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

This guide teaches P2.1 Motion (statements P2.1a to P2.1h) from the OCR GCSE (9-1) Gateway Science Physics A specification (J249), Version 5.0 (July 2026), for first assessment in 2018. Every P2.1 statement is assessed at both Foundation and Higher Tier. It sits inside Topic P2 Forces, so use it with the Forces study guide, the Forces revision notes and the Forces practice questions.

Quick links: revision notes for this unit, practice questions for this unit, the OCR GCSE Physics hub, the printable checklist and the free 10-minute diagnostics.

What this unit covers

Statement	You must be able to
P2.1a	Describe how to measure distance and time in a range of scenarios
P2.1b	Measure distance and time from graphs and use them to calculate speed
P2.1c	Use ratios and proportional reasoning to convert units (including non-SI to SI) and compute rates
P2.1d	Explain the vector-scalar distinction: displacement and distance, velocity and speed
P2.1e	Relate changes and differences in motion to distance-time and velocity-time graphs; interpret lines and slopes
P2.1f	Interpret the enclosed area in velocity-time graphs
P2.1g	Calculate average speed for non-uniform motion
P2.1h	Apply the equations linking distance, time and speed for uniform motion, and for uniform acceleration

The equations to recall and apply are distance = speed × time, acceleration = change in velocity ÷ time, and (final velocity)² − (initial velocity)² = 2 × acceleration × distance.

Measuring distance and time (P2.1a, P2.1b)

Pick equipment that suits the size of the distance and the time.

• Short, fast motion (a trolley on a ramp): a metre rule for the distance, and light gates or a video with a frame counter for the time. A hand stopwatch is poor here because human reaction time is a large fraction of a short time.
• Walking or running a measured distance: a tape measure or trundle wheel for the distance, and a stopwatch. Use the longest practical distance so reaction time matters less.
• Vehicles: timing between two fixed marks a known distance apart gives an average speed between them.

To reduce error, repeat the timing and take a mean, start and stop the clock on the same signal, and measure the distance from the same point on the object each time.

Practical ideas in the specification include calculating the speeds of learners who walk and run a measured distance, and investigating whether the angle of a ramp affects the speed of a trolley (PAG P3). In a ramp investigation the angle is the independent variable, the speed is the dependent variable, and you control the trolley, the release point and the distance timed.

Units, ratios and rates (P2.1c)

A rate is a quantity per unit of something else, such as metres per second. Converting units is multiplying by a ratio that equals 1.

• km to m: multiply by 1000. Minutes to seconds: multiply by 60. Hours to seconds: multiply by 3600.
• km/h to m/s: divide by 3.6, because 1000 m ÷ 3600 s = 1 ÷ 3.6.
• m/s to km/h: multiply by 3.6.

Worked example 1. A train travels at 126 km/h. Give its speed in m/s.

126 km/h = 126 × 1000 m ÷ 3600 s
         = 126 000 ÷ 3600
         = 35 m/s

Worked example 2. A cyclist covers 5.4 km in 18 minutes. Find the speed in m/s.

distance = 5.4 km = 5400 m
time     = 18 min = 18 × 60 = 1080 s
speed    = distance ÷ time = 5400 ÷ 1080 = 5.0 m/s

Proportional reasoning: at constant speed, doubling the time doubles the distance. If a robot covers 450 m in 5 min, it covers 90 m each minute, so 7 min gives 630 m. Non-SI units such as miles per hour appear as well: 1 mile is about 1609 m, so 30 mph is about 13.4 m/s.

Scalars and vectors (P2.1d)

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

Scalar	Vector
distance (total path length)	displacement (straight-line change in position, with direction)
speed	velocity (speed in a stated direction)

Distance never decreases. Displacement can shrink or reverse. To show direction in one dimension, choose one direction as positive and write the opposite direction as negative. The specification warns that learners often think velocity must be positive: reversing direction means the sign changes.

Worked example 3. A walker goes 60 m east, then 25 m west. The whole trip takes 50 s. Find the distance, displacement, average speed and average velocity.

distance     = 60 + 25 = 85 m
displacement = 60 − 25 = 35 m east
average speed    = 85 ÷ 50 = 1.7 m/s
average velocity = 35 ÷ 50 = 0.70 m/s east

An object moving at a steady speed round a circle has a constant speed but a changing velocity, because its direction keeps changing.

Distance-time graphs (P2.1b, P2.1e)

Distance (or displacement) goes on the vertical axis and time on the horizontal axis.

Feature of the line	What it means
Straight, sloping line	Constant speed
Steeper line	Greater speed
Horizontal line	Stationary
Curve getting steeper	Speed increasing
Curve getting flatter	Speed decreasing

The gradient (slope) equals speed: gradient = change in distance ÷ change in time. Draw a large right-angled triangle on the straight section and read the two sides. For a curve, the speed at one instant is the gradient of the tangent drawn touching the curve at that point.

Worked example 4. A jogger’s graph passes through these points: (0 s, 0 m), (20 s, 30 m), (50 s, 30 m), (80 s, 90 m). Describe the motion and find the speeds.

0 to 20 s:  gradient = 30 ÷ 20 = 1.5 m/s
20 to 50 s: gradient = 0, so stationary
50 to 80 s: gradient = (90 − 30) ÷ 30 = 2.0 m/s

The jogger runs, stops for 30 s, then runs faster. If the graph is a displacement-time graph, a line sloping downwards means moving back towards the start, which is negative velocity.

Velocity-time graphs (P2.1e, P2.1f)

Velocity goes on the vertical axis. Here the slope means something different.

Feature of the line	What it means
Horizontal line above zero	Constant velocity (zero acceleration)
Straight upward slope	Constant (uniform) acceleration
Straight downward slope	Constant deceleration (negative acceleration)
Line on the time axis	Stationary
Curved line	Changing acceleration

Gradient = acceleration. Area between the line and the time axis = distance travelled. For a velocity-time graph this area is exact: split the shape into triangles and rectangles (or a trapezium), find each area and add them. Units: m/s × s = m.

Worked example 5. A car’s velocity rises steadily from 0 to 20 m/s in 8 s, stays at 20 m/s for 12 s, then falls steadily to 0 in 10 s. Find the accelerations, the distance and the average speed.

Acceleration, 0 to 8 s:     (20 − 0) ÷ 8   = 2.5 m/s²
Acceleration, 8 to 20 s:    0
Acceleration, 20 to 30 s:   (0 − 20) ÷ 10  = −2.0 m/s²

Areas: triangle  ½ × 8 × 20  = 80 m
       rectangle 12 × 20     = 240 m
       triangle  ½ × 10 × 20 = 100 m
Distance = 80 + 240 + 100 = 420 m
Average speed = 420 ÷ 30 = 14 m/s

If any part of a velocity-time graph goes below the axis, that part is motion in the opposite direction. Area above the axis counts as positive displacement and area below as negative; total distance adds both sizes.

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

When speed changes, use the whole journey:

average speed = total distance ÷ total time

Worked example 6. A runner covers the first 400 m in 80 s and the next 600 m in 150 s.

total distance = 400 + 600 = 1000 m
total time     = 80 + 150  = 230 s
average speed  = 1000 ÷ 230 = 4.3 m/s (2 s.f.)

Do not average the two speeds. The speeds were 5.0 m/s and 4.0 m/s, whose mean is 4.5 m/s, which is wrong because the runner spent different times at each speed. Include any rests in the total time.

Acceleration and the equations of motion (P2.1h)

Acceleration is the rate of change of velocity:

acceleration = change in velocity ÷ time, or a = (v − u) ÷ t

Here u is the initial velocity and v the final velocity. The unit is m/s². Slowing down gives a negative acceleration when velocity is taken as positive.

Worked example 7. A cyclist speeds up from 8 m/s to 26 m/s in 6 s.

a = (26 − 8) ÷ 6 = 18 ÷ 6 = 3.0 m/s²

For uniform acceleration the third equation links velocities, acceleration and distance without needing the time:

v² − u² = 2 × a × s

Worked example 8. A car brakes from 18 m/s to rest in 27 m. Find the acceleration.

v² − u² = 2as
0² − 18² = 2 × a × 27
−324 = 54a
a = −324 ÷ 54 = −6.0 m/s²

The car decelerates at 6.0 m/s². The negative sign shows the acceleration is opposite to the motion.

Worked example 9. A train at 4 m/s accelerates at 0.5 m/s² over 150 m. Find its final speed.

v² = u² + 2as = 4² + 2 × 0.5 × 150
   = 16 + 150 = 166
v  = √166 = 12.9 m/s (3 s.f.)

Always take the square root last, and remember a square root has a positive and a negative value: choose the one in the direction of motion. An investigation of acceleration (PAG P3) can use a trolley, light gates and a ramp: find the change in velocity from the gate readings and divide by the time between them.

Common errors

• Reading a distance-time graph as if it were velocity-time: a flat line means stationary on one and constant velocity on the other.
• Using the gradient of a velocity-time graph to find distance. Gradient gives acceleration; area gives distance.
• Forgetting to convert minutes or kilometres before using m/s.
• Averaging speeds instead of dividing total distance by total time.
• Answering with a speed where velocity is asked for, with no direction stated.
• Squaring the difference: v² − u² is not (v − u)². Square each velocity first.
• Dropping the negative sign on a deceleration.

Next steps

Try the practice questions, then condense with the revision notes. Tick progress off on the checklist and test yourself with the free diagnostics.

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.