Distance, Displacement, Speed and Velocity

Imagine running one full lap of a $400 \text{ m}$400 m athletics track. You finish exactly where you started, so in one sense you have gone nowhere — yet your legs have certainly carried you $400 \text{ m}$400 m. This puzzle is solved by realising that physicists use two different measures of how far you have travelled: the total ground covered (distance) and the straight-line shift from start to finish (displacement). The same split applies to how fast you move: speed ignores direction, while velocity includes it. This lesson, part of Topic P2 (Forces) of OCR Gateway Science A, defines all four quantities, gives a feel for everyday speeds, and works through the speed equation $v = \dfrac{s}{t}$v=ts​ in detail.

By the end of this lesson you should be able to distinguish distance from displacement and speed from velocity, recall some typical everyday speeds, use and rearrange the equation $v = \dfrac{s}{t}$v=ts​, and tell the difference between average and instantaneous speed.

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Distance and Displacement

Distance is the total length of the path travelled. It is a scalar — it has size only and takes no account of direction. If you walk $3 \text{ m}$3 m east and then $3 \text{ m}$3 m back west, the distance you have walked is $6 \text{ m}$6 m.

Displacement is the straight-line distance from the start to the finish, together with the direction. It is a vector. For the same there-and-back walk, the displacement is zero, because you end up exactly where you began. Displacement is often described as "how far out of place" you are.

For a runner completing one lap of a $400 \text{ m}$400 m track:

• the distance run is $400 \text{ m}$400 m (the whole path);
• the displacement is zero (start and finish are the same point).

The two are only equal when the motion is in a single straight line with no change of direction.

Exam Tip: Distance is the whole path (scalar); displacement is the straight-line gap from start to finish, with a direction (vector). If an object returns to its starting point, its displacement is zero even though the distance is not.

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Speed and Velocity

Speed is how fast an object is moving — the distance it travels each second. It is a scalar: it tells you the size of the motion but not the direction.

Velocity is the speed in a given direction. It is a vector: a velocity of $5 \text{ m/s}$5 m/s north is different from $5 \text{ m/s}$5 m/s south, even though the speeds are equal.

The distinction has real consequences. An object moving in a circle at a steady speed — say a car going round a roundabout at a constant $10 \text{ m/s}$10 m/s — has a constant speed but a changing velocity, because its direction is constantly changing. A changing velocity means the object is accelerating, even though its speed never changes; you will return to this idea when you study circular motion and acceleration.

Quantity	Scalar or vector?	What it tells you
Distance	scalar	total path length
Displacement	vector	straight-line shift + direction
Speed	scalar	how fast (size only)
Velocity	vector	how fast and in which direction

Exam Tip: Speed is a scalar and velocity is a vector. An object can move at constant speed but changing velocity if its direction changes — for example, anything moving in a circle.

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Typical Everyday Speeds

OCR expects you to have a rough feel for the speeds of everyday things, so that you can judge whether an answer is sensible. You do not need exact figures, but you should know the approximate order of magnitude. All speeds vary with conditions (a person's fitness, the wind, the type of road), so these are typical values:

Motion	Typical speed
Walking	about $1.5 \text{ m/s}$1.5 m/s
Running	about $3 \text{ m/s}$3 m/s
Cycling	about $6 \text{ m/s}$6 m/s
Car on a road	about $13$13–$30 \text{ m/s}$30 m/s
A commercial aeroplane	about $250 \text{ m/s}$250 m/s
Sound travelling through air	about $330 \text{ m/s}$330 m/s

The speed of a car spans a wide range because it depends on the road: roughly $13 \text{ m/s}$13 m/s in a town ($30 \text{ mph}$30 mph) up to about $30 \text{ m/s}$30 m/s on a motorway ($70 \text{ mph}$70 mph). The speed of sound in air of about $330 \text{ m/s}$330 m/s is worth fixing in memory, as it appears in questions about echoes and thunder.

Exam Tip: Learn the rough ladder: walking $\approx 1.5 \text{ m/s}$≈1.5 m/s, running $\approx 3 \text{ m/s}$≈3 m/s, cycling $\approx 6 \text{ m/s}$≈6 m/s, car $\approx 13$≈13–$30 \text{ m/s}$30 m/s, sound in air $\approx 330 \text{ m/s}$≈330 m/s. Use them to sanity-check answers — a "person walking at $50 \text{ m/s}$50 m/s" should ring alarm bells.

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The Speed Equation

The speed of an object is the distance travelled divided by the time taken:

$v = \dfrac{s}{t}$v=ts​

where $v$v is the speed (in $\text{m/s}$m/s), $s$s is the distance travelled (in $\text{m}$m) and $t$t is the time taken (in $\text{s}$s). The same equation works for velocity if $s$s is taken as the displacement.

The equation rearranges to make distance or time the subject:

$v = \dfrac{s}{t} \qquad s = v\,t \qquad t = \dfrac{s}{v}$v=ts​s=vtt=vs​

A formula triangle is a handy memory aid: cover the quantity you want and the triangle shows the calculation.

Distance sits at the top, so $s = v\,t$s=vt; speed and time sit side by side at the bottom, so $v = \dfrac{s}{t}$v=ts​ and $t = \dfrac{s}{v}$t=vs​.

Worked Example 1 — Finding speed

A cyclist travels $300 \text{ m}$300 m in $50 \text{ s}$50 s. Calculate their speed.

Step 1 — write the equation: $v = \dfrac{s}{t}$v=ts​.

Step 2 — substitute: $v = \dfrac{300}{50}$v=50300​.

Step 3 — calculate: $v = 6 \text{ m/s}$v=6 m/s.

Answer: the cyclist's speed is $6 \text{ m/s}$6 m/s, which is a typical cycling speed.

Worked Example 2 — Rearranging for distance

A train travels at a steady $45 \text{ m/s}$45 m/s for $120 \text{ s}$120 s. How far does it travel?

Step 1 — rearrange to make distance the subject: $s = v\,t$s=vt.

Step 2 — substitute: $s = 45 \times 120$s=45×120.

Step 3 — calculate: $s = 5400 \text{ m}$s=5400 m.

Answer: the train travels $5400 \text{ m}$5400 m (which is $5.4 \text{ km}$5.4 km).

Worked Example 3 — Rearranging for time

How long does it take sound to travel $1650 \text{ m}$1650 m through air? (Speed of sound in air $= 330 \text{ m/s}$=330 m/s.)