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.
Make a guess: how can a whole lap give zero velocity?
Mission 1How can a whole lap give zero velocity?
Mission 2How can one line tell the whole journey?
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 — measured in metres per second squared (m/s²). (At Higher tier, acceleration uses the same unit.)
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?
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
And for how quickly speed changes:
rate of change of speed = (final speed − initial speed) ÷ time takenrate of change of speed (m/s²), speeds (m/s), time taken (s)
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.
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 the size of the displacement are only equal when the motion is in a straight line without changing direction. 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
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.
Mission 1 check
How can a whole lap give zero velocity?
?A swimmer swims four lengths of a 25 m pool in 50 s, ending where she started. What are her average speed and her average velocity?
Mission 2 of 2
How can one line tell the whole journey?
Make a guess: what does the slope of a speed–time graph show?
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.
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 · 1.1.7 · 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 (on a speed–time graph, the rate of change of speed);
the area under the line is the displacement (on a speed–time graph, the distance moved).
Gradient = 8 ÷ 4 = 2 m/s² (acceleration). Shaded area = displacement: 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 displacement.
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).
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.
Mission 2 check
How can one line tell the whole journey?
?A cyclist's distance–time graph is a straight line rising from 0 m to 90 m in 15 s. Then it is a flat line for 10 s. What does the graph show?
Recap
The relationships to know
Average speed: distance moved ÷ time taken
Average speed (steady change): (initial speed + final speed) ÷ 2
Rate of change of speed: (final speed − initial speed) ÷ time taken
Average velocity: displacement ÷ time (Higher)
Acceleration: a = (v − u) ÷ t (Higher)
Distance–time graph: slope = speed
Speed–time graph: slope = rate of change of speed; area = distance moved
Velocity–time graph: slope = acceleration; area = displacement (Higher)
Scalars vs vectors: distance/speed = scalar; displacement/velocity/acceleration = vector (Higher)
You've covered the key ideas in 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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You've worked through Motion for CCEA GCSE Physics. 🎉