  3. Time to close 6 km: 6 ÷ 9 = ⅔ h = 40 min, so they meet at 09 40.
  4. Ann has walked 4 × ⅔ = 2⅔ km (about 2.7 km) from her house. Check: Ben walked 5 × ⅔ = 3⅓ km, and 2⅔ + 3⅓ = 6. ✔

✏️ Try it

A cyclist rides 12 km from home in 40 min, rests for 20 min, then rides home in 30 min.

  1. What is her speed on the way out, in km/h?
  2. What is her speed on the way home?
  3. What is her average speed for the whole trip, including the rest?
Answers

1. 12 ÷ ⅔ = 18 km/h · 2. 12 ÷ 0.5 = 24 km/h · 3. 24 km ÷ 1.5 h = 16 km/h

Next: speed–time graphs →

Speed–time graphs, acceleration and deceleration Extended

On the graphDistance–time graphSpeed–time graph
Horizontal lineStoppedConstant speed
Line sloping upMoving at constant speedSpeeding up (accelerating)
Line sloping downComing backSlowing down (decelerating)
GradientSpeed (m/s)Acceleration (m/s²)
Area underneath(no meaning)Distance travelled

💡 Always check the vertical axis label first. The same shape means different things on the two kinds of graph.

02468101214161820 048121620 accelerating: 16 ÷ 8 = 2 m/s² rise 16 m/s constant 16 m/s decelerating: 16 ÷ 2 = 8 m/s² Time (s) Speed (m/s)
The book's car over 16 seconds: speeding up for 8 s, steady at 16 m/s, then braking hard in the last 2 s.

🔑 Key ideas

  • Acceleration = , measured in m/s² (metres per second, every second).
  • A negative gradient is a deceleration. Say “decelerates at 3 m/s²”, not “−3 m/s² deceleration”.
  • “From rest” means the speed starts at 0.
  • Mixed units? Change km/h to m/s (÷ 3.6) and minutes to seconds (× 60) first.

📘 Example 11 · The car on the graph above

  1. Describe it: 0 to 8 s speeding up uniformly, 8 to 14 s steady at 16 m/s, 14 to 16 s slowing down uniformly.
  2. Acceleration in the first 8 s = gradient = 16 ÷ 8 = 2 m/s².
  3. From 14 to 16 s the speed falls by 16 m/s in 2 s: deceleration = 16 ÷ 2 = 8 m/s².

📘 Example 12 · A line from (0 s, 4 m/s) to (10 s, 24 m/s)

  1. It does not start from rest, so use the change in speed: 24 − 4 = 20 m/s.
  2. Acceleration = 20 ÷ 10 = 2 m/s².

📘 Example 13 · A bus at 18 m/s decelerates at 1.5 m/s². Sketch its graph.

  1. It loses 1.5 m/s every second, so the time to stop is 18 ÷ 1.5 = 12 s.
  2. Start the line at (0, 18) on the speed axis.
  3. End it at (12 s, 0) on the time axis and join with a straight line.

📘 Example 14 · A car goes from rest to 72 km/h in 10 s

  1. The time is in seconds, so change the speed to m/s: 72 ÷ 3.6 = 20 m/s.
  2. Acceleration = 20 ÷ 10 = 2 m/s².

✏️ Try it

  1. A car goes from 0 to 30 m/s in 12 s. Find its acceleration.
  2. A cyclist slows from 25 m/s to 10 m/s in 5 s. Find the deceleration.
  3. A plane accelerates from rest at 4 m/s² for 15 s. What speed does it reach?
Answers

1. 30 ÷ 12 = 2.5 m/s² · 2. 15 ÷ 5 = 3 m/s² · 3. 4 × 15 = 60 m/s

Next: area under a speed–time graph →

Area under a speed–time graph Extended

🔑 Key ideas

  • The area under a speed–time graph is the distance travelled (speed × time = distance).
  • Split the area into rectangles, triangles and trapeziums, then add them.
  • Rectangle = base × height · Triangle = ½ × base × height · Trapezium = ½ × (a + b) × width.
  • The time units must match the speed units: minutes on the axis with m/s on the other means × 60 first.
051015202530 05101520253035 A: 30 × 20 = 600 m B: ½ × 20 × 10 = 100 m Time (s) Speed (m/s)
The book's train: steady at 20 m/s for 10 s, then speeding up to 30 m/s. Rectangle A plus triangle B gives 700 m.

📘 Example 15 · The train above (book)

  1. Acceleration in the last 20 s: speed rises 30 − 20 = 10 m/s, so 10 ÷ 20 = ½ m/s².
  2. Split the area: rectangle A under 20 m/s for all 30 s, triangle B on top.
  3. A = 30 × 20 = 600 m.
  4. B = ½ × 20 × 10 = 100 m.
  5. Total distance = 600 + 100 = 700 m.

📘 Example 16 · The car from the last tab: (0, 0), (8, 16), (14, 16), (16, 0)

  1. Triangle while speeding up: ½ × 8 × 16 = 64 m.
  2. Rectangle at constant speed: 6 × 16 = 96 m.
  3. Triangle while braking: ½ × 2 × 16 = 16 m.
  4. Total = 64 + 96 + 16 = 176 m. Check with one trapezium: ½ × (6 + 16) × 16 = 176. ✔

📘 Example 17 · Time axis in minutes

A train reaches 25 m/s in 1 min, holds that speed for 4 min, then stops in 1 min. How far does it go, in km?

  1. The speed is per second, so change the times: 1 min = 60 s, 4 min = 240 s.
  2. Speeding up: ½ × 60 × 25 = 750 m.
  3. Steady: 240 × 25 = 6000 m.
  4. Stopping: ½ × 60 × 25 = 750 m.
  5. Total = 7500 m = 7.5 km.

📘 Example 18 · When has it gone half the distance?

A car accelerates uniformly from rest to 20 m/s in 10 s.

  1. Total distance = ½ × 10 × 20 = 100 m, so half is 50 m.
  2. Acceleration = 20 ÷ 10 = 2 m/s², so after t seconds the speed is 2t.
  3. Area of the triangle up to time t: ½ × t × 2t = t².
  4. Solve t² = 50: t = √50 = 7.07 s (3 s.f.). Not 5 s: the car covers less ground early on, when it is slow.

🧪 Try it: speed–time graph builder

Type the corner points of a speed–time graph as time,speed pairs (seconds and m/s), for example 0,0 8,16 14,16 16,0. The builder draws the graph, finds the acceleration of each part from its gradient, and adds up the area to get the distance.

    ✏️ Try it

    1. A bike speeds up from rest to 12 m/s in 6 s. How far does it go?
    2. A van holds 8 m/s for 15 s, then stops uniformly in 4 s. Total distance?
    3. A runner speeds up from 5 m/s to 15 m/s in 10 s. How far does she run in that time?
    Answers

    1. ½ × 6 × 12 = 36 m · 2. 8 × 15 + ½ × 4 × 8 = 120 + 16 = 136 m · 3. Trapezium ½ × (5 + 15) × 10 = 100 m

    Next: curved travel graphs →

    Non-linear travel graphs Extended

    Real objects speed up and slow down gradually, so real travel graphs curve. The gradient of a curve keeps changing, so you find it at one moment with a tangent: a straight line that just touches the curve at that point.

    1. Markthe point 2. Tangentruler touching 3. Two pointsfar apart 4. Gradientrise ÷ run Distance–time: gradient = speed · Speed–time: gradient = acceleration
    The tangent method. It is an estimate, because the tangent is drawn by eye.
    012345678910 0102030405060 rise 36 m run 9.5 − 2 = 7.5 s tangent at 4 s: 36 ÷ 7.5 = 4.8 m/s (4, 9.6) Time (s) Distance (m)
    The book's cyclist, D = 0.6t². The tangent at t = 4 s passes through (2, 0) and (9.5, 36), so her speed then is about 4.8 m/s.

    📘 Example 19 · Cyclist with D = 0.6t² (book)

    1. Make a table: t = 0, 1, 2, …, 10 gives D = 0, 0.6, 2.4, 5.4, 9.6, 15, 21.6, 29.4, 38.4, 48.6, 60.
    2. Plot the points and join them with a smooth curve.
    3. At t = 4 s draw a tangent. It passes through (2, 0) and (9.5, 36).
    4. Gradient = (36 − 0) ÷ (9.5 − 2) = 36 ÷ 7.5 = 4.8, so the speed is about 4.8 m/s.
    5. At t = 8 s the tangent passes through (4, 0) and (9, 48): 48 ÷ 5 = 9.6 m/s. She is going twice as fast.

    📘 Example 20 · Acceleration on a curved speed–time graph, v = t²

    1. Plot v = t² for t = 0 to 5 (v = 0, 1, 4, 9, 16, 25) and draw a smooth curve.
    2. At t = 3 (v = 9) draw a tangent. It passes through about (1.5, 0) and (4, 15).
    3. Gradient = (15 − 0) ÷ (4 − 1.5) = 15 ÷ 2.5 = 6.
    4. On a speed–time graph the gradient is the acceleration: about 6 m/s².

    💡 A negative tangent gradient on a speed–time graph is a deceleration. In the book's example the tangent at 4 s gives −6.4, so the object is decelerating at 6.4 m/s².

    ✏️ Try it

    1. A tangent to a distance–time curve passes through (1, 2) and (5, 14). Estimate the speed.
    2. A tangent to a speed–time curve passes through (0, 20) and (8, 4). What is happening, and at what rate?
    3. Why are tangent answers only estimates?
    Answers

    1. 12 ÷ 4 = 3 m/s · 2. Gradient −16 ÷ 8 = −2: decelerating at 2 m/s² · 3. The tangent is drawn by eye, so the readings are not exact.

    Next: summary and quiz →

    Chapter summary

    • Conversion graph: a straight line from two known points. Read up then across, or across then down.
    • Speed = distance ÷ time; average speed = total distance ÷ total time. Match the units first (1 m/s = 3.6 km/h).
    • Distance–time graph: gradient = speed, horizontal = stopped, sloping down = returning.
    • Speed–time graph: gradient = acceleration (m/s²), horizontal = constant speed, sloping down = deceleration.
    • Area under a speed–time graph = distance. Split it into rectangles, triangles and trapeziums.
    • Curved graph: draw a tangent and find its gradient. The answer is an estimate.

    📄 Practice questions

    The book's Exercises 17.1 to 17.8 and Student assessments 1 to 3 are in the printable assignment for this chapter.

    Mini-quiz

    1. A conversion graph uses €1 = 40 baht. About how many baht is €6?
    2. A bus travels 180 km in 2 h 15 min. What is its average speed?
    3. How far does a runner go in 40 s at 7.5 m/s?
    4. On a distance–time graph, what does a horizontal line show?
    5. A distance–time line rises 120 m in 8 s. What is the speed?
    6. What does the gradient of a speed–time graph give?
    7. A car goes from 0 to 24 m/s in 6 s at a steady rate. What is its acceleration?
    8. A speed–time line goes down from 15 m/s to 0 in 5 s. What is happening?
    9. What does the area under a speed–time graph give?
    10. A bike speeds up from 0 to 10 m/s in 8 s. How far does it go in that time?
    11. A van holds 12 m/s for 20 s, then stops in 6 s. Total distance?
    12. How do you find the speed at one moment on a curved distance–time graph?

    Source: Cambridge IGCSE Mathematics Core and Extended (4th edition, Pimentel and Wall), Chapter 17 (Graphs in practical situations). Explanations, the Try it answers, the tool and the quiz are Guru Jazzy's own. This page is not made or endorsed by Cambridge or Hodder Education.