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CCEA GCSE Physics · Motion
Mini-Lesson

Motion

This mini-lesson walks you through the whole of CCEA Unit 1.1 — Motion: speed, distance and time, the difference between scalars and vectors, displacement, velocity and acceleration, and how to read motion graphs.

start finish distance moved 🏃
How far (distance) and how fast (speed) — measured over a span of time.

Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Press Start when you're ready.

Quantities & SI units · 1.1.1

The quantities of motion

Every motion measurement is a physical quantity with a number and a unit. CCEA uses these SI units:

  • Distance (and displacement) — measured in metres (m).
  • Speed (and velocity) — measured in metres per second (m/s).
  • Rate of change of speed (acceleration) — measured in metres per second squared (m/s²).
  • Time — measured in seconds (s).

Everyday speeds: a person walks at roughly 1.5 m/s, a sprinter runs at around 10 m/s, and a car on a town road travels near 13 m/s (about 30 mph). Knowing these helps you sense-check an answer.

Quick check

Match the unit

?In CCEA's SI units, which is the correct unit for the rate of change of speed (acceleration)?
Speed & average speed · 1.1.1

Speed = distance ÷ time

Speed tells you how quickly an object covers distance. CCEA wants you to recall and use the average speed equation:

average speed = distance moved ÷ time takenaverage speed (m/s) = distance moved (m) ÷ time taken (s)

There is a second form CCEA lists for steady changes of speed — the average of the start and end speeds:

average speed = (initial speed + final speed) ÷ 2useful when speed changes at a steady rate
Worked example

A cyclist travels 300 m in 25 s.

average speed = 300 ÷ 25 = 12 m/s

Calculate

Your turn — average speed

1A train covers a distance of 1500 m in 60 s. Calculate its average speed.
m/s
Hint: average speed = distance ÷ time = 1500 ÷ 60.
Scalars & vectors · 1.1.3 · Higher tier

Scalars vs vectors

A scalar has size only. A vector has size and direction. CCEA pairs them up:

  • Distance is a scalar; displacement is a vector — both in metres (m).
  • Speed is a scalar; velocity is a vector — both in m/s.
  • Rate of change of speed is a scalar; acceleration is a vector — both in m/s².
distance = path length (scalar) displacement = straight line + direction (vector) start end
Walk a curving path and your distance is the whole route; your displacement is the straight arrow from start to end.

Watch out: distance and displacement are only equal when the motion is in a perfectly straight line. Walk a full lap of a track and your distance is 400 m, but your displacement is zero — you finished where you started.

Sort it

Scalar or vector?

Tap a quantity, then tap the box it belongs in.

📏 Scalar (size only)

🧭 Vector (size + direction)

Velocity · 1.1.4 · Higher tier

Velocity is rate of change of displacement

Where speed only says "how fast", velocity says "how fast and in which direction". CCEA defines it through displacement:

average velocity = displacement ÷ timeaverage velocity (m/s) = displacement (m) ÷ time (s)

For a steady change of velocity in one direction, you can also use the average of the start and end velocities:

average velocity = (initial velocity + final velocity) ÷ 2CCEA only sets problems on motion in one direction

Watch out: two cars can have the same speed (30 m/s) but different velocities if they travel in different directions. Velocity is a vector — change the direction and you change the velocity, even if the speed stays the same.

Calculate

Your turn — average velocity

2A runner has a displacement of 180 m due north in a time of 24 s. Calculate her average velocity.
m/s
Hint: average velocity = displacement ÷ time = 180 ÷ 24.
Acceleration · 1.1.4–1.1.5 · Higher tier

Acceleration = change in velocity ÷ time

Acceleration is how quickly velocity changes. For motion in one direction:

a = (v − u) ÷ tacceleration (m/s²) = (final velocity − initial velocity) ÷ time taken
v u (initial) v (final) velocity rising time t →

Watch out: acceleration is the change in velocity, not the velocity itself. An object can be moving fast yet have zero acceleration (steady velocity). A negative acceleration — slowing down — is called retardation in CCEA.

Worked example

A car speeds up from u = 8 m/s to v = 20 m/s in t = 4 s.

a = (20 − 8) ÷ 4 = 12 ÷ 4 = 3 m/s²

Calculate

Your turn — acceleration

3A motorbike accelerates from rest (u = 0) to a final velocity v = 18 m/s in a time of 6 s. Calculate its acceleration.
m/s²
Hint: a = (v − u) ÷ t = (18 − 0) ÷ 6.
Distance–time graphs · 1.1.6

Distance–time graphs: slope = speed

On a distance–time graph the gradient (slope) is the speed. The steeper the line, the faster the object. A flat line means it is stationary.

distance (m) time (s) 0 40 60 10 20 Δdistance = 40 m Δtime = 10 s slope = 40 ÷ 10 = 4 m/s flat = stationary
Gradient of the sloped part = 40 m ÷ 10 s = 4 m/s. The flat section means the object has stopped.
Read the graph

Reading a distance–time graph

?On a distance–time graph, an object's line is perfectly horizontal (flat) for 8 seconds. What is happening during that time?
Velocity–time graphs · 1.1.6–1.1.7 · Higher tier

Velocity–time graphs: slope & area

A velocity–time graph packs in two pieces of information CCEA expects you to extract:

  • the slope (gradient) is the acceleration;
  • the area under the line is the distance moved (displacement).
velocity (m/s) time (s) 0 12 6 10 Δv = 12 m/s Δt = 6 s slope = 12 ÷ 6 = 2 m/s² (acceleration) area = distance moved
Gradient = 12 ÷ 6 = 2 m/s² (acceleration). Shaded area = distance: triangle (½×6×12 = 36 m) + rectangle (4×12 = 48 m) = 84 m.

Watch out: don't confuse the two graphs. On a distance–time graph the slope is the speed; on a velocity–time graph the slope is the acceleration and the area underneath is the distance.

Read the graph

Your turn — gradient of a v–t graph

4On a velocity–time graph, a straight line rises from 4 m/s to 28 m/s over a time of 8 s. Calculate the acceleration (the gradient).
m/s²
Hint: gradient = Δvelocity ÷ Δtime = (28 − 4) ÷ 8.
Read the graph

Your turn — area under a v–t graph

5A car moves at a steady velocity of 15 m/s for 12 s. On its velocity–time graph this is a flat line. Find the distance moved (the area under the line).
m
Hint: the area is a rectangle = velocity × time = 15 × 12.
Calculate

Your turn — rearranging speed

6A jogger runs at an average speed of 4 m/s for 90 s. Calculate the distance moved.
m
Hint: rearrange speed = distance ÷ time → distance = speed × time = 4 × 90.
Match it

Which graph tells you what?

Tap a feature on the left, then its meaning on the right.

Recap

The relationships to know

Average speed: distance moved ÷ time taken

Average speed (steady change): (initial speed + final speed) ÷ 2

Average velocity: displacement ÷ time (Higher)

Acceleration: a = (v − u) ÷ t (Higher)

Distance–time graph: slope = speed

Speed–time / velocity–time graph: slope = acceleration; area = distance moved

Scalars vs vectors: distance/speed = scalar; displacement/velocity/acceleration = vector (Higher)

You've covered the whole of CCEA Unit 1.1 — Motion: quantities and units, speed and average speed, scalars and vectors, displacement, velocity and acceleration, and motion graphs. Press Finish to see your score.

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Mini-lesson complete!

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