OCR GCSE Physics A: Estimating stopping distances, forces and crash dangers (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 follow the OCR GCSE (9-1) Gateway Science Physics A specification (J249), Version 5.0, July 2026, statements P8.1f to P8.1i. Statements P8.1f, P8.1h and P8.1i are Higher Tier only, and questions on them are labelled. P8.1g is for both tiers. Revise first with the study guide or the revision notes. Other links: the course hub, the checklist and the free diagnostics. Take g as 9.8 N/kg where weight is needed.

Questions

1. (Higher Tier only) A car travels at 72 km/h. A driver’s reaction time is 0.75 s and the car then decelerates steadily at 6.0 m/s². Estimate the stopping distance. [4]

2. (Higher Tier only) A car’s braking distance on a dry road is 12 m from 15 m/s.

(a) Estimate the braking distance from 30 m/s on the same road. [2] (b) State how the thinking distance changes when the speed doubles. [1]

3. (Higher Tier only) Explain why a graph of stopping distance against speed is a curve and not a straight line. [3]

4. Explain why a passenger who is not wearing a seat belt is likely to be more seriously hurt when a car stops suddenly. [3]

5. A passenger of mass 60 kg is travelling at 12 m/s.

(a) Calculate the average force needed to stop them in 0.060 s. [2] (b) Calculate the average force when a seat belt and airbag extend the stop to 0.30 s. [2] (c) Explain how the airbag reduces the chance of injury. [1]

6. (Higher Tier only) A car of mass 1200 kg slows from 18 m/s to rest in 36 m. Calculate the average braking force. [4]

7. (Higher Tier only) Estimate the average force needed to bring a cyclist and bicycle to rest from a typical cycling speed in a typical time. State the values you assume. [4]

8. (Higher Tier only) An electric car of mass 1600 kg accelerates from rest to 27 m/s in 5.4 s.

(a) Calculate its acceleration. [2] (b) Calculate the resultant force. [1] (c) Calculate the distance travelled, assuming constant acceleration. [2]

9. (Higher Tier only) Skid marks are 36 m long on a wet road. Assume the deceleration was 4.0 m/s².

(a) Estimate the car’s speed at the start of the skid, in m/s. [3] (b) Give one reason the estimate may be unreliable. [1]

10. (Higher Tier only) A car of mass 1000 kg travelling at 15 m/s hits a barrier. The front of the car crumples by 0.75 m as the car stops.

(a) Calculate the car’s kinetic energy before the impact. [2] (b) Calculate the average force on the car while it crumples. [2] (c) Calculate the deceleration, and state how many times larger it is than 9.8 m/s². [2] (d) Explain why designers want a longer crumple distance. [2]

11. (Higher Tier only) A car of mass 1500 kg is driven at 22 m/s. The driver reacts after 0.50 s and the brakes then give a steady deceleration of 6.0 m/s².

(a) Estimate the stopping distance. [3] (b) Calculate the braking force. [2] (c) A hazard is 50 m ahead when the driver first sees it. Calculate the speed of the car when it reaches the hazard. [3] (d) Explain why a collision at this speed (about 9 mph) could still seriously injure a pedestrian. [2]

Answers

1. 72 km/h ÷ 3.6 = 20 m/s [1]. Thinking distance = 20 × 0.75 = 15 m [1]. Braking distance = 20² ÷ (2 × 6.0) = 33.3 m [1]. Stopping distance = 48 m [1]. [4] Examiner insight: Convert units before anything else. Using 72 in the thinking distance undermines every later step, and showing each of the two distances separately makes your reasoning clear even if the total slips.

2. (a) The speed is multiplied by 2 [1], so the braking distance is multiplied by 4: 12 × 4 = 48 m [1]. (b) The thinking distance doubles [1]. [3] Examiner insight: Say which quantity is squared. Answers that simply double the braking distance have applied the wrong proportionality, and writing the factor (2² = 4) shows the reasoning.

3. Thinking distance is proportional to speed, so it forms a straight line [1]. Braking distance is proportional to speed squared [1]. The sum rises faster at high speed, so the graph curves upwards [1]. [3] Examiner insight: Give each part its own relationship. A bare statement that the distance “increases faster” earns less than naming the squared dependence of braking distance.

4. Inertia means the passenger keeps moving at the original speed when the car stops [1]. They then stop suddenly against the dashboard or windscreen, in a very short time, so the deceleration is very large [1]. A large deceleration needs a large force (F = ma), which causes injury [1]. [3] Examiner insight: A strong answer links inertia, a short stopping time and a large force. Writing only that the passenger “flies forward” describes what happens without explaining why it hurts.

5. (a) a = 12 ÷ 0.060 = 200 m/s² [1]. F = 60 × 200 = 12 000 N [1]. (b) a = 12 ÷ 0.30 = 40 m/s² [1]. F = 60 × 40 = 2400 N [1]. (c) It makes the stop take longer, so the deceleration and force are smaller (and it spreads the force over a larger area) [1]. [5] Examiner insight: Quote the unit with each force. In part (c), write about time and force together; saying the airbag is “soft” gives no physics.

6. v² = 2as, so a = 18² ÷ (2 × 36) [1] = 4.5 m/s² [1]. F = ma = 1200 × 4.5 [1] = 5400 N [1]. [4] Examiner insight: Show the rearranged equation. Using 18 ÷ 36 as the acceleration confuses speed and distance with acceleration, and a clear substitution makes errors easy to see.

7. Sensible values stated, for example mass 90 kg, speed 6 m/s, time 3 s [1]. a = 6 ÷ 3 = 2 m/s² [1]. F = 90 × 2 [1] = 180 N, with an answer of tens to a few hundred newtons being reasonable [1]. [4] Examiner insight: An estimate needs stated assumptions and sensible rounding. Choosing a cyclist mass of 9 kg or a speed of 60 m/s should prompt you to stop and check.

8. (a) a = 27 ÷ 5.4 [1] = 5.0 m/s² [1]. (b) F = 1600 × 5.0 = 8000 N [1]. (c) s = 27² ÷ (2 × 5.0) [1] = 729 ÷ 10 = 73 m [1]. [5] Examiner insight: Part (c) can also use average speed × time. Whichever route you take, use consistent units and carry the unrounded acceleration into later steps.

9. (a) v² = 2as = 2 × 4.0 × 36 [1] = 288 [1]. v = 17 m/s [1]. (b) The deceleration was assumed, not measured, and the real value depends on the road surface and tyres [1]. [4] Examiner insight: Take the square root as the final step. For an unreliability point, name the assumption that the estimate depends on, not “human error”.

10. (a) Kinetic energy = ½ × 1000 × 15² [1] = 112 500 J [1]. (b) Work done = force × distance, so F = 112 500 ÷ 0.75 [1] = 150 000 N [1]. (c) a = 15² ÷ (2 × 0.75) = 150 m/s² [1]; 150 ÷ 9.8 = 15 times larger [1]. (d) A longer distance means a longer time to stop [1], so the deceleration is smaller, and the force on the occupants is smaller and they are less likely to be injured [1]. [8] Examiner insight: Part (b) is a good place to show the link between energy and force. Writing the equation, then the numbers, makes your method clear even if an arithmetic slip occurs.

11. (a) Thinking distance = 22 × 0.50 = 11 m [1]. Braking distance = 22² ÷ (2 × 6.0) = 40.3 m [1]. Stopping distance = 51 m [1]. (b) F = 1500 × 6.0 [1] = 9000 N [1]. (c) The thinking distance uses 11 m, leaving 50 − 11 = 39 m of braking [1]. v² = 22² − 2 × 6.0 × 39 = 484 − 468 = 16 [1]. v = 4.0 m/s [1]. (d) A pedestrian would be stopped in a very short time [1], so the deceleration and force on their body would be large enough to cause injury [1]. [10] Examiner insight: Part (c) is a multi-step question: split the distance first. In part (d), give the physics chain (short time, large deceleration, large force) rather than just “it would hurt”.

Where marks are usually lost

• Leaving speeds in mph or km/h.
• Forgetting to add thinking distance when stopping distance is asked for.
• Not squaring the speed in braking distance calculations.
• Giving a force without a unit, or confusing newtons and joules.
• Explaining injury as “a big force” with no reference to short time or distance.
• Writing that safety features reduce the speed of the car rather than the deceleration of the person.
• Stating estimates without any assumptions, or to too many significant figures.
• Using weight as the braking force.

Next steps

• Review the revision notes and the study guide.
• Tick off the topic on the OCR GCSE Physics checklist and see the course hub.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, GCSE (9-1) Gateway Science Physics A (J249), Version 5.0, July 2026, first assessment 2018. Topic P8.1, statements P8.1f to P8.1i.