You are viewing a free preview of this lesson.
Subscribe to unlock all 9 lessons in this course and every other course on LearningBro.
When a hazard appears in the road ahead — a child steps out, a car brakes sharply, a traffic light turns red — a driver cannot stop instantly. There is a delay while the brain notices the danger and decides to act, and then a further distance travelled while the brakes actually slow the car down. The total distance a car covers between the moment the driver first sees a hazard and the moment the car comes to a complete stop is called the stopping distance, and understanding what affects it is a matter of life and death on the road. This lesson opens Topic P8 (Global challenges) of OCR Gateway Science A by breaking the stopping distance into its two parts, examining the factors that lengthen each one, tracing where the kinetic energy goes during braking, and explaining the crucial fact that braking distance grows with the square of the speed.
By the end of this lesson you should be able to state that stopping distance is the sum of thinking distance and braking distance, list and explain the factors that affect each, describe the energy transfer that happens when a car brakes, and explain — using Ek=21mv2 — why doubling the speed makes the braking distance roughly four times as long.
The total stopping distance of a vehicle is made up of two separate distances added together:
stopping distance=thinking distance+braking distance
Both distances are measured in metres. The stopping distance matters because it tells a driver how much clear road they need in front of them to be able to stop safely — which is the whole basis of keeping a safe following distance.
Exam Tip: Learn the word equation exactly: stopping distance = thinking distance + braking distance. A very common exam slip is to give only the braking distance when the question asks for the stopping distance — always add the thinking distance too.
The thinking distance depends on how fast the car is moving and on the driver's reaction time.
Speed. During the reaction time the car keeps moving at a steady speed, so the faster it is going, the further it travels before the driver even touches the brake. Thinking distance is directly proportional to speed: double the speed and the thinking distance doubles.
Reaction time. Anything that slows the driver's reactions increases the thinking distance. Reaction time is lengthened by:
A typical reaction time for an alert, sober driver is around 0.7 s, but it can easily double or worse if the driver is impaired or distracted.
Exam Tip: Factors that affect thinking distance all act on the driver (reaction time) or on the speed. Do not confuse them with the road/vehicle factors that affect braking distance — examiners award marks for putting each factor in the correct box.
The braking distance depends on the speed of the car, on the mass of the car, and on the braking force available — which in turn depends on the condition of the road, the tyres and the brakes.
Speed. The faster the car is travelling, the greater the braking distance — and, as we will see below, this relationship is not a simple doubling: braking distance grows with the square of the speed.
Mass. A more heavily loaded vehicle has more kinetic energy at a given speed, so the brakes must do more work to stop it and the braking distance increases.
Road conditions. A wet or icy road, or a road covered in leaves, oil or loose gravel, reduces the friction (grip) between the tyres and the road. Less grip means a smaller braking force, so the braking distance increases.
Tyre condition. Worn or bald tyres — especially with too little tread to clear water — grip the road less well, increasing the braking distance. Correctly inflated tyres with good tread give the best grip.
Brake condition. Worn brake pads or brakes that have overheated ("brake fade") provide a smaller braking force, again lengthening the braking distance.
Exam Tip: Braking-distance factors act on the vehicle and the road (mass, brakes, tyres, road surface, speed). If a factor changes the grip or the braking force, it changes the braking distance — not the thinking distance.
To stop a moving car you have to get rid of its kinetic energy — the energy it has because it is moving. This does not simply disappear; it is transferred to another store.
When the brakes are applied, the brake pads are pressed hard against a disc or drum attached to the wheel. As the disc turns against the pads, friction between them does work. This friction transfers the car's kinetic energy into the thermal (internal) energy store of the brakes, which get hot. In a hard stop from motorway speed, the brake discs can reach several hundred degrees Celsius.
We can describe the transfer as:
kinetic energy of car⟶thermal energy of brakes (by friction)
The greater the speed, the more kinetic energy has to be transferred, and the more work the brakes must do — which is exactly why higher speeds mean longer braking distances. If the brakes get too hot they can no longer transfer energy effectively (brake fade), which is why lorries descending long hills use lower gears to help slow down and avoid overheating the brakes.
Exam Tip: Say kinetic energy is transferred to thermal energy in the brakes by friction. Naming the mechanism (friction) and the destination (thermal store of the brakes) is what earns full marks — "the energy is lost" is not enough.
This is the single most important idea in the whole topic, and it is a favourite of examiners. The kinetic energy of a moving car is given by:
Ek=21mv2
where m is the mass and v is the speed. The key feature is the v2: kinetic energy depends on the square of the speed.
When a car brakes to a stop, the braking force does work to remove all of that kinetic energy. The work done by the braking force is W=F×d, where F is the braking force and d is the braking distance. Setting the work done equal to the kinetic energy removed:
F×d=21mv2
Rearranging for the braking distance:
d=2Fmv2
For a given car and a given braking force, everything except v2 is constant, so the braking distance is proportional to the speed squared, d∝v2. This has a dramatic consequence:
A car has a braking distance of 12 m when travelling at 10 m/s. Estimate its braking distance at 20 m/s, assuming the same braking force.
Step 1 — the speed has doubled, from 10 m/s to 20 m/s, so the ratio of speeds is 2.
Step 2 — braking distance is proportional to speed squared, so multiply the distance by 22=4.
Step 3 — calculate: 12×4=48 m.
Answer: the braking distance is about 48 m — four times as far.
A car of mass 1200 kg travels at 15 m/s. The brakes provide a constant force of 6000 N. Calculate the braking distance.
Step 1 — find the kinetic energy: Ek=21mv2=21×1200×152.
Step 2 — evaluate: Ek=0.5×1200×225=135000 J.
Step 3 — the braking force does this much work: F×d=Ek, so d=FEk=6000135000.
Step 4 — calculate: d=22.5 m.
Answer: the braking distance is 22.5 m.
Exam Tip: If a question doubles or triples the speed and asks about braking distance, use d∝v2: multiply the distance by the square of the speed ratio, not just the ratio. This is the classic trap and marks are routinely lost by writing "twice as far" instead of "four times as far".
The chart below shows how the thinking distance and braking distance combine at different speeds. Notice that the thinking distance (in blue) grows steadily in proportion to speed, while the braking distance (in orange) grows much faster because of the v2 effect.
The table gives approximate figures. The exact numbers are not something you must memorise, but the pattern — that braking distance rises far more steeply than thinking distance as speed increases — is examinable.
| Speed | Thinking distance | Braking distance | Stopping distance |
|---|---|---|---|
| 13 m/s (about 30 mph) | 9 m | 14 m | 23 m |
| 22 m/s (about 50 mph) | 15 m | 38 m | 53 m |
| 31 m/s (about 70 mph) | 21 m | 75 m | 96 m |
Exam Tip: In a data question, remember thinking distance is proportional to speed (roughly doubles when speed doubles) but braking distance is proportional to speed squared (roughly quadruples). Comparing the two columns is a classic six-mark question.
| Misconception | The correct idea |
|---|---|
| "Stopping distance is just the braking distance" | Stopping distance = thinking distance + braking distance — you must include the thinking distance |
| "Doubling the speed doubles the braking distance" | Braking distance is proportional to speed squared, so doubling the speed makes it about four times as long |
| "Alcohol increases the braking distance" | Alcohol slows reactions, so it increases the thinking distance; it does not change the braking force |
| "Worn tyres increase the thinking distance" | Worn tyres reduce grip, increasing the braking distance; they have no effect on reaction time |
| "The lost kinetic energy just disappears" | It is transferred to the thermal store of the brakes by friction — the brakes get hot |
| "A heavier car and a lighter car have the same braking distance" | A heavier car has more kinetic energy, so its braking distance is greater for the same braking force |
Question (6 marks): A driver is travelling at 30 m/s instead of 15 m/s. Explain, in terms of energy, why the braking distance is much greater at the higher speed, and describe what happens to the car's kinetic energy when it brakes.
Mid-band response: "The car is going faster so it has more energy and takes longer to stop. When it brakes the energy goes into the brakes and they get hot."
Examiner-style commentary: The response has the right ideas — more speed means more energy and a longer stop, and the energy goes to the brakes — but it is vague. There is no mention of kinetic energy, no use of the v2 relationship, and "takes longer to stop" confuses time with distance. To improve, name the kinetic energy store, quote Ek=21mv2, and use the square relationship.
Stronger response: "The kinetic energy of the car is Ek=21mv2. Because it depends on v2, doubling the speed makes the kinetic energy four times as big. When the brakes are applied, the braking force does work to remove this energy, so a bigger braking distance is needed. The kinetic energy is transferred to thermal energy in the brakes."
Examiner-style commentary: A good answer that quotes the equation, uses the square relationship, and identifies the energy transfer. To reach the top band, state explicitly that the work done equals F×d, connect this to d∝v2, and say the transfer is by friction.
Top-band response: "The car's kinetic energy is given by Ek=21mv2, so it is proportional to the square of the speed. Doubling the speed from 15 m/s to 30 m/s therefore makes the kinetic energy 22=4 times greater. To stop the car, the braking force must do work equal to this kinetic energy: F×d=21mv2. For the same braking force F, the braking distance d is proportional to v2, so it too becomes about four times as long. During braking, friction between the brake pads and the discs transfers the car's kinetic energy into the thermal (internal) energy store of the brakes, which is why the brakes get hot."
Examiner-style commentary: Full marks. It quotes and applies Ek=21mv2, links work done (F×d) to the braking distance, states the fourfold increase with justification, and describes the kinetic-to-thermal transfer by friction naming both the mechanism and the destination store.
This content is aligned with OCR Gateway Science A GCSE Physics (J249), Topic P8 Global challenges (stopping distances; thinking and braking distance; factors affecting each; energy transfer in braking). Refer to the official OCR specification document for the exact wording.