OCR GCSE Physics A: Estimating stopping distances, forces and crash dangers (J249) – Revision Notes

These notes condense the full study guide for OCR GCSE (9-1) Gateway Science Physics A (J249), Version 5.0, July 2026, statements P8.1f to P8.1i. P8.1f, P8.1h and P8.1i are Higher Tier only; P8.1g is for both tiers. Test yourself with the practice questions, tick off topics on the checklist, see the course hub and take a free diagnostic.

Equations

Quantity	Equation
distance	distance = speed × time
acceleration	a = change in velocity ÷ time
constant acceleration	v² − u² = 2as
force	F = ma
kinetic energy	½ m v²
work done	W = F × d (along the line of the force)

Convert first: 1 mph is about 0.447 m/s. Divide km/h by 3.6 to get m/s.

P8.1f Stopping distances (Higher Tier only)

Part	Rule	Speed doubles
Thinking distance	speed × reaction time	doubles
Braking distance	v² ÷ (2a), final speed zero	four times larger
Stopping distance	sum of the two	between two and four times larger

Method in steps:

1. Convert the speed to m/s.
2. State an assumed reaction time and deceleration.
3. Calculate thinking distance, then braking distance.
4. Add them, and give a sensible number of significant figures.

Shortcut: if the deceleration stays the same, new braking distance = old braking distance × (new speed ÷ old speed)².

Reminder: with 0.8 s and 6.0 m/s², a speed of 20 m/s gives 16 m + 33 m = 49 m. A graph of stopping distance against speed curves upwards.

P8.1g Dangers of large decelerations (both tiers)

Chain of reasoning for a written answer:

1. A large deceleration is a large change in velocity in a short time.
2. F = ma, so a large deceleration needs a large force on the person.
3. Inertia means a person or loose object keeps moving at the original speed until a force acts.
4. A large force can cause broken bones, organ and brain injuries, and injury from loose objects.

Safety features and what they do:

Feature	How it reduces the danger
Crumple zone	deforms, so the vehicle stops over a longer distance and kinetic energy goes into deformation
Seat belt	stretches slightly, so stopping takes longer, and holds the person in the seat
Airbag	slows the head and chest over a longer time and spreads the force over a larger area
Cycle helmet	crushes, so the head stops over a longer time

Rule: the same change in velocity over a longer time or distance gives a smaller deceleration and a smaller force.

Reminder: a 75 kg passenger at 15 m/s stopped in 0.080 s feels about 14 000 N. Stopped in 0.40 s, it is about 2800 N.

P8.1h Forces on a public road (Higher Tier only)

Method: choose a mass, find a (from Δv ÷ t, or from v² = 2as), then use F = ma. Quote the answer to one or two significant figures with the assumptions.

Situation	Typical mass	Typical size of force
Cyclist slowing	90 kg	a few hundred newtons
Car braking hard	1000 to 1500 kg	several thousand newtons
Laden lorry braking	10 000 to 20 000 kg	tens of thousands of newtons
Car in a barrier crash	about 1000 kg	tens of thousands of newtons or more

Alternative route using energy: braking force = kinetic energy ÷ braking distance, because work done = force × distance. A 1000 kg car at 20 m/s has 200 000 J, so stopping in 40 m needs 5000 N.

P8.1i Large accelerations (Higher Tier only)

Situation	Estimated acceleration (size)
Normal pull away	2 to 3 m/s²
Emergency stop, dry road	6 to 8 m/s²
Crash into a solid barrier	100 m/s² or more

Useful rearrangements:

• speed gained = acceleration × time (from rest);
• speed from a skid: v = √(2as);
• force = mass × acceleration.

Example: a 1400 kg car reaches 25 m/s from rest in 10 s. a = 2.5 m/s² and F = 3500 N.

Worked reminders

Estimating a full stopping distance

Estimate the stopping distance at 25 m/s with a reaction time of 0.7 s and a deceleration of 5.0 m/s².

• Thinking distance = 25 × 0.7 = 17.5 m.
• Braking distance = 25² ÷ (2 × 5.0) = 625 ÷ 10 = 62.5 m.
• Stopping distance = 80 m (to 2 s.f.).

A wetter, slipperier road means a smaller deceleration, so the braking distance grows while the thinking distance does not change.

Comparing two safety designs

A 60 kg cyclist at 8.0 m/s hits a wall. Compare a head stop in 0.020 s with a stop in 0.10 s using a helmet that crushes.

• Without: a = 8.0 ÷ 0.020 = 400 m/s², F = 60 × 400 = 24 000 N.
• With: a = 8.0 ÷ 0.10 = 80 m/s², F = 60 × 80 = 4800 N.
• Five times the time gives one fifth of the force. The helmet works on the head and neck, so the figures are a model of the whole body, not an exact head force.

Force from a stopping distance

A van of mass 2000 kg slows from 15 m/s to rest in 45 m.

• a = 15² ÷ (2 × 45) = 225 ÷ 90 = 2.5 m/s².
• F = 2000 × 2.5 = 5000 N.
• Check with energy: ½ × 2000 × 15² = 225 000 J, and 225 000 ÷ 45 = 5000 N.

Command words and estimates

Command word	What the answer needs
Estimate	a reasoned approximate value; state the assumptions and use sensible rounding
Calculate	a numerical answer from given data, with the working and the unit
Explain	a cause-and-effect chain, not a list of facts
Compare	a similarity or difference, with a reason

Checks that an estimate is sensible:

• A car stopping at about 7 m/s² feels like a hard stop. A value of 70 m/s² for a braking car with no crash should make you check your working.
• A braking force on a car close to its weight (about 10 000 N) is of the right order for emergency braking.
• If a stopping distance for 30 m/s is shorter than for 20 m/s, a step has gone wrong.

Must-know distinctions

• Thinking distance depends on the driver and speed; braking distance depends on speed, the road, the brakes and the tyres.
• Weight is a force from gravity; the braking force is a different force on the vehicle.
• A deceleration is a size (positive) with a direction opposite to the motion.
• Doubling the time over which a stop happens halves the average force, for the same mass and change in speed.
• An estimate is a reasoned approximate value with assumptions, not an exact answer.

Quick self-test

1. Convert 54 km/h to m/s.
2. Find the thinking distance at 12 m/s with a reaction time of 0.9 s.
3. Find the braking distance at 12 m/s with a deceleration of 5.0 m/s².
4. A braking distance is 20 m from 8 m/s. Estimate it from 24 m/s on the same road.
5. A car of mass 1000 kg decelerates at 7.5 m/s². Find the braking force.
6. A person of mass 80 kg moving at 10 m/s is stopped in 0.25 s. Find the average force.
7. Skid marks are 25 m long and the deceleration was 8.0 m/s². Estimate the speed at the start of the skid.
8. A 1400 kg car reaches 25 m/s from rest in 10 s. Find the force.
9. A 1000 kg car at 20 m/s is stopped over 40 m. Use kinetic energy to find the average force.
10. Explain why a crumple zone reduces injury.

Answers

1. 54 ÷ 3.6 = 15 m/s.
2. 12 × 0.9 = 10.8 m.
3. 12² ÷ (2 × 5.0) = 14.4 m.
4. Speed ×3, so distance × 9 = 180 m.
5. 1000 × 7.5 = 7500 N.
6. a = 10 ÷ 0.25 = 40 m/s², so F = 80 × 40 = 3200 N.
7. v = √(2 × 8.0 × 25) = √400 = 20 m/s.
8. a = 25 ÷ 10 = 2.5 m/s², so F = 1400 × 2.5 = 3500 N.
9. Kinetic energy = ½ × 1000 × 20² = 200 000 J; F = 200 000 ÷ 40 = 5000 N.
10. It crushes, so the vehicle stops over a longer distance and time; the deceleration of the occupants is smaller, so the force on them (F = ma) is smaller.

Where marks are usually lost

• Using mph or km/h in an equation without converting.
• Forgetting to square the speed in braking distance.
• Leaving out the thinking distance when asked for stopping distance.
• Blaming alcohol or tiredness for a longer braking distance.
• Writing that a crumple zone “absorbs the force” instead of increasing the time or distance and so reducing the force.
• Treating an estimate as exact, or giving no assumptions.
• Missing units, or giving newtons where joules are needed.
• Using a stopping time of zero or the wrong direction for acceleration.

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.