OCR GCSE Physics A: Everyday speeds, reaction times and stopping distances (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 are aligned to the OCR GCSE (9-1) Gateway Science Physics A specification (J249), Version 5.0, July 2026, first assessment in 2018. They cover statements P8.1a to P8.1e, none of which is Higher Tier only. Where a question needs g, use 9.8 m/s².

Links: OCR GCSE Physics hub, printable checklist, study guide, revision notes and the free 10-minute diagnostics.

Questions

1. Give a typical speed in m/s for (a) a person walking and (b) sound in air. [2]

2. Convert (a) 126 km/h to m/s and (b) 4.5 m/s to km/h. [2]

3. A van travels at 45 mph. Calculate its speed in m/s to 2 significant figures. 1 mile = 1609 m. [3]

4. A cyclist rides 15 km in 50 minutes. Calculate the average speed in m/s. [3]

5. A motorbike accelerates from rest to 18 m/s in 6.0 s. (a) Calculate its acceleration. [2] (b) Compare this with the acceleration of a freely falling object, about 10 m/s². [1]

6. Estimate the acceleration of a person who starts from rest and reaches a normal walking speed. State the values you use. [3]

7. A student measures reaction time by catching a dropped ruler. The distances caught are 13, 17, 15, 16 and 14 cm. Part of the conversion table is: 10 cm = 0.14 s, 15 cm = 0.18 s, 20 cm = 0.20 s. (a) Calculate the mean distance. [2] (b) Use the table to give the reaction time. [1] (c) Suggest two ways to make the results more reliable. [2]

8. A computer test records the time between a symbol appearing and a mouse click. (a) Describe what the computer measures. [1] (b) Explain why this is better than one student timing another with a stopwatch. [2]

9. A student measures four reaction times without a phone: 0.20, 0.22, 0.24 and 0.22 s. With a phone call, the times are 0.30, 0.32, 0.29 and 0.33 s. (a) Calculate the mean for each condition. [2] (b) Calculate the percentage increase in mean reaction time when using the phone. [2] (c) A car moves at 20 m/s. Calculate the extra distance travelled during the increase in reaction time. [2]

10. For each change, state whether it affects thinking distance, braking distance or both. (a) a tired driver; (b) an icy road; (c) a lower speed. [3]

11. Explain why, for the same speed: (a) the braking distance is greater on a wet road than a dry road. [3] (b) a heavily loaded van has a greater braking distance than an empty van with the same brakes. [3]

12. A council plans to cut the speed limit outside a school from 30 mph to 20 mph. 1 mile = 1609 m. (a) Convert both limits to m/s, to 2 significant figures. [2] (b) A driver has a reaction time of 0.70 s. Calculate the thinking distance at each limit, using the speeds from part (a). [2] (c) Explain why the braking distance falls by more than the speed does in proportion. [2] (d) Suggest one other measure to reduce the risk to pedestrians, and state which distance it changes. [2]

Answers

1. (a) About 1.5 m/s, accept 1 to 2 m/s [1]. (b) About 340 m/s, accept 330 to 345 m/s [1]. [2] Examiner insight: Give the unit with each value. A walking speed quoted as 5 with no unit is ambiguous, as 5 km/h and 5 m/s are very different.

2. (a) 126 ÷ 3.6 = 35 m/s [1]. (b) 4.5 × 3.6 = 16.2 km/h [1]. [2] Examiner insight: Check the direction: a speed in km/h should be the bigger number. If it is not, the wrong operation has been used.

3. 45 × 1609 = 72,405 m per hour [1]. 72,405 ÷ 3600 = 20.1 m/s [1]. 20 m/s to 2 s.f. [1] [3] Examiner insight: Show both the distance and the time conversion. Giving an unrounded 20.1125 as well as the rounded answer is a safe way to show you followed the instruction on significant figures.

4. Distance = 15 × 1000 = 15,000 m [1]. Time = 50 × 60 = 3000 s [1]. Speed = 15,000 ÷ 3000 = 5.0 m/s [1] [3] Examiner insight: Dividing 15 by 50 gives a number in km per minute. Convert both quantities before dividing.

5. (a) a = Δv ÷ t = 18 ÷ 6.0 [1] = 3.0 m/s² [1]. (b) It is about 0.30 times, or less than a third of, the free-fall value [1]. [3] Examiner insight: Start from rest means the change in velocity is the final speed. A comparison is clearer with a number such as “0.30 times” than a bare “smaller”.

6. Final speed about 1.5 m/s, change in speed from rest 1.5 m/s [1]. Time about 2 s [1]. a = 1.5 ÷ 2 = about 0.75 m/s², so roughly 1 m/s² [1] (accept 0.5 to 1.5 m/s² from sensible values). [3] Examiner insight: An estimate is judged on whether the inputs are sensible and the arithmetic is correct. Always state the values you assumed.

7. (a) (13 + 17 + 15 + 16 + 14) = 75 [1]; 75 ÷ 5 = 15 cm [1]. (b) 15 cm corresponds to 0.18 s [1]. (c) Repeat more times [1]; keep the starting position and the catching hand the same, or take the mean of a larger number of results [1]. [5] Examiner insight: Convert the mean distance to a time. Converting each result to a time and then averaging also works, but the table only has some values, so the mean first is simpler.

8. (a) The time between the stimulus appearing and the click (the response) [1]. (b) The timing is electronic [1], so it does not include the extra reaction time of the person with the stopwatch [1]. [3] Examiner insight: Good answers name the source of error. “More accurate” alone gives no reason.

9. (a) Without: 0.88 ÷ 4 = 0.22 s [1]. With: 1.24 ÷ 4 = 0.31 s [1]. (b) Increase = 0.31 − 0.22 = 0.09 s [1]; 0.09 ÷ 0.22 × 100 = 41% [1]. (c) Distance = speed × time = 20 × 0.09 [1] = 1.8 m [1]. [6] Examiner insight: Percentage change is divided by the original value, here 0.22 s, not the new one.

10. (a) Thinking distance [1]. (b) Braking distance [1]. (c) Both [1]. [3] Examiner insight: Tiredness changes the driver, so the reaction time; ice changes the road, so the friction. Decide which of the two is affected first.

11. (a) Less friction between the tyres and the wet road [1], so the maximum braking force is smaller [1], and the vehicle needs a greater distance to lose its kinetic energy [1]. (b) The loaded van has more mass, so more kinetic energy at the same speed [1]. With the same braking force, more work must be done to stop it [1], and work done = force × distance, so the distance is greater [1]. [6] Examiner insight: Link the cause to the force or the energy. A bare statement such as “it is heavier so it takes longer” does not show the physics.

12. (a) 30 mph = 30 × 1609 ÷ 3600 = 13.4, so 13 m/s [1]. 20 mph = 8.94, so 8.9 m/s [1]. (b) 13 × 0.70 = 9.1 m [1]; 8.9 × 0.70 = 6.2 m [1]. (c) Kinetic energy depends on the square of speed [1], so a smaller speed removes far more than a proportional share of the energy the brakes must remove, shortening the braking distance more than proportionally [1]. (d) Any sensible measure, such as a ban on phone use at the wheel or a crossing patrol [1], with the matching distance: phone use changes thinking distance; better tyres or road surface changes braking distance [1]. [8] Examiner insight: For an explain question, state the physical principle and apply it to the numbers in the context. A bare statement that “lower speed is safer” is usually too vague to score.

Where marks are usually lost

• Using 60 instead of 3.6 to convert between km/h and m/s.
• Leaving minutes or hours in a speed calculation.
• Failing to give the unit with a value, or the wrong unit.
• Averaging ruler-drop distances but forgetting to convert to a time.
• Dividing a percentage increase by the new value instead of the original.
• Putting tiredness, alcohol, phones or distraction under braking distance.
• Putting wet roads, worn tyres or worn brakes under thinking distance.
• Saying “friction” without linking it to the braking force that can be applied.
• Giving an estimate with no stated values.

Next steps

• Revise with the revision notes.
• Re-read 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.

Official syllabus

OCR, GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0, July 2026, first assessment in 2018. Statements P8.1a to P8.1e.