A distance-time graph rises from 0\ m to 90\ m over 30\ s in a straight line. Find the speed. [2 marks]

A speed-time graph shows constant speed 12\ m/s for 15\ s. Find the distance travelled. [2 marks]

On a speed-time graph, a line rises from 0 to 18\ m/s in 6\ s. Find the acceleration. [2 marks]

SEAB Original-style practice: A car accelerates uniformly from rest to 20\ m/s in 8 seconds, travels at 20\ m/s for 12 seconds, then decelerates to rest in 5 seconds. Using a speed-time graph, find the total distance travelled.

The speed-time graph is a trapezium made of three parts. The distance is the total area under the graph. Acceleration phase (triangle): 12 × 8 × 20 = 80\ m. Constant phase (rectangle): 12 × 20 = 240\ m. Deceleration phase (triangle): 12 × 5 × 20 = 50\ m. Total distance = 80 + 240 + 50 = 370\ m. Markers reward splitting the area into a triangle, rectangle and triangle, computing each, and summing to the total distance.

SingaporeMathsSyllabus dot point

What do the gradient and the area under a travel graph tell us about motion?

Interpret distance-time and speed-time graphs, using the gradient and the area under the graph to find speed, acceleration and distance

A focused answer to the O-Level E-Maths outcome on travel graphs. Reading speed from the gradient of a distance-time graph, acceleration from a speed-time graph, and distance from the area under a speed-time graph.

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 interpret distance-time and speed-time graphs, reading the gradient to find speed or acceleration, and the area under a speed-time graph to find distance travelled. These graphs turn motion into geometry, so the answers come from gradients and areas you can compute by hand.

The answer

Distance-time graphs

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

\text{speed} = \frac{\text{change in distance}}{\text{change in time}}

A straight sloping line means constant speed, a steeper line a faster speed, and a horizontal line means the object is at rest (distance not changing). A curve means the speed is changing.

Speed-time graphs: the gradient

On a speed-time graph, the gradient is the acceleration:

\text{acceleration} = \frac{\text{change in speed}}{\text{change in time}}

A line rising to the right is acceleration, a line falling is deceleration, and a horizontal line is constant speed. The units of acceleration are metres per second per second.

Speed-time graphs: the area

On a speed-time graph, the area between the graph and the time axis is the distance travelled. Split the region into triangles, rectangles and trapeziums, find each area, and add them. A triangle has area $\frac{1}{2} \times \text{base} \times \text{height}$ and a rectangle $\text{base} \times \text{height}$ .

Keeping units consistent

Convert all times and distances to consistent units before reading gradients or areas, usually seconds and metres so that speed is in metres per second. Mixing minutes and seconds is a frequent source of error.

Worked example

A speed-time graph shows a train accelerating uniformly from $5\ \text{m/s}$ to $25\ \text{m/s}$ over $10$ seconds. Find (a) the acceleration and (b) the distance travelled in this time.

Step 1: Find the acceleration from the gradient

\text{acceleration} = \frac{25 - 5}{10} = \frac{20}{10} = 2\ \text{m/s}^2

Step 2: Identify the area shape

The region under the graph is a trapezium with parallel sides $5$ and $25$ and width $10$ .

Step 3: Apply the trapezium area

\text{distance} = \frac{1}{2}(5 + 25) \times 10 = \frac{1}{2} \times 30 \times 10 = 150\ \text{m}

Step 4: State the answers

The acceleration is $2\ \text{m/s}^2$ and the distance travelled is $150\ \text{m}$ .

Examples in context

Example 1. A bus journey. A distance-time graph of a bus shows steep sections where it moves fast, gentle sections in traffic, and flat sections at stops. Reading gradients along the graph reconstructs the speed at each stage of the trip.

Example 2. Braking distance. A speed-time graph of a car braking to a stop is a triangle; its area is the distance covered while stopping. This is why a higher initial speed, giving a taller triangle, means a much greater braking distance.

Try this

Q1. A distance-time graph rises from $0\ \text{m}$ to $90\ \text{m}$ over $30\ \text{s}$ in a straight line. Find the speed. [2 marks]

Cue. Speed is the gradient: $\dfrac{90}{30} = 3\ \text{m/s}$ .

Q2. A speed-time graph shows constant speed $12\ \text{m/s}$ for $15\ \text{s}$ . Find the distance travelled. [2 marks]

Cue. Area is a rectangle: $12 \times 15 = 180\ \text{m}$ .

Q3. On a speed-time graph, a line rises from $0$ to $18\ \text{m/s}$ in $6\ \text{s}$ . Find the acceleration. [2 marks]

Cue. Gradient: $\dfrac{18 - 0}{6} = 3\ \text{m/s}^2$ .

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.

Original4 marksA car accelerates uniformly from rest to

20\ \text{m/s}

8

seconds, travels at

20\ \text{m/s}

for

12

seconds, then decelerates to rest in

5

seconds. Using a speed-time graph, find the total distance travelled.

Show worked answer →

The speed-time graph is a trapezium made of three parts. The distance is the total area under the graph.

Acceleration phase (triangle): $\dfrac{1}{2} \times 8 \times 20 = 80\ \text{m}$ .

Constant phase (rectangle): $12 \times 20 = 240\ \text{m}$ .

Deceleration phase (triangle): $\dfrac{1}{2} \times 5 \times 20 = 50\ \text{m}$ .

Total distance $= 80 + 240 + 50 = 370\ \text{m}$ .

Markers reward splitting the area into a triangle, rectangle and triangle, computing each, and summing to the total distance.

Original3 marksA distance-time graph shows a cyclist covering

1200\ \text{m}

in the first

4

minutes at constant speed, then resting. (a) Find the speed during the first

4

minutes in metres per second. (b) State the gradient of the graph during the rest.

Show worked answer →

(a) Speed is the gradient of a distance-time graph. Convert $4$ minutes to $240$ seconds.

Speed $= \dfrac{1200}{240} = 5\ \text{m/s}$ .

(b) During the rest the cyclist is stationary, so the distance does not change and the gradient is $0$ .

Markers reward speed as the gradient of distance over time with consistent units, the value $5\ \text{m/s}$ , and a zero gradient at rest.