Cambridge IGCSE Mathematics 0580: Graphs in practical situations – Study Guide

This study guide covers subtopic 2.9, Graphs in practical situations, from the Cambridge IGCSE Mathematics 0580 syllabus for examination in 2025, 2026 and 2027. It teaches both the Core outcomes (C2.9) and the Extended outcomes (E2.9). Core candidates sit Papers 1 (non-calculator) and 3 (calculator); Extended candidates sit Papers 2 (non-calculator) and 4 (calculator). Sections marked Extended only are not examined on the Core papers.

For the whole of Topic 2, see the Algebra and graphs study guide and revision notes.

What this subtopic covers

Syllabus reference	What you must be able to do	Tier
C2.9 / E2.9 (1)	Use and interpret graphs in practical situations, including travel graphs and conversion graphs; interpret the gradient of a straight-line graph as a rate of change	Core and Extended
E2.9 (1), notes	Estimate and interpret the gradient of a tangent at a point on a curve	Extended only
C2.9 / E2.9 (2)	Draw graphs from given data, e.g. a distance–time graph for a journey	Core and Extended
E2.9 (3)	Apply rate of change to simple kinematics: distance–time and speed–time graphs, acceleration and deceleration	Extended only
E2.9 (4)	Calculate distance travelled as the area under a speed–time graph (linear sections only)	Extended only

The speed, gradient and trapezium area formulas are not on the 0580 list of formulas printed on the papers (only the triangle area formula is), so learn them.

Gradient as a rate of change

The gradient of a straight line is

gradient = change in y / change in x  =  rise / run

In a practical graph the axes have units, so the gradient has units too: (units of the vertical axis) per (unit of the horizontal axis). That is why a gradient is a rate of change.

Graph	Gradient means	Units
Distance against time	speed	km/h, m/s
Cost against number of items	price per item	$ per item
Volume against time	rate of flow	litres per minute
Speed against time (Extended)	acceleration	m/s²

The vertical intercept (where the line meets the vertical axis) is the value when the horizontal quantity is zero, such as a fixed charge or a starting depth.

To measure a gradient from a drawn graph, pick two points far apart on grid lines and read the scales carefully.

Conversion graphs

A conversion graph changes one unit or currency into another. If zero of one matches zero of the other, the line passes through the origin and the quantities are in direct proportion.

Worked example 1: a graph through the origin

A conversion graph for miles (horizontal axis) and kilometres (vertical axis) is a straight line through (0, 0) and (50, 80).

(a) Convert 30 miles to kilometres. (b) Convert 200 km to miles. (c) State the gradient and what it means.

(a) Go up from 30 on the miles axis to the line, then across:
    the line gives 48 km.
    Check: 80/50 = 1.6 km per mile, 30 x 1.6 = 48 km
(b) 200 km is beyond the top of the graph. Because the line
    passes through the origin, use a value you can read and scale:
    80 km = 50 miles, so 200 km = 50 x (200/80) = 125 miles
(c) gradient = 80/50 = 1.6 -> 1 mile is 1.6 km

Scaling only works when the line goes through the origin.

Worked example 2: a line that does not pass through the origin

A graph shows the cost, $C, of a taxi ride against distance, d km. It is a straight line through (0, 3) and (10, 18).

(a) Find the gradient and say what it represents. (b) What does the intercept represent? (c) A ride cost $24. How far was it?

(a) gradient = (18 - 3)/(10 - 0) = 15/10 = 1.5
    The fare rises by $1.50 for every extra kilometre.
(b) At d = 0 the cost is $3: a fixed charge paid on every ride.
(c) C = 1.5d + 3
    24 = 1.5d + 3  ->  1.5d = 21  ->  d = 14 km

You cannot scale here. A 20 km ride costs 33 dollars, not twice the 18 dollars for 10 km, because the fixed charge is paid once.

Travel graphs (distance–time)

On a distance–time graph, time is on the horizontal axis and distance from a fixed point (often “home”) is on the vertical axis.

• The gradient is the speed. A steeper line means a faster speed.
• A horizontal section means the object is stationary (distance not changing).
• A line sloping down means moving back towards the start; the speed is still positive.
• Average speed = total distance travelled / total time taken, including any time spent stopped.

Change minutes into hours before calculating a speed in km/h: 45 minutes = 45/60 = 0.75 hours.

Worked example 3: interpreting a journey

A delivery van’s distance–time graph is made of straight lines joining these points:

Time	09 00	09 45	10 15	11 15
Distance from depot (km)	0	36	36	0

(a) Find the speed on the way out. (b) How long did the van stop? (c) Find the speed on the way back. (d) Find the average speed for the whole trip.

(a) 36 km in 45 min = 36 / 0.75 = 48 km/h
(b) horizontal from 09 45 to 10 15 = 30 minutes
(c) 36 km in 1 hour = 36 km/h
(d) total distance = 36 + 36 = 72 km
    total time = 09 00 to 11 15 = 2.25 hours
    average speed = 72 / 2.25 = 32 km/h

In (d), the van ends where it started but still travelled 72 km, and the stop counts in the time.

Drawing graphs from given data

To draw a practical graph:

1. Put time (or the quantity you control) on the horizontal axis.
2. Use the scale given, or choose an easy one (1, 2, 5 or 10 per square) that fills the grid.
3. Label both axes with the quantity and its unit.
4. Plot points accurately with small crosses.
5. Join points with a ruler when the rate is constant (constant speed, fixed price per unit). Use a smooth curve only when the quantity changes smoothly and the rate is not constant.

Worked example 4: drawing a distance–time graph from a description

Amira leaves home at 10 00 and cycles at 16 km/h for 30 minutes. She rests for 20 minutes, then cycles straight home at 12 km/h. Work out the points needed for her distance–time graph and find her average speed.

Leg 1: distance = 16 x 0.5 = 8 km     -> she is 8 km away at 10 30
Rest:  20 minutes, still 8 km away    -> 10 50
Leg 3: time = 8 / 12 h = 40 minutes   -> home at 11 30

Points to plot and join with straight lines:
(10 00, 0)  (10 30, 8)  (10 50, 8)  (11 30, 0)

Average speed = 16 km / 1.5 h = 10.7 km/h (3 s.f.)

The return leg is less steep than the first because 12 km/h is slower than 16 km/h.

Gradient of a tangent to a curve (Extended only)

On a curved graph the rate of change varies. The rate of change at one point is the gradient of the tangent there; on a distance–time graph it is the speed at that instant.

Method:

1. Use a ruler to draw a straight line that touches the curve at the point and does not cut across it nearby.
2. Make the tangent long, then pick two points on it far apart.
3. Gradient = change in y / change in x, using those two points (not points on the curve).
4. Give the units and say what the value means.

Worked example 5: estimating a speed

A ball rolls down a slope. Its distance–time graph is a curve through (0, 0), (1, 1), (2, 4), (3, 9), (4, 16) and (5, 25), with distance in metres and time in seconds. Estimate the speed of the ball after 3 seconds.

Tangent drawn at (3, 9). Suppose it passes through about
(1.5, 0) and (5, 21).
gradient = (21 - 0) / (5 - 1.5) = 21 / 3.5 = 6
Speed after 3 s is about 6 m/s.

A tangent drawn by eye gives an estimate. A line joining two points on the curve (a chord) gives an average speed over an interval, not the speed at 3 s.

Speed–time graphs, acceleration and deceleration (Extended only)

On a speed–time graph, speed is on the vertical axis.

• Gradient = acceleration (rate of change of speed), in m/s² when speed is in m/s and time in s.
• A line sloping down shows deceleration. State deceleration as a positive number: a gradient of −2 m/s² is a deceleration of 2 m/s².
• A horizontal line means constant speed (acceleration 0). It does not mean the object has stopped. That only happens where the graph is on the time axis.
• Area under the graph = distance travelled. The syllabus only uses straight-line sections, so split the area into rectangles, triangles and trapeziums.

triangle  = 1/2 x base x height
trapezium = 1/2 x (a + b) x h     (a, b = parallel sides)

Keep units matched: to convert km/h to m/s, divide by 3.6 (72 km/h = 20 m/s); to convert m/s to km/h, multiply by 3.6.

Feature	Distance–time	Speed–time
Gradient	speed	acceleration
Horizontal line	stationary	constant speed
Area under graph	no meaning	distance travelled

Worked example 6: acceleration, deceleration and distance

A cyclist starts from rest and reaches 12 m/s after 8 s. She rides at 12 m/s for 20 s, then slows steadily to rest in 6 s.

(a) Find the acceleration. (b) Find the deceleration. (c) Find the total distance. (d) Find the average speed.

(a) acceleration = 12 / 8 = 1.5 m/s^2
(b) gradient = -12 / 6 = -2, so deceleration = 2 m/s^2
(c) Whole shape is a trapezium: parallel sides 20 s (top)
    and 8 + 20 + 6 = 34 s (base), height 12 m/s
    distance = 1/2 x (20 + 34) x 12 = 324 m
    Check by parts: 48 + 240 + 36 = 324 m
(d) average speed = 324 / 34 = 9.53 m/s (3 s.f.)

Worked example 7: working backwards from a distance

A car’s speed rises steadily from 5 m/s to 15 m/s in 4 s. It then travels at 15 m/s for T seconds. The total distance is 340 m. Find T and the acceleration.

First section is a trapezium:
  1/2 x (5 + 15) x 4 = 40 m
Second section is a rectangle: 15T
  40 + 15T = 340  ->  15T = 300  ->  T = 20 s
acceleration = (15 - 5) / 4 = 2.5 m/s^2

The first section is a trapezium because the car does not start from rest.

Common errors

• Reading a scale as 1 unit per square when it is 2, 5 or 10.
• Dividing by minutes when the answer must be in km/h (36 km in 45 minutes is 48 km/h, not 0.8).
• Leaving out stationary time when finding an average speed.
• Calling the gradient of a distance–time graph an acceleration. It is speed.
• Scaling a conversion graph that does not pass through the origin.
• (Extended) Measuring a tangent’s gradient from points on the curve instead of points on the tangent.
• (Extended) Treating the first part of a speed–time graph as a triangle when the starting speed is not zero.
• (Extended) Giving a deceleration as a negative number when the question asks for the deceleration.

Next steps

• Practise with the graphs in practical situations practice questions.
• Revise gradients and straight lines in the Coordinate geometry study guide.
• Wider Topic 2 practice: Algebra and graphs practice questions and Algebra and graphs (Extended) practice.
• Find your weak areas with the free Core diagnostic or Extended diagnostic.
• See every topic on the Cambridge IGCSE Mathematics hub and tick off outcomes on the printable checklist.

Official syllabus

Cambridge IGCSE Mathematics 0580 syllabus for examination in 2025, 2026 and 2027 (Version 3), Cambridge University Press & Assessment. Subtopic C2.9 / E2.9 Graphs in practical situations.