SingaporeMathsSyllabus dot point

How does a distance-time graph describe a journey, and how do we read speed from its gradient?

Interpret and draw distance-time graphs, find speed from the gradient, and describe stationary and return stages of a journey

A focused answer to the N(A)-Level Mathematics outcome on travel graphs. Reading distance-time graphs, finding speed from the gradient, and describing rest stops and return journeys.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to read and draw distance-time graphs, find speed from the gradient of the graph, and describe the stages of a journey including moving, resting and returning. A travel graph turns a journey into a picture, so learning to read its slopes is the whole skill.

The answer

What a distance-time graph shows

A distance-time graph has time on the horizontal axis and distance (from a starting point) on the vertical axis. Each point shows how far the traveller is from the start at a given time. The shape of the line tells the story of the journey.

Speed is the gradient

On a distance-time graph, speed is the gradient of the line:

\text{speed} = \frac{\text{distance travelled}}{\text{time taken}}

A steeper line means a faster speed. A gentle slope means a slow speed. This is the same gradient idea as for a straight-line graph, applied to a journey.

Reading the stages of a journey

Different line shapes mean different things:

An upward slope means moving away from the start.
A horizontal (flat) line means staying still - the distance is not changing, so the traveller is at rest.
A downward slope means returning toward the start.
A steeper line means a higher speed than a gentler one.

Average speed for a whole journey

For a journey with several stages (including any rests), the average speed for the whole trip is the total distance travelled divided by the total time taken, including the time spent stationary.

Units and conversions

Take care with units. If distance is in kilometres and time in minutes, convert the time to hours before dividing to get a speed in km/h. To convert minutes to hours, divide by $60$ .

Worked example

A motorist drives $60\ \text{km}$ in $1$ hour, stops for $30$ minutes, then drives a further $30\ \text{km}$ in $30$ minutes. Find the speed of each driving stage and the average speed for the whole journey.

Step 1: Speed of the first stage

\text{speed} = \frac{60\ \text{km}}{1\ \text{h}} = 60\ \text{km/h}.

Step 2: Speed of the second driving stage

$30$ minutes is $0.5$ hour, so:

\text{speed} = \frac{30\ \text{km}}{0.5\ \text{h}} = 60\ \text{km/h}.

Step 3: Total distance and total time

Total distance $= 60 + 30 = 90\ \text{km}$ . Total time includes the rest: $1 + 0.5 + 0.5 = 2\ \text{h}$ .

Step 4: Average speed for the whole journey

\text{average speed} = \frac{90\ \text{km}}{2\ \text{h}} = 45\ \text{km/h}.

The average is lower than each driving speed because the $30$ -minute rest counts in the total time.

Examples in context

Example 1. Comparing two cyclists. On the same axes, a steep line and a gentle line both start at the origin. The steeper line belongs to the faster cyclist, because a bigger gradient means more distance covered per unit time. Comparing slopes is the quickest way to compare speeds without any calculation.

Example 2. A delivery round. A van's graph rises, flattens while a parcel is dropped off, rises again, then slopes back down to the depot. Each flat section is a delivery stop and the final downward slope is the return. Describing a real journey from its graph shape is exactly what exam questions ask for.

Try this

Cue. A runner covers $4\ \text{km}$ in $20$ minutes. Convert: $20$ min $= \dfrac{1}{3}$ h, so speed $= 4 \div \dfrac{1}{3} = 12\ \text{km/h}$ .
Cue. On a distance-time graph, what does a flat horizontal line mean? The distance is unchanged, so the traveller is at rest.
Cue. A car travels $90\ \text{km}$ in total over $2$ hours including a stop. Its average speed is $\dfrac{90}{2} = 45\ \text{km/h}$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original3 marksOn a distance-time graph, a cyclist travels from home to a park

12\ \text{km}

away in

40

minutes. Find the cyclist's average speed in km/h.

Show worked answer →

Speed is the gradient of a distance-time graph: distance divided by time.

Convert $40$ minutes to hours: $40 \div 60 = \dfrac{2}{3}$ hour.

Speed $= \dfrac{12}{\frac{2}{3}} = 12 \times \dfrac{3}{2} = 18\ \text{km/h}$ .

What markers reward: identifying speed as the gradient, converting the time to hours before dividing, and the correct unit km/h. Forgetting to convert the minutes is the usual error.

Original4 marksA distance-time graph shows a walker going

3\ \text{km}

1

hour, then resting for

30

minutes, then returning home in

1

hour. Describe what each stage of the graph looks like and find the speed of the return stage.

Show worked answer →

Stage 1: a line sloping upward from $0$ to $3\ \text{km}$ over $1$ hour (moving away from home).

Stage 2: a flat horizontal line for $30$ minutes at $3\ \text{km}$ (resting, no change in distance).

Stage 3: a line sloping downward from $3\ \text{km}$ back to $0$ over $1$ hour (returning home).

Return speed: distance $3\ \text{km}$ in $1$ hour gives $3\ \text{km/h}$ .

What markers reward: an upward slope for moving away, a flat line for rest, a downward slope for returning, and the correct return speed with units. Saying the flat section means "stopped" earns the rest mark.