OCR GCSE Physics A: Distance, speed, velocity, acceleration and motion graphs (J249) – Revision Notes

These notes condense P2.1 Motion (P2.1a to P2.1h) from OCR GCSE (9-1) Gateway Science Physics A (J249), Version 5.0 (July 2026), for first assessment in 2018. All of P2.1 is assessed at both tiers. For full explanations read the study guide; to test yourself use the practice questions. Related pages: the OCR GCSE Physics hub, the checklist, the Forces revision notes and the free 10-minute diagnostics.

Equations to recall and apply

Quantity	Equation	Unit
Distance	distance = speed × time	m
Speed	speed = distance ÷ time	m/s
Acceleration	a = (v − u) ÷ t	m/s²
Uniform acceleration	v² − u² = 2as	m²/s²
Average speed	total distance ÷ total time	m/s

Symbols: u initial velocity, v final velocity, a acceleration, s distance, t time.

Measuring (P2.1a, P2.1b)

• Short, fast motion: light gates or video, not a hand stopwatch.
• Walking or running: tape measure or trundle wheel, plus a stopwatch over the longest practical distance.
• Repeat and take a mean. Start and stop timing on the same signal.
• Ramp investigation (PAG P3): vary the angle, measure the speed, control the trolley, release point and distance.

Units and rates (P2.1c)

Convert	Do this
km to m	× 1000
min to s	× 60
h to s	× 3600
km/h to m/s	÷ 3.6
m/s to km/h	× 3.6
mph to m/s	roughly × 0.447 (1 mile is about 1609 m)

Method in steps

1. Write the quantity with its old unit.
2. Multiply by the ratio that equals 1 (for example 1000 m per 1 km).
3. Check the size: a larger unit gives a smaller number.

Worked reminder. A runner covers 4.5 km in 25 min. Distance 4500 m, time 1500 s, speed 4500 ÷ 1500 = 3.0 m/s, which is 10.8 km/h.

Rates by proportion. 540 m in 3 min 20 s: time 200 s, rate 540 ÷ 200 = 2.7 m/s.

Scalar and vector (P2.1d)

Scalar	Vector
distance	displacement
speed	velocity

• Vectors need a direction; in one dimension, give opposite directions opposite signs.
• Displacement is the straight-line change in position from start to finish.
• Reversing direction makes velocity change sign.
• Constant speed round a circle means changing velocity.

Graph rules (P2.1e, P2.1f)

Graph	Gradient	Area under line	Flat line means
Distance-time	speed	no meaning needed	stationary
Velocity-time	acceleration	distance travelled	constant velocity

Distance-time: straight slope is constant speed, steeper is faster, curve steepening is speeding up, curve flattening is slowing down. On a curve, use the gradient of a tangent at that point.

Velocity-time: upward straight slope is uniform acceleration, downward straight slope is uniform deceleration, a curve is changing acceleration. Below the time axis means motion the other way.

Method for area: split into rectangles and triangles, work out each, add. Triangle ½ × base × height. Trapezium ½ × (sum of parallel sides) × width.

Worked reminder. Velocity rises from 0 to 12 m/s in 4 s, then stays at 12 m/s for 6 s. Distance = ½ × 4 × 12 + 6 × 12 = 24 + 72 = 96 m.

Average speed for non-uniform motion (P2.1g)

• Use total distance ÷ total time for the whole journey.
• Include stops in the time.
• Never average the separate speeds.

Uniform acceleration (P2.1h)

Method in steps for v² − u² = 2as

1. List u, v, a, s and mark the unknown.
2. Write the equation, then substitute with units.
3. Rearrange, for example v² = u² + 2as.
4. Square root last.
5. Check the sign: braking gives negative a.

Worked reminders

• Start at rest, a = 3.2 m/s², s = 40 m: v² = 2 × 3.2 × 40 = 256, so v = 16 m/s.
• Brakes to rest with a = −5.0 m/s² over 22.5 m: u² = 2 × 5.0 × 22.5 = 225, so u = 15 m/s.
• Velocity 6 m/s to 14 m/s in 5 s: a = 8 ÷ 5 = 1.6 m/s², and distance = ½ × (6 + 14) × 5 = 50 m (area of a trapezium).

Describing a journey from a graph

Use a fixed pattern for each section of the graph: say what the line does, say what that means for the motion, then give a number.

Worked reminder (distance-time). Points (0 s, 0 m), (10 s, 25 m), (25 s, 25 m), (35 s, 5 m).

• 0 to 10 s: straight upward slope, constant speed 25 ÷ 10 = 2.5 m/s.
• 10 to 25 s: horizontal, stationary for 15 s.
• 25 to 35 s: straight downward slope, returning towards the start at 20 ÷ 10 = 2.0 m/s, so velocity is −2.0 m/s.
• Distance travelled = 25 + 20 = 45 m. Displacement = 5 m from the start. Average speed = 45 ÷ 35 = 1.3 m/s (2 s.f.).

Worked reminder (velocity-time). A lift speeds up from 0 to 2.4 m/s in 3 s, moves at 2.4 m/s for 5 s, then slows to rest in 2 s.

• Accelerations: 2.4 ÷ 3 = 0.80 m/s², then 0, then −2.4 ÷ 2 = −1.2 m/s².
• Distance = ½ × 3 × 2.4 + 5 × 2.4 + ½ × 2 × 2.4 = 3.6 + 12 + 2.4 = 18 m.

Choosing the right equation

You are given	You want	Use
distance and time	speed	speed = distance ÷ time
speed and time	distance	distance = speed × time
two velocities and a time	acceleration	a = (v − u) ÷ t
u, v and a, no time	distance	v² − u² = 2as
u, a and distance, no time	v	v² = u² + 2as
a graph with a sloping line	speed or acceleration	gradient
a velocity-time graph	distance	area

Rearranging: t = s ÷ v, v = s ÷ t, v = u + at, and a = (v² − u²) ÷ 2s. Practise cover-up triangles only if you can also rearrange without them.

Quick checks on your answer

• A walking speed of about 1 to 2 m/s is sensible; 20 m/s on foot is not.
• Distances from a velocity-time graph must have units of metres when the axes are m/s and s.
• If the object slows down, the final speed must be smaller than the initial speed.
• A positive and a negative answer from a square root: pick the one in the direction of travel.

Must-know distinctions

• Speed has size only (scalar); velocity also has direction (vector).
• Gradient of distance-time is speed; gradient of velocity-time is acceleration.
• Area of velocity-time is distance; area of distance-time has no use here.
• Constant speed does not mean constant velocity if direction changes.
• Deceleration means a negative acceleration when forward is positive.

Quick self-test

1. Convert 72 km/h to m/s.
2. Convert 0.25 m/s to km/h.
3. How long does it take to travel 2.4 km at 8 m/s?
4. State the difference between distance and displacement.
5. What does the gradient of a velocity-time graph give?
6. A car at rest accelerates at 3.2 m/s² over 40 m. State its final speed.
7. A velocity-time graph is a horizontal line at 5 m/s. Describe the motion.
8. A cyclist goes 300 m in 40 s, rests for 20 s, then goes 200 m in 40 s. Find the average speed.
9. A trolley’s velocity rises uniformly from 6 m/s to 14 m/s in 5 s. Find the acceleration.
10. What does a downward-sloping straight line on a displacement-time graph show?

Answers

1. 72 ÷ 3.6 = 20 m/s.
2. 0.25 × 3.6 = 0.90 km/h.
3. Time = 2400 ÷ 8 = 300 s (5 min).
4. Distance is the total path length (scalar). Displacement is the straight-line change in position with direction (vector).
5. Acceleration.
6. v² = 2 × 3.2 × 40 = 256, so 16 m/s.
7. Constant velocity of 5 m/s, zero acceleration.
8. Total distance 500 m, total time 100 s, so 5.0 m/s.
9. (14 − 6) ÷ 5 = 1.6 m/s².
10. Moving back towards the start at constant speed, which is negative velocity.

Where marks are usually lost

• Leaving a speed in km/h or a time in minutes when the unit asked for is m/s.
• Reading a gradient from a tiny triangle, which makes it inaccurate.
• Quoting a size only where velocity or displacement is asked for.
• Treating a flat line on a velocity-time graph as stationary.
• Counting squares for area without checking what one square represents.
• Dividing the distance by only the moving time and missing the rest period.
• Forgetting that a negative acceleration is a slowing down, and quoting a deceleration with the wrong sign.
• Taking the square root too early in a v² − u² = 2as question.

Official syllabus

OCR, GCSE (9-1) Gateway Science Physics A (J249), Version 5.0 (July 2026), for first assessment in 2018, Topic P2 Forces, P2.1 Motion.