Velocity is speed in a stated direction. A car moving at 15\ m s^-1 has a speed; a car moving at 15\ m s^-1 due north has a velocity. The size is the same, but velocity also tells you the direction.

What is wrong unit for acceleration?

It is m s^-2, not m s^-1. The extra "per second" is because it is a change of velocity each second.

SingaporePhysicsSyllabus dot point

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

Define speed, velocity and acceleration, and calculate each using simple one-step formulas

Define speed, velocity and acceleration, tell the difference between speed and velocity, and use the formulas for average speed and acceleration with simple N(A)-Level numbers.

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

SEAB wants you to define speed, velocity and acceleration, to tell the difference between speed and velocity, and to calculate each using simple one-step formulas. The key idea is that speed says how fast, velocity says how fast and in which direction, and acceleration says how quickly the velocity is changing.

The answer

Speed

Speed tells you how far something travels in each second. The formula is:

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

The unit is metres per second, written $\text{m s}^{-1}$ . Speed is a size only, with no direction.

Average speed uses the total distance over the total time, even if the object sped up and slowed down along the way.

Velocity

Velocity is speed in a stated direction. A car moving at $15\ \text{m s}^{-1}$ has a speed; a car moving at $15\ \text{m s}^{-1}$ due north has a velocity. The size is the same, but velocity also tells you the direction.

Because velocity has a direction, it can change even when the speed stays the same. A car going round a bend at a steady $15\ \text{m s}^{-1}$ has a changing velocity because its direction is changing.

Acceleration

Acceleration tells you how quickly the velocity changes each second. The formula is:

\text{acceleration} = \frac{\text{change in velocity}}{\text{time taken}} = \frac{v - u}{t}

where $u$ is the starting velocity and $v$ is the final velocity. The unit is metres per second squared, $\text{m s}^{-2}$ .

A positive acceleration means speeding up. A negative acceleration (sometimes called deceleration) means slowing down. An object moving at a steady velocity has zero acceleration.

Scalars and vectors

Speed and distance are scalars: they have size only. Velocity, displacement and acceleration are vectors: they have size and direction. This is why a question that gives a direction is asking about a vector.

Worked example

A train starts from rest and speeds up steadily to $30\ \text{m s}^{-1}$ in $12\ \text{s}$ . It then travels at this steady speed for $200\ \text{m}$ . Find (i) the acceleration during the speeding up, and (ii) the time spent at the steady speed.

Step 1: Identify the values for the acceleration

Starting velocity $u = 0$ (from rest), final velocity $v = 30\ \text{m s}^{-1}$ , time $t = 12\ \text{s}$ .

Step 2: Use the acceleration formula

a = \frac{v - u}{t} = \frac{30 - 0}{12}

Step 3: Calculate the acceleration

a = 2.5\ \text{m s}^{-2}

Step 4: Find the time at the steady speed

Now use speed $= \dfrac{\text{distance}}{\text{time}}$ , rearranged to time $= \dfrac{\text{distance}}{\text{speed}} = \dfrac{200}{30} = 6.7\ \text{s}$ .

So the acceleration is $2.5\ \text{m s}^{-2}$ and the train spends about $6.7\ \text{s}$ at the steady speed.

Examples in context

Example 1. A bus journey. A bus covers $9.0\ \text{km}$ in $15$ minutes. Converting to SI units gives $9000\ \text{m}$ in $900\ \text{s}$ , so the average speed is $9000 \div 900 = 10\ \text{m s}^{-1}$ . The bus stops and starts many times, so its actual speed varies, but the average smooths this out.

Example 2. A lift starting to move. A lift starts from rest and reaches $1.5\ \text{m s}^{-1}$ in $3.0\ \text{s}$ . Its acceleration is $\dfrac{1.5 - 0}{3.0} = 0.5\ \text{m s}^{-2}$ . When it slows to a stop at the top floor, the acceleration is negative, which is why you feel lighter for a moment.

Try this

Cue. A runner covers $400\ \text{m}$ in $50\ \text{s}$ . Find the average speed. [2 marks] Speed $= \dfrac{400}{50} = 8.0\ \text{m s}^{-1}$ .
Cue. Explain why a car going round a roundabout at a steady speed has a changing velocity. [2 marks] Velocity includes direction; going round a bend changes the direction, so the velocity changes even though the speed is constant.
Cue. A motorbike slows from $24\ \text{m s}^{-1}$ to $6\ \text{m s}^{-1}$ in $3.0\ \text{s}$ . Find its acceleration and say what the sign means. [3 marks] $a = \dfrac{6 - 24}{3.0} = -6.0\ \text{m s}^{-2}$ ; the negative sign means it is slowing down.

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 marksA cyclist travels

600\ \text{m}

along a straight road in

40\ \text{s}

. (a) Calculate the average speed. (b) State the unit of speed. (c) Is this a speed or a velocity? Explain.

Show worked answer →

(a) Average speed $= \dfrac{\text{distance}}{\text{time}} = \dfrac{600}{40} = 15\ \text{m s}^{-1}$ .

(b) The unit of speed is metres per second ( $\text{m s}^{-1}$ ).

(c) Because a direction (along a straight road) is given as well as the size, this is a velocity. Speed is the size only; velocity is the size with a direction.

What markers reward: the formula distance over time, the correct unit, and a clear statement that velocity includes direction while speed does not.

Original4 marksA car starts from rest and reaches

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

8.0\ \text{s}

. (a) Calculate the acceleration. (b) State the unit. (c) The car then slows from

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

to rest in

4.0\ \text{s}

. Find this acceleration and say what the sign means.