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Cambridge IGCSE Mathematics 0580 · Algebra and graphs

Graphs in practical situations

Reading and drawing graphs that describe real things: conversion graphs, distance-time graphs of journeys, and the gradient of a line as a rate of change. Extended students also use speed-time graphs for acceleration and distance, and estimate the gradient of a curve by drawing a tangent.

C2.9E2.9Core and ExtendedFree lesson

What the syllabus says

C2.9 (Core)

1 Use and interpret graphs in practical situations including travel graphs and conversion graphs.
2 Draw graphs from given data.

Notes and examples

e.g. interpret the gradient of a straight-line graph as a rate of change.
e.g. draw a distance-time graph to represent a journey.

E2.9 (Extended)

1 Use and interpret graphs in practical situations including travel graphs and conversion graphs.
2 Draw graphs from given data.
3 Apply the idea of rate of change to simple kinematics involving distance-time and speed-time graphs, acceleration and deceleration.
4 Calculate distance travelled as area under a speed-time graph.

Notes and examples

Includes estimation and interpretation of the gradient of a tangent at a point.
Areas will involve linear sections of the graph only.

Quoted from the Cambridge IGCSE Mathematics 0580 syllabus for 2025 to 2027.

Reading a graph in context

A graph in a real situation is a picture of how one quantity changes with another. Before you read anything from it, read the axes: what each one measures, its units, and how much one small square is worth. Most marks lost on these questions are lost on the scale, not the idea.

To read a value, go across from the axis you know to the line, then straight to the other axis. Use a ruler on the real paper: a line drawn by eye drifts by a small square, which is often exactly the tolerance the mark scheme allows.

Work out the value of one small square first. If 10 units on an axis are split into 5 small squares, each small square is worth $10 \div 5 = 2$ units. A reading of three small squares past 40 is then $46$ , not $43$ .

Conversion graphs

A conversion graph changes a quantity in one unit into another: miles and kilometres, kilograms and pounds, one currency and another. When the two units are directly proportional (0 miles is 0 km), the graph is a straight line through the origin, and its gradient is the conversion rate.

Two skills are tested:

Reading both ways. Start on whichever axis holds the value you are given.
Going beyond the graph. If the graph stops at 50 miles and you need 150 miles, read 50 miles and multiply by 3. This works only because the line passes through the origin: doubling one quantity doubles the other.

Not every conversion goes through the origin. Temperature is the classic case: $0^{°} C$ is $32^{°} F$ , so doubling the Celsius value does not double the Fahrenheit value. A graph with a fixed charge, such as a taxi fare or a phone bill, is the same: the intercept on the vertical axis is the cost before you have travelled or used anything.

A conversion graph

Across: miles. Up: kilometres. Each small square across is 2 miles and each small square up is 2 km. The line passes through $(50, 80)$ , so 50 miles is 80 km and the conversion rate is $80 \div 50 = 1.6$ km per mile.

Worked example 1

Use the conversion graph above to convert (a) 35 miles to kilometres, (b) 40 km to miles and (c) 200 miles to kilometres.

(a) Go up from 35 on the miles axis to the line, then across to the kilometres axis. The reading is $56$ km.
(b) Start this time on the kilometres axis: across from 40 to the line, then down. The reading is $25$ miles.
(c) 200 miles is off the graph. Read a value that divides into 200: 50 miles is 80 km.
$200 = 4 \times 50$ , and the line goes through the origin, so 200 miles is $4 \times 80 = 320$ km.
A sensible check: a kilometre is shorter than a mile, so the number of kilometres must be bigger. It is.

Answer: (a) $56$ km (b) $25$ miles (c) $320$ km

Distance-time graphs

A distance-time graph (a travel graph) shows how far something is from a starting point as time passes. Time goes across and distance goes up. Each straight section is part of the journey at a constant speed, and the graph tells you everything about it:

The gradient is the speed. Gradient is $\frac{change in distance}{change in time}$ , and that is exactly what speed means.
Steeper means faster. A shallow line is slow travel.
A horizontal line means stopped. Time passes but the distance does not change, so the speed is 0.
A line going down means coming back towards the starting point. The speed is still positive; the downward slope shows the direction.

Watch the units. If time is in minutes and you want km/h, either change the minutes to hours first (40 minutes is $\frac{40}{60} = \frac{2}{3}$ of an hour) or work out the distance in one minute and multiply by 60.

The average speed for a whole journey is $\frac{total distance}{total time}$ , and the total time includes the stops. It is not the average of the separate speeds.

A journey on a distance-time graph

A cyclist leaves home and rides 15 km in 30 minutes, rests for 20 minutes, rides 10 km further in 30 minutes, then rides home in 40 minutes. One small square across is 2 minutes and one small square up is 1 km.

Worked example 2

For the cycle ride in the graph above, work out (a) the speed for the first 30 minutes, in km/h, (b) the speed on the way home, in km/h, and (c) the average speed for the whole ride, including the rest.

(a) 15 km in 30 minutes. 30 minutes is $\frac{1}{2}$ hour, so the speed is $15 \div \frac{1}{2} = 30$ km/h.
(b) The way home is the last section: from 25 km back to 0, between 80 and 120 minutes. That is 25 km in 40 minutes.
40 minutes is $\frac{40}{60} = \frac{2}{3}$ hour, so the speed is $25 \div \frac{2}{3} = 25 \times \frac{3}{2} = 37.5$ km/h.
(c) Total distance: 25 km out and 25 km back, so 50 km. Total time: 120 minutes, which is 2 hours, including the rest.
Average speed $= \frac{50}{2} = 25$ km/h.

Answer: (a) $30$ km/h (b) $37.5$ km/h (c) $25$ km/h

Gradient as a rate of change

The gradient of a straight line is $\frac{change up}{change across}$ . In a practical graph both changes have units, so the gradient has units too: the units up per unit across.

Distance (km) against time (hours): gradient in km per hour, a speed.
Volume (litres) against time (minutes): gradient in litres per minute, a rate of flow.
Cost (£) against distance (km): gradient in £ per km, a price per kilometre.

To find a gradient, choose two points on the line that are far apart and easy to read, ideally where the line crosses grid lines. A line sloping downwards has a negative gradient: the quantity is decreasing. In words you say it as a positive rate: a gradient of $- 15$ litres per minute means the water level is falling by 15 litres every minute.

The intercept on the vertical axis also has a meaning: it is the value when the quantity across is zero, such as the starting volume or a fixed charge.

Check yourself

A distance-time graph is a straight line from $(0, 0)$ to $(15, 6)$ , with time in minutes and distance in km. Work out the speed in km/h.

Answer it in the portal, where your answer is checked as you type →

Check yourself

Use the miles and kilometres conversion graph in this lesson to convert 120 km to miles.

Answer it in the portal, where your answer is checked as you type →

Drawing a graph from given data

You may be given a table of values or a description of a journey and asked to draw the graph. The method:

Use the scales on the grid you are given. If you choose your own, make one small square a simple amount (1, 2, 5 or 10 units) and use as much of the grid as you can.
Plot each point with a small, neat cross. A mark is often given for the points alone.
Join points with a ruler when the situation changes at a steady rate (a journey at constant speed, a fare that rises by the same amount each kilometre).

For a journey described in words, turn each stage into a time and a distance before you draw it. Time is distance divided by speed: 1.5 km at 4.5 km/h takes $\frac{1.5}{4.5} = \frac{1}{3}$ hour, which is 20 minutes. Then each stage starts where the last one ended.

Worked example 3

Sara leaves home and walks 1.5 km to a friend's house at 4.5 km/h. She stays there for 10 minutes. She then walks 1.5 km further to a park, taking 15 minutes. She jogs straight home from the park, arriving 1 hour after she first left. Draw the distance-time graph of her journey.

First stage: time $= \frac{1.5}{4.5} = \frac{1}{3}$ hour $= 20$ minutes. Draw a line from $(0, 0)$ to $(20, 1.5)$ .
The stay: 10 minutes at the same distance. Draw a horizontal line from $(20, 1.5)$ to $(30, 1.5)$ .
To the park: 1.5 km further, so 3 km from home, 15 minutes later. Draw a line from $(30, 1.5)$ to $(45, 3)$ .
Home: she arrives at 60 minutes, at distance 0. Draw a line from $(45, 3)$ down to $(60, 0)$ .
The finished graph is shown above. Her jogging speed home is $3$ km in 15 minutes, which is $12$ km/h: the steepest section is the fastest.

Answer: The graph shown, through $(0, 0)$ , $(20, 1.5)$ , $(30, 1.5)$ , $(45, 3)$ and $(60, 0)$

Speed-time graphs Extended

A speed-time graph has speed going up and time going across. It looks like a travel graph, but it means something different, and confusing the two is the most expensive mistake in this topic.

The gradient is the acceleration. Acceleration is $\frac{change in speed}{change in time}$ , measured in ${m/s}^{2}$ when speed is in m/s and time in seconds.
A horizontal line means constant speed, not stopped. The acceleration is 0.
A line going down means slowing down. The gradient is negative; its size is the deceleration. A gradient of $- 2$ is a deceleration of $2 {m/s}^{2}$ .
The area under the graph is the distance travelled.

Why the area? At a constant speed, distance $=$ speed $\times$ time, and on the graph that is the height times the width of a rectangle. When the speed changes steadily from $u$ to $v$ , the average speed over that stretch is $\frac{u + v}{2}$ , so the distance is $\frac{u + v}{2} \times t$ , which is exactly the area of a trapezium (or a triangle when one speed is 0).

Split the area into triangles, rectangles and trapeziums, find each one, and add. In this syllabus the graph is made of straight sections only, so these shapes are all you need.

A speed-time graph Extended

The bus accelerates for 10 seconds, travels at a constant 15 m/s for 30 seconds, then slows to a stop in 15 seconds.

Worked example 4 Extended

For the bus in the speed-time graph above, work out (a) the acceleration in the first 10 seconds, (b) the deceleration in the last 15 seconds, (c) the total distance travelled and (d) the average speed for the 55 seconds.

(a) Gradient of the first section: $\frac{15 - 0}{10 - 0} = 1.5 {m/s}^{2}$ .
(b) Gradient of the last section: $\frac{0 - 15}{55 - 40} = - 1$ . The deceleration is the size of this: $1 {m/s}^{2}$ .
(c) The shape under the graph is a trapezium with parallel sides 30 (the top, from 10 to 40) and 55 (the base), and height 15.
Area $= \frac{1}{2} (30 + 55) \times 15 = \frac{1}{2} \times 85 \times 15 = 637.5$ m.
Check by pieces: triangle $\frac{1}{2} \times 10 \times 15 = 75$ , rectangle $30 \times 15 = 450$ , triangle $\frac{1}{2} \times 15 \times 15 = 112.5$ . Total $75 + 450 + 112.5 = 637.5$ m.
(d) Average speed $= \frac{637.5}{55} = 11.6$ m/s (3 s.f.).

Answer: (a) $1.5 {m/s}^{2}$ (b) $1 {m/s}^{2}$ (c) $637.5$ m (d) $11.6$ m/s

Worked example 5 Extended

A car starts from rest and accelerates steadily to $V$ m/s in 12 seconds. It then travels at $V$ m/s for 20 seconds. The total distance travelled in the 32 seconds is 520 m. Find $V$ .

Sketch the graph: a triangle from 0 to 12 seconds, then a rectangle from 12 to 32 seconds, both of height $V$ .
Area of the triangle: $\frac{1}{2} \times 12 \times V = 6 V$ .
Area of the rectangle: $20 \times V = 20 V$ .
The total area is the distance: $6 V + 20 V = 520$ , so $26 V = 520$ and $V = 20$ .

Answer: $V = 20$

Check yourself Extended

A cyclist slows steadily from 9 m/s to 3 m/s in 4 seconds. Work out the distance travelled in those 4 seconds.

Answer it in the portal, where your answer is checked as you type →

Check yourself Extended

On a speed-time graph, a straight line goes from $(5, 18)$ to $(13, 6)$ , with time in seconds and speed in m/s. Work out the deceleration.

Answer it in the portal, where your answer is checked as you type →

Curved graphs: the gradient of a tangent Extended

When a travel graph or a speed-time graph is curved, the speed or acceleration is changing all the time. The gradient at one instant is the gradient of the tangent at that point: the straight line that just touches the curve there, without cutting across it.

To estimate it:

Mark the point on the curve. Lay a ruler against the curve so that it touches at that point and the curve bends away equally on both sides.
Draw a long tangent. A long line makes the reading far more accurate.
Choose two points on the tangent (not on the curve) far apart, and work out $\frac{change up}{change across}$ .

On a distance-time graph the tangent gradient is the speed at that instant. On a speed-time graph it is the acceleration at that instant. Because the tangent is drawn by eye, the answer is an estimate, and the mark scheme accepts a range.

A tangent to a distance-time curve Extended

The dashed tangent touches the curve at $P (4, 8)$ . It passes through $(2, 0)$ and $(8, 24)$ , so its gradient is $\frac{24 - 0}{8 - 2} = 4$ . The object is moving at about 4 m/s when $t = 4$ .

Worked example 6 Extended

Use the curve above to compare the speed at $t = 4$ with the average speed over the first 4 seconds.

Average speed over the first 4 seconds: distance 8 m in 4 s, so $\frac{8}{4} = 2$ m/s. This is the gradient of the chord from $(0, 0)$ to $(4, 8)$ , not of the tangent.
Speed at $t = 4$ : from the tangent, about 4 m/s.
The speed at the instant is greater than the average so far, which makes sense: the curve gets steeper, so the object is speeding up.

Answer: About $4$ m/s at $t = 4$ , against an average of $2$ m/s

Key points

Read the scale first: work out what one small square is worth on each axis.
A conversion graph through the origin can be scaled: read a value that divides in, then multiply.
Distance-time: gradient is speed, horizontal means stopped, downward means returning.
The gradient of any practical graph is a rate, in units up per unit across; the intercept is the starting value.
Average speed is total distance over total time, stops included.
Extended, speed-time: gradient is acceleration, horizontal means constant speed, area under the graph is distance.
Extended: the gradient of a curve at a point is the gradient of the tangent there; draw it long and read two points on the tangent.

Where marks are lost

Misreading the scale, for example counting a small square as 1 when it is worth 2.
Leaving time in minutes when the speed is asked for in km/h. Divide the minutes by 60 first.
Averaging two speeds to get an average speed. Always use total distance over total time.
Leaving the stops out of the total time for an average speed.
Treating a speed-time graph as a distance-time graph: a horizontal line is constant speed, not a stop.
Forgetting the $\frac{1}{2}$ in the area of a triangle under a speed-time graph.
Giving a deceleration as a negative number, or an acceleration with the units m/s instead of ${m/s}^{2}$ .
Reading two points on the curve instead of on the tangent, which finds an average rate, not the rate at the point.

Related IGCSE Mathematics 0580 lessons

C2.10 · E2.10

All IGCSE Mathematics 0580 lessons →

An original lesson written to the Cambridge IGCSE Mathematics 0580 syllabus (2025 to 2027). The worked examples and questions are our own, not Cambridge questions. Back to every IGCSE Mathematics 0580 lesson.

Graphs in practical situations

Reading a graph in context

Conversion graphs

A conversion graph

Worked example 1

Distance-time graphs

A journey on a distance-time graph

Worked example 2

Gradient as a rate of change

Check yourself

Check yourself

Drawing a graph from given data

Worked example 3

Speed-time graphs Extended

A speed-time graph Extended

Worked example 4 Extended

Worked example 5 Extended

Check yourself Extended

Check yourself Extended

Curved graphs: the gradient of a tangent Extended

A tangent to a distance-time curve Extended

Worked example 6 Extended

Key points

Where marks are lost

Related IGCSE Mathematics 0580 lessons

Graphs of functions

Proportion

Sketching curves

Sequences