What are travel graphs and how are they used in Cambridge IGCSE Mathematics?

Travel graphs are a type of line graph used to represent the movement of an object over time, typically showing distance against time. In Cambridge IGCSE Mathematics, they help students understand concepts of speed, distance, and time, and how these quantities relate to each other. By analyzing travel graphs, students can interpret real-world scenarios and solve problems involving motion.

How do you interpret a travel graph in terms of speed?

In a travel graph, speed is represented by the gradient (slope) of the line. A steeper slope indicates a higher speed, while a flatter slope indicates a lower speed. If the line is horizontal, it means the object is stationary. By calculating the gradient, which is the change in distance divided by the change in time, students can determine the speed of the object at different intervals.

What is the significance of a horizontal line on a travel graph?

A horizontal line on a travel graph indicates that the object is not moving; its distance remains constant over time. This means the speed is zero. Understanding this helps students identify periods of rest or stops in a journey, which is crucial for solving problems related to travel and motion in the Cambridge IGCSE Mathematics curriculum.

How can you calculate the total distance traveled using a travel graph?

To calculate the total distance traveled using a travel graph, you need to sum up the distances covered during each segment of the journey. This can be done by identifying the vertical changes between points on the graph. Each segment's distance is the difference in the y-values (distance) between two points, and adding these differences gives the total distance traveled.

What does a curve on a travel graph indicate about the object's motion?

A curve on a travel graph indicates that the object's speed is changing over time. If the curve is getting steeper, the object is accelerating, meaning its speed is increasing. Conversely, if the curve is becoming less steep, the object is decelerating, or slowing down. Understanding curves is important for analyzing non-uniform motion in Cambridge IGCSE Mathematics.

Why is "Mixing up distance-time and speed-time graphs" a common mistake in Travel graphs?

Why it happens: Both have time on the horizontal axis; students forget which y-axis they're reading. How to avoid it: Always check the y-axis label. Distance-time → gradient is speed. Speed-time → gradient is acceleration, area is distance. Source: 0580 Extended — recurring

Why is "Excluding stationary periods from total time" a common mistake in Travel graphs?

Why it happens: Students focus on the moving sections. How to avoid it: Average speed uses **total** distance over **total** time, including stops.

Why is "Mixing m/s with km/h or seconds with minutes" a common mistake in Travel graphs?

Why it happens: Reading off a graph in km/h then computing area in seconds. How to avoid it: Convert units BEFORE doing the calculation. m/s ↔ km/h: multiply or divide by 3.6.

Why is "Computing area as just $v \times t$ for a non-rectangular region" a common mistake in Travel graphs?

Why it happens: Treating an acceleration phase as constant speed. How to avoid it: If the graph is sloped, the region is a triangle or trapezium — use the proper formula.

Travel graphs – Study Notes & Past Paper Style Questions | Cambridge IGCSE Maths 0580 Extended

Short Notes - Travel graphs

Detailed Study Notes

Detailed notes on Algebra for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Travel Graphs (Distance-Time and Speed-Time) — Cambridge IGCSE 0580 Maths Extended (2026)

Two graphs, two interpretations. Distance-time gradient = speed; speed-time area = distance. Master both and motion problems on Paper 4 become a 5-mark gift.

At a glance

Distance-time graph: $y$ = distance, $x$ = time.
Gradient of $d$ - $t$ graph = SPEED.
Horizontal segment = stationary.
Steeper line = faster.
Speed-time graph: $y$ = speed, $x$ = time.
Gradient of $s$ - $t$ graph = ACCELERATION.
AREA under $s$ - $t$ graph = DISTANCE.
Horizontal $s$ - $t$ segment = constant speed.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.16 — Construct, interpret and use travel graphs (distance-time and speed-time).

Distance-time graphs

Time on the x-axis, distance on the y-axis. Gradient gives speed.

Reading a distance-time graph.

Feature	Meaning
Straight line, positive gradient	Constant speed (away from start)
Straight line, negative gradient	Constant speed (returning toward start)
Horizontal line	Stationary
Steeper slope	Faster

The gradient between any two points is the AVERAGE SPEED over that interval: $\text{speed} = \dfrac{\Delta d}{\Delta t} = \dfrac{\text{change in distance}}{\text{change in time}}.$

Worked. A car travels $400$ m from $t = 0$ to $t = 80$ s, then stays parked from $80$ to $100$ s, then returns $400$ m from $100$ to $200$ s.

Outbound speed: $\dfrac{400}{80} = 5$ m/s.
Stationary: $0$ m/s.
Return speed: $\dfrac{400}{100} = 4$ m/s.

A horizontal segment means stationary; a falling segment means returning toward the start.

Total distance vs displacement. A $d$ - $t$ graph showing the runner returning to start has TOTAL distance $= 800$ m but final DISPLACEMENT $= 0$ .

Time on x, distance on y.
Gradient = speed.
Horizontal = stationary.
Returning curve has negative gradient (or graph drops back to 0).

Speed-time graphs

Time on the x-axis, speed on the y-axis. Gradient = acceleration; AREA UNDER = distance.

Reading a speed-time graph.

Feature	Meaning
Horizontal line	Constant speed
Positive gradient (rising)	Accelerating
Negative gradient (falling)	Decelerating
Returning to zero	Coming to a stop

Two formulas.

Gradient: $\text{acceleration} = \dfrac{\Delta v}{\Delta t}$ .
Area under the line: $\text{distance} = \int v \, dt$ (geometrically, sum of trapezia/rectangles/triangles).

Worked. A vehicle accelerates from rest to $20$ m/s over $5$ seconds, holds $20$ m/s for $10$ seconds, then decelerates uniformly to rest in $4$ seconds. Find the total distance.

The graph has three regions:

Triangle: $\dfrac{1}{2} \times 5 \times 20 = 50$ m.
Rectangle: $20 \times 10 = 200$ m.
Triangle: $\dfrac{1}{2} \times 4 \times 20 = 40$ m.

Total: $50 + 200 + 40 = 290$ m.

Split the area into a triangle, rectangle and triangle — their sum is the total distance.

Acceleration in each phase.

Phase 1: $\dfrac{20 - 0}{5} = 4$ m/s $^{2}$ .
Phase 2: $0$ (constant speed).
Phase 3: $\dfrac{0 - 20}{4} = -5$ m/s $^{2}$ (deceleration).

Time on x, speed on y.
Gradient = acceleration.
Area under the curve = distance.
Use rectangles, triangles, and trapezia to compute areas.

Average speed and average acceleration

Total distance / total time. Watch units.

Average speed for a complete journey: $\bar{v} = \dfrac{\text{total distance}}{\text{total time}}.$

This is NOT the arithmetic mean of segment speeds — see the Speed/Distance/Time notes for the trap.

Average acceleration between two times: $\bar{a} = \dfrac{v_{\text{final}} - v_{\text{initial}}}{t_{\text{final}} - t_{\text{initial}}}.$

On a distance-time graph read the gradient; on a speed-time graph read the area underneath.

Unit conversions (covered in the Speed/Distance/Time notes):

km/h to m/s: divide by $3.6$ .
m/s to km/h: multiply by $3.6$ .
min to seconds: $\times 60$ .

Always check that the question's units match your computation.

Average speed = TOTAL distance / TOTAL time.
Average acceleration = $\Delta v / \Delta t$ .
Convert units BEFORE plugging in.
km/h to m/s: $\div 3.6$ .

Quick recap

$d$ - $t$ graph: gradient is speed; horizontal is stationary.
$s$ - $t$ graph: gradient is acceleration; AREA is distance.
Sum trapezia, rectangles and triangles to find areas.
Average speed = total distance / total time.
Always check units before computing.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Distance-time graph — gradient gives speed; horizontal segment is stationary.
Speed-time graph — gradient gives acceleration; area under the curve gives distance.
Acceleration — rate of change of velocity, units m/s $^{2}$ .
Average speed — total distance divided by total time.

How it’s examined

Travel graphs appear most years on Paper 4 as a 5-7 mark question, often combining $d$ - $t$ and $s$ - $t$ interpretation. Paper 2 has simpler 2-3 mark items. Examiner reports flag confusing $d$ - $t$ with $s$ - $t$ (mixing area and gradient), and computing simple-mean instead of total-distance-over-total-time for average speed.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.16); 0580/42 Oct/Nov 2024 — Q11 (multi-leg speed-time); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Travel graphs

Step-by-step solutions to past-paper-style questions on travel graphs, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Travel graphs

Almost every travel graphs exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Graph or diagram

Recognise it by

The question gives — or asks you to draw — a distance-time or speed-time graph, quoting segments, plateaus or corner coordinates.

How to approach it

Check the $y$ -axis label first. On a distance-time graph the gradient is speed; on a speed-time graph the gradient is acceleration and the area under it is distance — decompose the area into triangles, rectangles and trapezia.

Common trap

Mixing up the two graph types, or plotting a return journey on a distance-time graph as an upward segment. Examiner reports flag the return as downward — distance is measured from the start.

Word problem

Recognise it by

A real-world journey described in words — stages, stops and rests — with no graph drawn, asking for total distance or average speed.

How to approach it

Compute each stage's distance with $d = vt$ , total the distances and total the times, then average speed $= \dfrac{\text{total distance}}{\text{total time}}$ .

Common trap

Leaving the stationary rest period out of the total time. Examiner reports flag this — excluding stops inflates the average speed.

Multi-step problem

Recognise it by

Several quantities chained — an acceleration phase, a constant-speed phase and a stated total distance — with an unknown time or speed to find.

How to approach it

Break the journey into its phases, write each phase's distance as an area (triangle or rectangle), add them, set the sum equal to the given total and solve for the unknown.

Common trap

Treating an acceleration phase as constant speed and using $d = v \times t$ on a sloped region. Examiner reports flag the sign and unit slips this causes — use the area method.

1Read a distance-time graph
CoreGraph or diagram• distance-time
Question
On a distance-time graph, a runner covers $400$ metres in $80$ seconds (constant speed). What is her speed in m/s and km/h?
Step-by-step solution
1. Step 1
  Speed = gradient = $\dfrac{\Delta d}{\Delta t}$ .
  $v = \dfrac{400}{80} = 5\,\text{m/s}$
2. Step 2
  Convert m/s → km/h by multiplying by $3.6$ .
  $5 \times 3.6 = 18\,\text{km/h}$
Answer
$5\,\text{m/s} = 18\,\text{km/h}$
2Average speed for a multi-stage journey
ExtendedWord problem• Adapted from 0580/42 May/Jun 2024 Q7• average speed
Question
A car travels $60\,\text{km}$ in $1\,\text{h}$ , stops for $30\,\text{min}$ , then travels $90\,\text{km}$ in $1.5\,\text{h}$ . Find the average speed for the whole journey.
Step-by-step solution
1. Step 1
  Total distance.
  $60 + 90 = 150\,\text{km}$
2. Step 2
  Total time, including the stop.
  $1 + 0.5 + 1.5 = 3\,\text{h}$
3. Step 3
  Average speed.
  $\bar{v} = \dfrac{150}{3} = 50\,\text{km/h}$
Answer
$50\,\text{km/h}$
Examiner tip
Include the stationary period in total time. Excluding it gives a higher (wrong) average speed.
3Find acceleration from a speed-time graph
ExtendedGraph or diagram• speed-time, acceleration
Question
A car accelerates uniformly from $5\,\text{m/s}$ to $25\,\text{m/s}$ in $4\,\text{seconds}$ . Find the acceleration.
Step-by-step solution
1. Step 1
  Acceleration = gradient of speed-time graph.
  $a = \dfrac{\Delta v}{\Delta t} = \dfrac{25 - 5}{4} = 5\,\text{m/s}^{2}$
Answer
$a = 5\,\text{m/s}^{2}$
4Distance from a speed-time graph
ExtendedGraph or diagram• Adapted from 0580/42 Oct/Nov 2024 Q12• area, distance
Question
A car accelerates from rest to $20\,\text{m/s}$ in $4\,\text{s}$ , then travels at $20\,\text{m/s}$ for $10\,\text{s}$ , then decelerates to rest in $6\,\text{s}$ . Find the total distance travelled.
Step-by-step solution
1. Step 1
  Distance = area under the speed-time graph. The shape is a trapezium with parallel sides $10$ and $4 + 10 + 6 = 20$ .
2. Step 2
  Trapezium area.
  $\text{Area} = \tfrac{1}{2}(a + b) \times h = \tfrac{1}{2}(10 + 20)(20) = 300$
Answer
$300\,\text{m}$
Examiner tip
The trapezium method works whenever the graph has a constant-speed plateau between acceleration and deceleration. Decompose into triangles + rectangle if the shape is awkward.
5Read two speeds from a distance-time graph
CoreGraph or diagram• distance-time, two stages
Question
A walker's distance-time graph has two straight segments: from $(0, 0)$ to $(20\,\text{min}, 1.2\,\text{km})$ , then $(20\,\text{min}, 1.2\,\text{km})$ to $(50\,\text{min}, 3.6\,\text{km})$ . Find her speed during each stage in km/h.
Step-by-step solution
1. Step 1
  Stage 1: gradient $= \dfrac{1.2 - 0}{20 - 0} = 0.06\,\text{km/min}$ .
2. Step 2
  Convert: $0.06 \times 60 = 3.6\,\text{km/h}$ .
3. Step 3
  Stage 2: gradient $= \dfrac{3.6 - 1.2}{50 - 20} = \dfrac{2.4}{30} = 0.08\,\text{km/min}$ .
4. Step 4
  Convert: $0.08 \times 60 = 4.8\,\text{km/h}$ .
Answer
Stage 1: $3.6\,\text{km/h}$ ; Stage 2: $4.8\,\text{km/h}$ .
6Total distance and average speed
CoreWord problem• average speed, total
Question
A cyclist travels for $2\,\text{h}$ at $15\,\text{km/h}$ , rests for $0.5\,\text{h}$ , then travels for $1\,\text{h}$ at $25\,\text{km/h}$ . Find the total distance and her average speed for the whole journey.
Step-by-step solution
1. Step 1
  Distance covered in each moving stage.
  $d_{1} = 15 \times 2 = 30\,\text{km};\ d_{2} = 25 \times 1 = 25\,\text{km}$
2. Step 2
  Total distance.
  $30 + 25 = 55\,\text{km}$
3. Step 3
  Total time = moving + rest.
  $2 + 0.5 + 1 = 3.5\,\text{h}$
4. Step 4
  Average speed.
  $\bar{v} = \dfrac{55}{3.5} \approx 15.7\,\text{km/h}$
Answer
Total distance $55\,\text{km}$ ; average speed $\approx 15.7\,\text{km/h}$ .
7Draw a distance-time graph from a description
ExtendedGraph or diagram• Adapted from 0580/42 May/Jun 2023 Q5• draw, distance-time
Question
A boy walks $400\,\text{m}$ to school in $5$ minutes, stays at school for $10$ minutes, then runs back home covering $400\,\text{m}$ in $4$ minutes. Describe the distance-time graph: state coordinates of the corner points.
Step-by-step solution
1. Step 1
  Starting point $(0, 0)$ .
2. Step 2
  After $5$ min he is $400\,\text{m}$ from home: $(5, 400)$ .
3. Step 3
  Stationary for $10$ min — horizontal segment to $(15, 400)$ .
4. Step 4
  Returns home in $4$ min — line down to $(19, 0)$ .
Answer
Plot $(0, 0) \to (5, 400) \to (15, 400) \to (19, 0)$ , joining with straight line segments.
Examiner tip
The examiner report flags candidates who plot the return journey as ANOTHER positive segment (treating distance as a total). Distance-time means distance from START — the return is downward.
8Deceleration from a speed-time graph
ExtendedGraph or diagram• deceleration, gradient
Question
A train decelerates uniformly from $30\,\text{m/s}$ to a stop over $24\,\text{seconds}$ . Find the deceleration.
Step-by-step solution
1. Step 1
  Acceleration $= \dfrac{\Delta v}{\Delta t} = \dfrac{0 - 30}{24} = -1.25\,\text{m/s}^{2}$ .
2. Step 2
  Magnitude of deceleration is $1.25\,\text{m/s}^{2}$ .
Answer
Deceleration $= 1.25\,\text{m/s}^{2}$ .
Examiner tip
State deceleration as a POSITIVE value (the magnitude). The examiner report flags students who leave the negative sign in the final 'deceleration' answer.
9Combined distance-time and speed-time (Challenge)
ChallengeMulti-step problem• Adapted from 0580/42 Oct/Nov 2023 Q9• combined, challenge
Question
A car accelerates uniformly from rest for $10\,\text{s}$ , reaching $20\,\text{m/s}$ . It then travels at $20\,\text{m/s}$ for $T$ seconds. The total distance travelled in this $(10 + T)$ -second period is $500\,\text{m}$ . Find $T$ .
Step-by-step solution
1. Step 1
  Distance during acceleration = triangle area.
  $d_{1} = \tfrac{1}{2} \times 10 \times 20 = 100\,\text{m}$
2. Step 2
  Distance at constant speed = rectangle area.
  $d_{2} = 20T$
3. Step 3
  Set total equal to $500\,\text{m}$ .
  $100 + 20T = 500$
4. Step 4
  Solve.
  $20T = 400 \Rightarrow T = 20\,\text{s}$
Answer
$T = 20\,\text{seconds}$ .
Examiner tip
The examiner report flags candidates who use $d = v \times t$ with the average speed for the acceleration phase but make sign / unit slips. Stick to the AREA-UNDER-GRAPH method: triangles, rectangles, trapezia.

Key Formulae — Travel graphs

The formulae you need to memorise for travel graphs on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Speed from distance-time graph
$v = \dfrac{\Delta d}{\Delta t}$
When to use
Gradient of a distance-time graph gives the speed at that interval.
Acceleration from speed-time graph
$a = \dfrac{\Delta v}{\Delta t}$
When to use
Gradient of a speed-time graph gives the acceleration.
Distance from speed-time graph
$d = \text{area under the graph}$
When to use
Area = total distance for that section. Decompose into triangles, rectangles or trapezia.
Trapezium area
$A = \dfrac{1}{2}(a + b) \times h$
$a, b$
lengths of the two parallel sides
$h$
perpendicular distance between them
When to use
Common shape under speed-time graphs with a plateau.

Key Definitions and Keywords — Travel graphs

Definitions to memorise and the exact keywords mark schemes credit for travel graphs answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Distance-time graph
Examiner keyword
Graph with time on the horizontal axis and distance from start on the vertical axis. Gradient = speed.
Speed-time graph
Examiner keyword
Graph with time on the horizontal axis and speed on the vertical axis. Gradient = acceleration; area = distance.
Acceleration
Examiner keyword
Rate of change of speed (m/s²). Constant gradient on a speed-time graph means uniform acceleration.
Deceleration
Negative acceleration — speed is decreasing over time. Shown as a downward-sloping segment on a speed-time graph.
Stationary section
On a distance-time graph: horizontal line (no distance covered). On a speed-time graph: zero speed segment.

Common Mistakes and Misconceptions — Travel graphs

The traps other students keep falling into on travel graphs questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Mixing up distance-time and speed-time graphs
0580 Extended — recurring
Why it happens
Both have time on the horizontal axis; students forget which y-axis they're reading.
How to avoid it
Always check the y-axis label. Distance-time → gradient is speed. Speed-time → gradient is acceleration, area is distance.
✕Excluding stationary periods from total time
Why it happens
Students focus on the moving sections.
How to avoid it
Average speed uses total distance over total time, including stops.
✕Mixing m/s with km/h or seconds with minutes
Why it happens
Reading off a graph in km/h then computing area in seconds.
How to avoid it
Convert units BEFORE doing the calculation. m/s ↔ km/h: multiply or divide by $3.6$ .
✕Computing area as just $v \times t$ for a non-rectangular region
Why it happens
Treating an acceleration phase as constant speed.
How to avoid it
If the graph is sloped, the region is a triangle or trapezium — use the proper formula.

Travel graphs — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Travel graphs

Short Notes - Travel graphs

Detailed Study Notes

At a glance

What you’ll learn

Distance-time graphs

Speed-time graphs

Average speed and average acceleration

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Travel graphs

1Read a distance-time graph

2Average speed for a multi-stage journey

3Find acceleration from a speed-time graph

4Distance from a speed-time graph

5Read two speeds from a distance-time graph

6Total distance and average speed

7Draw a distance-time graph from a description

8Deceleration from a speed-time graph

9Combined distance-time and speed-time (Challenge)

Speed from distance-time graph

Acceleration from speed-time graph

Distance from speed-time graph

Trapezium area

Distance-time graph

Speed-time graph

Acceleration

Deceleration

Stationary section

✕Mixing up distance-time and speed-time graphs

✕Excluding stationary periods from total time

✕Mixing m/s with km/h or seconds with minutes

✕Computing area as just v×tv \times tv×t for a non-rectangular region

Travel graphs — frequently asked questions

Travel graphs

Short Notes - Travel graphs

Detailed Study Notes

At a glance

What you’ll learn

Distance-time graphs

Speed-time graphs

Average speed and average acceleration

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Travel graphs

1Read a distance-time graph

2Average speed for a multi-stage journey

3Find acceleration from a speed-time graph

4Distance from a speed-time graph

5Read two speeds from a distance-time graph

6Total distance and average speed

7Draw a distance-time graph from a description

8Deceleration from a speed-time graph

9Combined distance-time and speed-time (Challenge)

Speed from distance-time graph

Acceleration from speed-time graph

Distance from speed-time graph

Trapezium area

Distance-time graph

Speed-time graph

Acceleration

Deceleration

Stationary section

✕Mixing up distance-time and speed-time graphs

✕Excluding stationary periods from total time

✕Mixing m/s with km/h or seconds with minutes

✕Computing area as just v×tv \times tv×t for a non-rectangular region

Travel graphs — frequently asked questions

✕Computing area as just $v \times t$ for a non-rectangular region

✕Computing area as just $v \times t$ for a non-rectangular region