Factors affecting braking distance 2 — AQA GCSE Physics
Test yourself on Factors affecting braking distance 2 with AQA GCSE practice questions.
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Factors affecting braking distance 2 explained
Pressing the brake pedal forces the pads against the discs or drums attached to the wheels.
Read the full explanation
As the wheel turns against the pads, a friction force acts and work is done. That work transfers energy out of the kinetic energy store of the vehicle and into the thermal energy store of the brakes, so the vehicle slows and the brakes get hot. The amount is given by work done = force × distance, and it equals the kinetic energy removed, ½ m v². This is why brake discs glow after a long descent and why repeated heavy braking can overheat them. Saying only 'friction' is not enough in an examination: you must say that work is done. For a 1200 kg car at 20 m/s the brakes must do 240 000 J of work, which is why brake discs can reach several hundred degrees Celsius.
The greater the speed of a vehicle the greater the braking force needed to stop the vehicle in a certain distance.
A vehicle's kinetic energy is ½ m v², so speed matters far more than anything else in the equation. The brakes must do work equal to that energy: braking force × braking distance = ½ m v². Fixing the distance and rearranging gives F = ½ m v² ÷ d, which shows the force needed rises with the square of the speed. A 1200 kg car stopping in 30 m from 15 m/s needs 1200 × 15² ÷ (2 × 30) = 4500 N; from 30 m/s in the same 30 m it needs 18 000 N, four times as much. Real brakes and tyres cannot supply an unlimited force, which is why the distance stretches instead at higher speeds. It is also why a modest speed limit matters: dropping from 30 m/s to 20 m/s removes more than half the kinetic energy the brakes must deal with.
The greater the braking force the greater the deceleration of the vehicle. Large decelerations may lead to brakes overheating and/or loss of control.
Deceleration follows from Newton's Second Law: a = F ÷ m, so for a given mass a larger braking force gives a larger deceleration. Braking hard therefore stops the vehicle in less distance, but it brings two dangers. First, the energy removed from the kinetic store is transferred into the thermal store of the brakes in a shorter time, so the temperature rises steeply and the brakes can overheat and fade, losing their grip. Second, if the braking force demanded exceeds the friction the tyres can get from the road, the wheels lock and the vehicle skids; a skidding tyre gives less friction and no steering, so the driver loses control. Anti-lock braking systems exist for this reason, releasing and reapplying the brakes rapidly so the wheels keep turning and the tyres keep their grip.
Students should be able to explain the dangers caused by large decelerations.
When a vehicle stops quickly, its occupants and the vehicle itself must lose momentum in a short time. A large deceleration means a large rate of change of velocity, so a large resultant force acts. The force on a passenger is exerted through the seat belt, and the pressure on the body can cause injury. Brakes convert kinetic energy into thermal energy; very large decelerations can overheat brakes and reduce their effectiveness, increasing stopping distance. The relationship F = m × a shows that for a fixed mass, a larger deceleration requires a larger force. For example, a car of mass 1200 kg decelerating at 8 m/s² experiences a resultant force of 9600 N, which is dangerous without a seat belt.
(HT only) estimate the forces involved in the deceleration of road vehicles in typical situations on a public road.
Higher-tier students should be able to estimate the resultant force on a road vehicle during typical decelerations. The method uses F = m × a, where a is the deceleration. A typical car of mass about 1200 kg braking from 13 m/s to rest in 2.0 s has a deceleration of 6.5 m/s², giving a force of about 7800 N. Students should choose sensible values for mass, initial speed and stopping time, convert units if needed, and calculate the deceleration using a = Δv ÷ t. The estimate should be given to a reasonable number of significant figures and the direction of the force should be opposite to the motion. This skill is assessed by checking that the method is physically correct and the numerical answer is plausible for a public-road situation.
Your focus
- Explain why the temperature of the brakes rises during a stop, naming the energy stores involved.
- Calculate the work done by the brakes from the braking force and the braking distance.
- Determine the braking force needed to stop a vehicle of stated mass and speed in a given distance.
Show all 15 objectives
- Calculate the braking force needed to stop a vehicle of given mass and speed within a stated distance.
- Show that doubling the speed quadruples the braking force needed over the same distance.
- Explain why a heavily laden vehicle needs a larger braking force than an empty one at the same speed.
- Calculate the deceleration of a vehicle from the braking force acting on it and its mass.
- Explain why a large deceleration can cause the brakes to overheat.
- Describe how locked wheels lead to a skid and the loss of steering control.
- Explain why a large deceleration produces a large resultant force on a vehicle and its occupants.
- Describe how safety features reduce the force on passengers during a collision.
- Relate overheating of brakes to large decelerations and increased stopping distance.
- Select realistic values for mass, speed and stopping time for a road vehicle.
- Calculate deceleration and resultant force using a = Δv ÷ t and F = m × a.
- Evaluate whether an estimated force is reasonable for a typical public-road situation.
Factors affecting braking distance 2 exam tips
Marking Points
- one mark for stating that work is done against friction between the brake pads and the wheel or disc
- one mark for stating that the kinetic energy store of the vehicle decreases
- one mark for stating that the internal, or thermal, energy of the brakes increases so their temperature rises
- one mark for using work done = force × distance
- one mark for equating the work done to the kinetic energy ½ m v²
- one mark for stating that the kinetic energy is greater at the greater speed
- one mark for using ½ m v² to show that doubling the speed quadruples the kinetic energy
- one mark for equating work done, force × distance, to the kinetic energy
- one mark for concluding that the braking force must be four times larger for the same braking distance
- one mark for using a = F ÷ m to link a larger braking force to a larger deceleration
- one mark for stating that energy is transferred to the thermal store of the brakes at a greater rate, so they overheat
- one mark for stating that the tyres may lose grip so the vehicle skids
- one mark for the consequence that the driver loses steering control of the vehicle
- State that large deceleration means velocity changes greatly in a short time, so the rate of change of velocity is large.
- Use F = m × a to explain that a larger deceleration produces a larger resultant force on the vehicle and its occupants for a given mass.
- Explain that a seat belt or crumple zone increases the time taken to decelerate, reducing the deceleration and therefore the force on the passenger.
- Describe that brakes transfer kinetic energy to thermal energy, and very large decelerations can cause brakes to overheat and fade, reducing braking force.
- Link injury to the pressure or force exerted on the body when a passenger is stopped by a hard surface rather than by a belt or airbag.
- Select realistic values for the mass of the vehicle, its initial speed and the time taken to stop, for example a car of mass 1200 kg stopping from 13 m/s in 2.0 s.
- Calculate the deceleration using a = Δv ÷ t, ensuring the change in velocity is in m/s and the time is in s.
- Substitute the mass and deceleration into F = m × a to estimate the resultant force, and state the unit as N.
- Comment on whether the estimated force is reasonable for the situation, and note that the force acts opposite to the direction of motion.
Examiner Tips
- 💡Use the phrase 'work is done against friction' rather than 'friction slows it down'.
- 💡Name both stores in the transfer: kinetic store down, thermal store of the brakes up.
- 💡Two-mark questions here usually want one mark for the work done and one for the temperature rise, so give both.
- 💡Square the speed before you multiply by the mass; doing it in the other order invites slips.
- 💡Quote the ratio as well as the number: 'four times the force' makes the physics visible.
- 💡Check the distance in the question is the braking distance, not the total stopping distance.
- 💡Give two distinct dangers when a question asks about large decelerations; overheating and skidding are the standard pair.
- 💡Say that deceleration is an acceleration in the opposite direction to the motion rather than dropping the sign.
- 💡Link every danger to a physical cause, not just to 'it is unsafe'.
- 💡Always link the size of the deceleration to the size of the resultant force using F = m × a, and name the force on the passenger.
- 💡When explaining safety features, state clearly that they increase the time of deceleration, which reduces the deceleration and the force.
- 💡Use the correct units for deceleration (m/s²) and force (N), and show any substitution clearly if a calculation is required.
- 💡Write down the values you choose for mass, speed and time before calculating, so the examiner can follow your estimate.
- 💡Show the deceleration calculation separately from the force calculation to make the method clear.
- 💡Check that your final force is in newtons and is plausible; a force of a few thousand newtons is typical for a car braking hard.
Common Mistakes
- writing only the word friction, which the mark scheme explicitly says is insufficient
- saying heat is created rather than energy being transferred to a thermal store
- describing the kinetic energy as lost or used up
- substituting the whole stopping distance into work done = force × distance rather than the braking distance
- saying the required force doubles when the speed doubles
- leaving the mass out of the kinetic energy calculation
- substituting the stopping distance rather than the braking distance
- rearranging F = ½ m v² ÷ d incorrectly and dividing by the speed instead
- saying the brakes 'stop working' instead of describing overheating and brake fade
- forgetting that the deceleration depends on the mass as well as on the braking force
- claiming a skidding tyre has more friction because it is sliding
- confusing the braking force with the weight of the vehicle
- Thinking that a large deceleration means a large speed; correct this by stating that deceleration is the rate of change of velocity, so a vehicle can have a small speed but still decelerate rapidly.
- Believing that the force on a passenger is the same whatever the stopping time; correct this by explaining that for a fixed change in momentum, a shorter stopping time gives a larger force.
- Assuming brakes always work perfectly; correct this by explaining that overheating can reduce braking force and increase stopping distance.
- Using the speed in km/h directly in the calculation; correct this by converting to m/s before finding the deceleration.
- Forgetting to divide by the stopping time and using the change in velocity as the deceleration; correct this by applying a = Δv ÷ t.
- Giving an answer with unrealistic precision, such as many decimal places; correct this by rounding to a sensible number of significant figures for an estimate.