What is wrong units for acceleration?

Acceleration is in m s^-2, not m s^-1.

A runner covers 400\ m in 50\ s. Calculate the average speed. [2 marks]

A car accelerates from 8.0\ m s^-1 to 20\ m s^-1 in 6.0\ s. Calculate the acceleration. [2 marks]

SingaporePhysicsSyllabus dot point

What is the difference between speed, velocity, and acceleration, and how are they calculated?

Define speed, velocity, and acceleration, and calculate them for objects moving in a straight line

A focused answer to the O-Level Physics outcome on speed, velocity, and acceleration. Definitions, average versus instantaneous speed, the link between velocity and acceleration, and straight-line calculations with correct units.

Generated by Claude Opus 4.87 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, velocity, and acceleration, to know which are scalars and which are vectors, and to calculate each for an object moving in a straight line. The key relationships are speed as distance over time, velocity as displacement over time, and acceleration as the rate of change of velocity.

The answer

Speed and velocity

Speed is how fast an object moves, with no direction. Velocity is speed in a stated direction.

\text{speed} = \frac{\text{distance}}{\text{time}}, \qquad \text{velocity} = \frac{\text{displacement}}{\text{time}}

Both have the unit metre per second, written $\text{m s}^{-1}$ . Speed is a scalar; velocity is a vector because it includes direction.

Average versus instantaneous speed

Average speed uses the total distance over the total time for a whole journey. Instantaneous speed is the speed at one instant, such as the value shown on a car's speedometer right now. They are equal only when the speed is constant.

Acceleration

Acceleration is how quickly velocity changes. It is a vector, with unit metre per second squared, $\text{m s}^{-2}$ :

a = \frac{v - u}{t}

where $u$ is the initial velocity, $v$ the final velocity, and $t$ the time taken. A positive value means speeding up in the chosen positive direction; a negative value (deceleration) means slowing down.

A worked relationship

If a body starts from rest and accelerates uniformly, its velocity after time $t$ is $v = u + at$ . With $u = 0$ this becomes $v = at$ , so velocity grows in proportion to time when acceleration is constant.

Worked example

A train accelerates uniformly from rest, reaching $30\ \text{m s}^{-1}$ in $20\ \text{s}$ . (a) Find its acceleration. (b) Find its velocity after $12\ \text{s}$ .

Step 1: Find the acceleration

Use $a = \dfrac{v - u}{t}$ with $u = 0$ , $v = 30\ \text{m s}^{-1}$ , $t = 20\ \text{s}$ :

a = \frac{30 - 0}{20} = 1.5\ \text{m s}^{-2}

Step 2: Find the velocity at 12 s

Use $v = u + at$ with $u = 0$ , $a = 1.5\ \text{m s}^{-2}$ , $t = 12\ \text{s}$ :

v = 0 + 1.5 \times 12 = 18\ \text{m s}^{-1}

Step 3: State the answers

The acceleration is $1.5\ \text{m s}^{-2}$ and the velocity after $12\ \text{s}$ is $18\ \text{m s}^{-1}$ . Because the acceleration is constant, velocity rises steadily in step with time.

Examples in context

Example 1. The speedometer. A car's speedometer shows instantaneous speed, the value at this instant. Over a trip the average speed is usually lower, because the total distance is divided by the total time including stops at traffic lights.

Example 2. A lift starting up. When a lift sets off it accelerates from rest, so its velocity rises from zero; near the top floor it decelerates, a negative acceleration, until it stops. The same numbers describe both phases, only the sign of the acceleration changes.

Try this

Q1. A runner covers $400\ \text{m}$ in $50\ \text{s}$ . Calculate the average speed. [2 marks]

Cue. Average speed $= \dfrac{400}{50} = 8.0\ \text{m s}^{-1}$ .

Q2. A car accelerates from $8.0\ \text{m s}^{-1}$ to $20\ \text{m s}^{-1}$ in $6.0\ \text{s}$ . Calculate the acceleration. [2 marks]

Cue. $a = \dfrac{20 - 8.0}{6.0} = \dfrac{12}{6.0} = 2.0\ \text{m s}^{-2}$ .

Q3. Explain why an object moving at constant speed around a circle still has a changing velocity. [2 marks]

Cue. Velocity includes direction; going round a circle the direction changes constantly, so the velocity changes even though the speed stays the same.

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 travels

150\ \text{m}

10\ \text{s}

. (a) Calculate its average speed. (b) The car then speeds up from

15\ \text{m s}^{-1}

25\ \text{m s}^{-1}

4.0\ \text{s}

. Calculate its acceleration.

Show worked answer →

(a) Average speed $= \dfrac{\text{distance}}{\text{time}} = \dfrac{150}{10} = 15\ \text{m s}^{-1}$ .

(b) Acceleration $= \dfrac{\text{change in velocity}}{\text{time}} = \dfrac{25 - 15}{4.0} = \dfrac{10}{4.0} = 2.5\ \text{m s}^{-2}$ .

Markers reward speed as distance over time, acceleration as change in velocity over time, and both answers with correct units.

Original3 marksA cyclist moving at

12\ \text{m s}^{-1}

brakes and stops in

3.0\ \text{s}

. (a) Calculate the acceleration. (b) State what the sign of your answer means.