SEAB Original-style practice: A speed-time graph shows a car at a constant 20\ m/s for 5\ s. (a) Describe the motion. (b) Find the distance travelled in that time.

(a) The speed does not change, so the car moves at constant speed (zero acceleration). (b) On a speed-time graph the distance is the area under the line: distance = speed × time = 20 × 5 = 100\ m. What markers reward: identifying constant speed from a flat line, knowing the area under a speed-time graph gives distance, and the correct value with the unit m.

SingaporeCombined ScienceSyllabus dot point

How do we describe how fast something moves and how its motion changes over time?

Define speed and acceleration, calculate them from distance and time, and interpret distance-time and speed-time graphs

A focused N(A)-Level answer on motion. Defining and calculating speed and acceleration, and reading distance-time and speed-time graphs including gradient and area meaning.

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 define speed and acceleration, calculate each from simple data, and read the two motion graphs: the distance-time graph and the speed-time graph. The key idea is that speed tells you how fast you cover distance, while acceleration tells you how quickly your speed is changing.

The answer

Speed

Speed is the distance travelled in each second. In symbols:

\text{speed} = \frac{\text{distance}}{\text{time}} \qquad v = \frac{d}{t}

The SI unit is metres per second ( $\text{m/s}$ ). For a journey where the speed changes, the average speed is the total distance divided by the total time.

Acceleration

Acceleration is how much the speed changes each second. In symbols:

\text{acceleration} = \frac{\text{change in speed}}{\text{time taken}} \qquad a = \frac{v - u}{t}

where $u$ is the starting speed and $v$ is the final speed. The unit is metres per second squared ( $\text{m/s}^2$ ). If the object slows down, the change in speed is negative, so the acceleration is negative (this is called deceleration).

The distance-time graph

A distance-time graph plots distance on the vertical axis against time on the horizontal axis:

A straight sloping line means constant speed.
The steeper the line, the faster the object.
A flat (horizontal) line means the object is not moving.
The gradient (slope) of the line equals the speed.

The speed-time graph

A speed-time graph plots speed against time:

A flat line means constant speed.
A sloping line means the object is accelerating (going up) or decelerating (going down).
The gradient equals the acceleration.
The area under the line equals the distance travelled.

Reading a speed-time graph in two parts

A bus starts from rest, accelerates steadily to $12\ \text{m/s}$ in $4\ \text{s}$ , then travels at that constant speed for $6\ \text{s}$ . Find the acceleration and the total distance.

Step 1: Find the acceleration in the first part

The speed changes from $0$ to $12\ \text{m/s}$ in $4\ \text{s}$ :

a = \frac{12 - 0}{4} = 3\ \text{m/s}^2

Step 2: Find the distance in the first part

This is the area of the triangle under the sloping line:

d_1 = \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 4 \times 12 = 24\ \text{m}

Step 3: Find the distance in the second part

This is the area of the rectangle under the flat line:

d_2 = \text{speed} \times \text{time} = 12 \times 6 = 72\ \text{m}

Step 4: Add the two distances

d = 24 + 72 = 96\ \text{m}

So the acceleration was $3\ \text{m/s}^2$ and the bus travelled $96\ \text{m}$ in total.

Examples in context

Example 1. A car braking to a stop. A car moving at $20\ \text{m/s}$ brakes to rest in $4\ \text{s}$ . The acceleration is $\dfrac{0 - 20}{4} = -5\ \text{m/s}^2$ . The negative sign shows it is slowing down (deceleration), and the magnitude tells you it loses $5\ \text{m/s}$ of speed every second.

Example 2. Comparing two runners on a distance-time graph. Two lines on a distance-time graph start together. The steeper line belongs to the faster runner, because a steeper gradient means more distance covered each second. Where the lines cross, both runners are at the same distance at the same time.

Try this

Cue. A train covers $1500\ \text{m}$ in $60\ \text{s}$ . Find its average speed: $\dfrac{1500}{60} = 25\ \text{m/s}$ .
Cue. A car speeds up from $10\ \text{m/s}$ to $30\ \text{m/s}$ in $5\ \text{s}$ . Find its acceleration: $\dfrac{30 - 10}{5} = 4\ \text{m/s}^2$ .
Cue. Explain how to find the distance travelled from a speed-time graph. Calculate the area enclosed between the line and the time axis.

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 cyclist travels

600\ \text{m}

40\ \text{s}

at a steady pace. (a) Calculate the speed. (b) The cyclist then speeds up from

15\ \text{m/s}

21\ \text{m/s}

3\ \text{s}

. Calculate the acceleration.

Show worked answer →

(a) Speed = distance divided by time:

$v = \dfrac{600}{40} = 15\ \text{m/s}$ .

(b) Acceleration = change in speed divided by time:

$a = \dfrac{21 - 15}{3} = \dfrac{6}{3} = 2\ \text{m/s}^2$ .

What markers reward: the correct formula each time, the change in speed (not just the final speed) for acceleration, and the units $\text{m/s}$ and $\text{m/s}^2$ .

Original3 marksA speed-time graph shows a car at a constant

20\ \text{m/s}

for

5\ \text{s}

. (a) Describe the motion. (b) Find the distance travelled in that time.

Show worked answer →

(a) The speed does not change, so the car moves at constant speed (zero acceleration).

(b) On a speed-time graph the distance is the area under the line:

$\text{distance} = \text{speed} \times \text{time} = 20 \times 5 = 100\ \text{m}$ .

What markers reward: identifying constant speed from a flat line, knowing the area under a speed-time graph gives distance, and the correct value with the unit $\text{m}$ .