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Distance, displacement, speed and velocity · Fathom

Distance, displacement, speed and velocity

7 minAS Level4.2.1

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can tell scalars from vectors, compute average speed and average velocity, and explain why a journey can have a large distance but a small (or zero) displacement.

Key to the symbols

Symbol	Meaning	Unit
$s$	displacement	metres, m
$t$	time	seconds, s
$u$	velocity at the start	$\text{m s}^{-1}$ (metres per second)
$v$	velocity at the end	$\text{m s}^{-1}$
$a$	acceleration	$\text{m s}^{-2}$ (metres per second, every second)

A scalar has size only. A vector has size and direction.

One lap, two answers

A runner goes once around a 400 m track. Someone watching says "she ran 400 metres". Someone else says "she is back where she started, so she did not move at all".

Both statements are about motion, and both can be useful. They are answering different questions:

How much ground did she cover? 400 m — the distance.
How far is she from the start, in a particular direction? 0 m — the displacement.

Mechanics keeps the two ideas separate, and the words tell you which one is meant.

Definition

Distance and displacement

Distance is the total length of the path travelled. It is a scalar: it has size but no direction, and it can never be negative.

Displacement is the straight-line distance from the starting point to the finishing point, together with the direction. It is a vector.

In one dimension, direction is handled by a sign: choose one direction as positive, and motion the other way has negative displacement.

Distance adds up along the path; displacement is the single arrow from start to finish. Here the finish is 12 − 6 = 6 m east and 8 m north of the start, so by Pythagoras the displacement is √(6² + 8²) = 10 m. In this topic all motion is along one line, so you will only need signs, not Pythagoras.

Speed and velocity

Average speed is distance divided by time:

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

Average velocity is displacement divided by time:

$\text{average velocity} = \frac{\text{displacement}}{\text{time}}$

Velocity is a vector, so it carries a sign. A velocity of $-3\ \text{m s}^{-1}$ means $3\ \text{m s}^{-1}$ in the negative direction.

Instantaneous speed or velocity means the value at one particular moment. The speedometer in a car shows instantaneous speed.

A useful fact: for a round trip, the displacement is zero, so the average velocity is zero — even though the average speed is definitely not.

Worked example: A walk with a change of direction

A shopper walks $400\ \text{m}$ east, then turns round and walks $300\ \text{m}$ west. The whole trip takes $100\ \text{s}$ . Take east as the positive direction.

Find (a) the distance travelled, (b) the displacement, (c) the average speed and (d) the average velocity.

Have a go first?

Acceleration

Acceleration is the rate of change of velocity:

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

where $u$ is the velocity at the start of a time interval of length $t$ , and $v$ is the velocity at the end. This gives the average acceleration over the interval (and the acceleration itself when it is constant).

Acceleration is a vector. In one dimension its sign tells you whether it points along the positive direction or against it.

That gives a result people often find strange: when a car is slowing down while moving in the positive direction, its acceleration is negative. Negative acceleration is not a special kind of motion — it just points the other way.

Confusingly, a negative acceleration can also make a particle speed up, if it's already moving in the negative direction.

The test is not the sign of $a$ on its own. Compare the signs of $a$ and $v$ :

same sign: speeding up;
opposite signs: slowing down.

Common mistakes

Displacement and distance are the same thing.

Distance is the total path length (scalar); displacement is the straight-line change in position with direction (vector). They are equal only for motion in a straight line without reversing.

Velocity and speed are interchangeable.

Speed is the magnitude of velocity. Average velocity uses displacement; average speed uses distance, so they can differ a lot.

Negative acceleration always means slowing down.

It means acceleration points in the negative direction. A particle moving in the negative direction with negative acceleration is speeding up.

Displacement cannot be negative.

In the exam

Read the words carefully: "speed" wants distance over time; "velocity" wants displacement over time, with a direction or a sign.

When a question says "find the displacement", give a direction as well as a size (for example, " $12\ \text{m}$ to the left"), or state your positive direction and give a signed number.

For an average velocity over an interval, use the net displacement, even if the particle turned round inside the interval.

Check yourself

CalculateQuestion 1 · Apply

Multiple choiceQuestion 2 · Recall

Answer the 2 questions above to continue.