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

This guide is aligned to the OCR GCSE (9-1) Gateway Science Physics A specification (J249), Version 5.0, July 2026. It covers statements P8.1f to P8.1i in topic P8.1, Physics on the move. P8.1f, P8.1h and P8.1i are Higher Tier only. P8.1g is assessed in both tiers. The sibling unit on everyday speeds, reaction times and the factors that change stopping distance covers P8.1a to P8.1e, so this guide builds on those ideas rather than repeating them.

Useful links: the OCR GCSE Physics hub, the printable checklist, the revision notes, the practice questions and the free diagnostics. For the underlying equations, see Newton’s second law, inertia and momentum, work done and stored energy and motion graphs and acceleration.

What this unit covers

Statement	What you must be able to do	Tier
P8.1f	Estimate how the distances needed for road vehicles to stop in an emergency vary over a range of typical speeds	Higher Tier only
P8.1g	Explain the dangers caused by large decelerations	Both tiers
P8.1h	Estimate the forces involved in typical situations on a public road	Higher Tier only
P8.1i	Estimate, for everyday road transport, the speed, accelerations and forces involved in large accelerations	Higher Tier only

The specification’s practical suggestions are to research stopping distances using the Highway Code, and to research and build casings for eggs on trolleys to investigate crumple zones and safety features in cars.

Equations used here

Quantity	Equation
distance travelled	distance (m) = speed (m/s) × time (s)
acceleration	a = Δv ÷ t, with a in m/s², Δv in m/s and t in s
constant acceleration	(final velocity)² − (initial velocity)² = 2 × acceleration × distance
force	force (N) = mass (kg) × acceleration (m/s²)
kinetic energy	kinetic energy (J) = ½ × mass × (speed)²
work done	work done (J) = force (N) × distance (m), along the line of action of the force

A deceleration is an acceleration in the opposite direction to the motion. In calculations, use its size as a positive number and say that it acts against the motion.

Estimating stopping distances (P8.1f, Higher Tier only)

Stopping distance = thinking distance + braking distance. Each part depends on speed in a different way.

• Thinking distance = speed × reaction time. The vehicle moves at its original speed, so doubling the speed doubles this distance.
• Braking distance comes from (final velocity)² − (initial velocity)² = 2 × acceleration × distance. The final velocity is zero, so braking distance = v² ÷ (2a). For a steady deceleration, doubling the speed multiplies braking distance by four.

An estimate needs three stated assumptions: a typical reaction time, a typical deceleration for the road conditions, and a speed in m/s. Convert miles per hour first: 1 mph is about 0.447 m/s, so 30 mph is about 13.4 m/s and 70 mph is about 31.3 m/s.

Worked example 1: building an estimate

Estimate the stopping distance of a car at 20 m/s, assuming a reaction time of 0.8 s and a steady deceleration of 6.0 m/s².

Thinking distance = 20 × 0.8 = 16 m. Braking distance = 20² ÷ (2 × 6.0) = 400 ÷ 12 = 33.3 m. Stopping distance = 16 + 33.3 = 49 m (about 50 m).

Applying the same method across typical speeds gives this table:

Speed (m/s)	Thinking distance (m)	Braking distance (m)	Stopping distance (m)
10	8.0	8.3	16
15	12	19	31
20	16	33	49
25	20	52	72
30	24	75	99

Going from 10 m/s to 30 m/s triples the speed. The thinking distance triples (8 to 24 m), the braking distance goes up by a factor of nine (8.3 to 75 m), and the stopping distance rises by a factor of about six. A plot of stopping distance against speed therefore curves upwards: it is not a straight line.

The Highway Code gives overall stopping distances of 23 m at 30 mph and 96 m at 70 mph. Applying the same assumptions to those speeds gives about 26 m and 107 m. The estimate is a little cautious, which is reasonable, and it has the right shape. Real values depend on the driver, the tyres, the brakes and the road.

Worked example 2: scaling from one known value

A car’s braking distance on a dry road is 9.0 m from 10 m/s. Estimate it from 25 m/s.

Braking distance is proportional to speed squared, and the speed rises by a factor of 25 ÷ 10 = 2.5. New distance = 9.0 × 2.5² = 9.0 × 6.25 = 56 m.

This shortcut only works if the deceleration is the same in both cases. On a wet road the deceleration is smaller, so every braking distance is larger.

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

A large deceleration means a large change in velocity in a short time. By F = ma, a large deceleration of a person needs a large force on them. The force comes from whatever stops them: a seat belt, an airbag, the dashboard or the road. Because of inertia, an unrestrained person carries on at the original speed until something exerts that force.

Large forces on the body can cause:

• bruising and broken bones where the force is applied;
• damage to internal organs and the brain, which are not fixed rigidly to the skeleton and keep moving after the body is stopped;
• injury from loose objects in the vehicle, which also keep moving at the original speed.

The force depends on the time or distance over which the change in motion happens. Safety features work by making that time or distance longer:

• Crumple zones deform in a collision, so the passenger compartment stops over a longer distance. The vehicle’s kinetic energy is transferred in doing work to deform the structure.
• Seat belts stretch slightly, which lengthens the stopping time for the wearer, and they hold the person to the seat.
• Airbags let the head and chest slow over a longer time and spread the force over a larger area.
• Cycle helmets contain material that crushes, which lengthens the time for the head to stop.

Worked example 3: comparing two stops

A passenger of mass 75 kg travels at 15 m/s. Compare the average force needed to stop them in 0.080 s (a rigid obstacle) and in 0.40 s (crumple zone and seat belt).

Rigid: a = 15 ÷ 0.080 = 188 m/s², so F = 75 × 188 = 14 000 N (to 2 s.f.). Protected: a = 15 ÷ 0.40 = 37.5 m/s², so F = 75 × 37.5 = 2800 N (to 2 s.f.).

The weight of the passenger is 75 × 9.8 = 735 N, so the first force is about 19 times their weight and the second about 3.8 times. Five times longer gives one fifth of the force.

Worked example 4: using distance

A car of mass 900 kg travelling at 12 m/s crumples by 0.80 m as it stops against a barrier. Estimate the average force on the car.

Kinetic energy = ½ × 900 × 12² = 64 800 J. Work done = force × distance, so F = 64 800 ÷ 0.80 = 81 000 N.

Check with acceleration: a = 12² ÷ (2 × 0.80) = 90 m/s², and F = 900 × 90 = 81 000 N. If the crumple distance were only 0.20 m, the force would be four times larger, 324 000 N.

Estimating forces on a public road (P8.1h, Higher Tier only)

To estimate a force, pick a sensible mass, estimate the acceleration from a speed change and a time (or from v² = 2as), then use F = ma. State your assumptions. Answers to one or two significant figures are appropriate for estimates.

Typical masses to assume: an adult about 70 to 80 kg, a cyclist with bicycle about 90 kg, a car about 1000 to 1500 kg, a laden lorry in the order of 10 000 to 20 000 kg.

Worked example 5: a car braking hard

A car of mass 1250 kg brakes with a deceleration of 6.4 m/s². The resultant braking force is 1250 × 6.4 = 8000 N. Its weight is 1250 × 9.8 = 12 250 N, so the braking force is about two thirds of the weight.

Worked example 6: a lorry and a cyclist

A lorry of mass 16 000 kg slows from 22 m/s to rest in 60 m. Its deceleration is 22² ÷ (2 × 60) = 4.03 m/s², so the force is 16 000 × 4.03 = 65 000 N (2 s.f.).

A cyclist and bicycle of total mass 90 kg slow from 6.0 m/s to rest in 2.0 s. a = 6.0 ÷ 2.0 = 3.0 m/s², so F = 90 × 3.0 = 270 N.

The lorry needs a force about 240 times larger than the cyclist, mainly because its mass is about 180 times larger.

Large accelerations in road transport (P8.1i, Higher Tier only)

Large accelerations are accelerations well above the everyday values. Examples are an emergency stop, a very fast sports car pulling away, and a crash. Typical sizes to quote as estimates:

Situation	Estimated acceleration (size)
Car pulling away normally	2 to 3 m/s²
Emergency stop on a dry road	6 to 8 m/s²
Crash into a solid barrier	100 m/s² or more

Worked example 7: a fast start

A sports car of mass 1500 kg reaches 26.8 m/s (60 mph) from rest in 4.0 s.

a = 26.8 ÷ 4.0 = 6.7 m/s². F = 1500 × 6.7 = 10 000 N (to 2 s.f.). Distance = 26.8² ÷ (2 × 6.7) = 54 m.

Worked example 8: estimating speed from skid marks

Skid marks 30 m long suggest a steady deceleration of about 7 m/s² on a dry road. The car’s speed at the start of the skid was v = √(2 × 7 × 30) = 20 m/s, about 46 mph. On a wet road with a deceleration of 3.5 m/s² the same skid would imply only 14 m/s. The estimate is only as good as the assumed deceleration.

Common errors

• Writing “braking distance doubles when speed doubles”. It quadruples, if the deceleration is constant.
• Putting speed in mph into an equation that needs m/s.
• Saying alcohol or tiredness lengthens braking distance. These affect the thinking distance.
• Explaining crash injuries as “too much force” with no mention of how short the time or distance is.
• Using the weight of the car as the braking force.
• Giving an estimate to five significant figures, or with no stated assumptions.

Next steps

Try the practice questions, then check yourself with the revision notes and the diagnostics.

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.