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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover P2.1 Motion (P2.1a to P2.1h) in OCR GCSE (9-1) Gateway Science Physics A (J249), Version 5.0 (July 2026), for first assessment in 2018. All P2.1 statements are assessed at both tiers, so every question here is suitable for both. Learn the content first from the study guide and the revision notes. See also the OCR GCSE Physics hub, the checklist and the Forces practice questions.

Questions

1. Convert units.

(a) Change 54 km/h into m/s. [1] (b) Change 8 m/s into km/h. [1] (c) A drone flies 0.36 km in 90 s. Calculate its speed in m/s. [2]

2. A trolley rolls down a ramp. Describe how you would measure its average speed over the last 1.2 m of the ramp, and how you would make the result more reliable. [4]

3.

(a) A sprinter runs 100 m in 12.5 s. Calculate the average speed. [2] (b) A hiker walks at a steady 1.4 m/s for 25 minutes. Calculate the distance walked in metres. [2]

4. A taxi drives 8 km north, then 3 km south. The journey takes 20 minutes.

(a) State the distance travelled. [1] (b) State the displacement, including its direction. [2] (c) Calculate the average speed in km/h. [2] (d) State the average velocity in km/h. [1]

5. A displacement-time graph for a runner is made of straight lines joining these points: (0 s, 0 m), (8 s, 16 m), (14 s, 16 m), (20 s, 4 m).

(a) Calculate the speed in the first 8 s. [2] (b) Describe the motion between 8 s and 14 s. [1] (c) Calculate the velocity between 14 s and 20 s and explain the sign of your answer. [3] (d) Calculate the average speed over the whole 20 s. [2]

6. The velocity-time graph for a rider is made of straight lines joining: (0 s, 0 m/s), (5 s, 10 m/s), (15 s, 10 m/s), (18 s, 0 m/s).

(a) Calculate the acceleration during the first 5 s. [2] (b) Calculate the total distance travelled using the area under the graph. [3] (c) Calculate the acceleration between 15 s and 18 s. Give your answer to 2 significant figures. [2] (d) Calculate the average speed over the 18 s, to 2 significant figures. [1]

7.

(a) A scooter changes velocity from 3 m/s to 15 m/s in 4 s. Calculate its acceleration. [2] (b) A car starts from rest and accelerates uniformly at 1.6 m/s² for 9 s. Calculate its final velocity. [2]

8. A train travelling at 30 m/s brakes uniformly to rest with an acceleration of −1.5 m/s². Calculate the distance it travels while braking. [4]

9. A student runs 1200 m in 300 s, then 800 m in 100 s.

(a) Calculate the average speed for the whole run. [3] (b) The student says the average speed is the mean of the two stage speeds. Explain why this is wrong. [1]

10. A skateboarder moving at 2 m/s accelerates uniformly at 1.5 m/s² over a distance of 39 m. Calculate the final speed. [4]

11. A car is travelling at 14 m/s. The driver reacts after 0.6 s (a value supplied for this question), during which the car keeps its speed. The car then brakes uniformly at 4.0 m/s² until it stops.

(a) Calculate the distance travelled during the 0.6 s. [2] (b) Calculate the distance travelled while braking. [3] (c) State the total distance from the moment the driver sees the hazard to stopping, to 2 significant figures. [1] (d) Calculate the time taken to brake to rest. [2]

12.

(a) From this list, state the two vector quantities: speed, displacement, distance, velocity. [1] (b) A car drives round a roundabout at a constant 12 m/s. Explain why its velocity is not constant. [2]

Answers

1. (a) 54 ÷ 3.6 = 15 m/s [1] (b) 8 × 3.6 = 28.8 km/h [1] (c) 0.36 km = 360 m [1]; speed = 360 ÷ 90 = 4.0 m/s [1] Examiner insight: Show the conversion step (km to m) separately, so that a slip in the final division does not hide that you knew what to do.

2. Mark two points 1.2 m apart on the ramp with a metre rule. [1] Time the trolley between the marks with light gates (or a video with a timer), since a hand stopwatch has a reaction time that is large compared with the short time. [1] Speed = distance ÷ time. [1] Repeat the run several times and take a mean, releasing from the same point each time. [1] Examiner insight: A reason attached to your choice of equipment (why light gates beat a stopwatch here) is stronger than naming the equipment alone.

3. (a) speed = distance ÷ time = 100 ÷ 12.5 [1] = 8.0 m/s [1] (b) time = 25 × 60 = 1500 s [1]; distance = 1.4 × 1500 = 2100 m [1] Examiner insight: Part (b) is lost most often by multiplying 1.4 by 25 and getting 35 m; check that the time is in seconds when the speed is in m/s.

4. (a) 8 + 3 = 11 km [1] (b) 8 − 3 = 5 km [1], north [1] (c) time = 20 min = 1/3 h [1]; average speed = 11 ÷ (1/3) = 33 km/h [1] (d) 5 ÷ (1/3) = 15 km/h north [1] Examiner insight: Displacement and velocity answers need a direction; a number alone describes only the size.

5. (a) speed = 16 ÷ 8 [1] = 2.0 m/s [1] (b) The runner is stationary (horizontal line) for 6 s. [1] (c) velocity = (4 − 16) ÷ 6 [1] = −2.0 m/s [1]; the negative sign shows motion back towards the start, opposite to the original direction [1] (d) distance = 16 + 12 = 28 m [1]; average speed = 28 ÷ 20 = 1.4 m/s [1] Examiner insight: In part (d), using the final displacement of 4 m instead of the total distance of 28 m would give an average velocity, not an average speed.

6. (a) a = (10 − 0) ÷ 5 [1] = 2.0 m/s² [1] (b) triangle ½ × 5 × 10 = 25 m [1]; rectangle 10 × 10 = 100 m [1]; triangle ½ × 3 × 10 = 15 m, total 140 m [1] (c) a = (0 − 10) ÷ 3 [1] = −3.3 m/s² [1] (d) 140 ÷ 18 = 7.8 m/s [1] Examiner insight: Splitting the shape into labelled rectangles and triangles makes your method clear and helps you spot a slip in the total; writing only 140 m hides your working.

7. (a) a = (15 − 3) ÷ 4 [1] = 3.0 m/s² [1] (b) v = u + at, rearranged from a = (v − u) ÷ t, so v = 0 + 1.6 × 9 [1] = 14.4 m/s [1] Examiner insight: Write the starting equation before substituting, and carry the unit m/s² through your working so the final unit is easy to check.

8. v² − u² = 2as [1]; 0² − 30² = 2 × (−1.5) × s [1]; −900 = −3.0s [1]; s = 300 m [1] Examiner insight: A negative acceleration with a negative left-hand side gives a positive distance; getting a negative distance usually signals a sign slip you should catch.

9. (a) total distance = 1200 + 800 = 2000 m [1]; total time = 300 + 100 = 400 s [1]; average speed = 2000 ÷ 400 = 5.0 m/s [1] (b) The stage speeds are 4.0 m/s and 8.0 m/s, whose mean is 6.0 m/s, but the student spent three times as long at the slower speed, so the mean of speeds is not the average speed. [1] Examiner insight: For an explain question, give the reason (different times at each speed) as well as the correct value.

10. v² − u² = 2as [1]; v² = 2² + 2 × 1.5 × 39 [1]; v² = 4 + 117 = 121 [1]; v = 11 m/s [1] Examiner insight: Take the square root as the last step; stopping at 121 and quoting it as the speed is a common slip.

11. (a) distance = speed × time = 14 × 0.6 [1] = 8.4 m [1] (b) 0² − 14² = 2 × (−4.0) × s [1]; −196 = −8.0s [1]; s = 24.5 m [1] (c) 8.4 + 24.5 = 32.9, so 33 m [1] (d) t = (0 − 14) ÷ (−4.0) [1] = 3.5 s [1] Examiner insight: In a multi-part question, add the two distances only after both are found; showing each part separately keeps your working clear and makes a slip easy to trace.

12. (a) displacement and velocity [1] (b) Velocity has both size and direction. [1] The speed is constant, but the direction of travel keeps changing round the roundabout, so the velocity changes. [1] Examiner insight: Name both parts of a vector (size and direction) and say which one is changing; stating only that the car is turning is not enough to answer an explain question.

Where marks are usually lost

• Not converting minutes, hours or kilometres into seconds or metres before substituting.
• Giving a speed or distance when the question asks for a displacement or velocity with a direction.
• Reading the gradient on a velocity-time graph as a distance.
• Counting the total displacement instead of total distance when finding average speed.
• Missing out the stationary period when finding the total time.
• Squaring (v − u) instead of squaring v and u separately.
• Leaving off the negative sign on a deceleration, or giving a negative distance.
• Rounding too early and quoting an answer to the wrong number of significant figures.

Next steps

• Re-read the revision notes and the study guide.
• Go to the OCR GCSE Physics hub and the printable checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class: /trial/.

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.