5.21 Braking distance and large decelerations
9 exam questions worth 25 marks. Write your answer, open the mark scheme, then give yourself the marks you earned. Your answers are saved on this device.
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What happens to the braking distance of a car if its speed doubles (same braking force)?
Choose one answer.
Correct Not quite: the answer is It quadruples
Mark scheme
C: It quadruples
Model answer
It quadruples (C).
Give three factors that affect the braking distance of a car.
Mark scheme
any three from: speed (1); mass of the car (1); condition of the tyres (1); condition of the brakes (1); road surface / weather (1)
Model answer
The speed of the car, the condition of the tyres and whether the road is wet or icy.
Explain why the braking distance is longer on a wet road than on a dry road.
Mark scheme
there is less friction between the tyres and the road (1); so the maximum braking force is smaller, and the car takes longer to stop (1)
Model answer
There is less friction between the tyres and the road, so the braking force is smaller and the car takes longer to stop.
Describe the energy transfers that take place when a car brakes to a stop.
Mark scheme
the kinetic energy of the car decreases (1); friction between the brake pads and the disc does work (1); the energy is transferred to the thermal store of the brakes / surroundings (1)
Model answer
The car's kinetic energy is transferred by the friction of the brake pads on the discs to the thermal store of the brakes, which get hot.
A car of mass 1200 kg is travelling at 15 m/s. Calculate its kinetic energy.
Mark scheme
KE = ½ m v² (1); 0.5 × 1200 × 15² (1); 135 000 J (1)
Model answer
KE = ½ m v² = 0.5 × 1200 × 15² = 135 000 J.
The brakes of the car in the previous question apply a constant force of 6000 N. Calculate the braking distance.
Mark scheme
work done = kinetic energy = 135 000 J (1); d = 135 000 ÷ 6000 (1); 22.5 m (1)
Model answer
Work done = F d = 135 000 J, so d = 135 000 / 6000 = 22.5 m.
A car travelling at 15 m/s has a braking distance of 20 m. Estimate the braking distance when it is travelling at 30 m/s, with the same braking force. Explain your answer.
Mark scheme
80 m (1); speed doubled so the kinetic energy is four times bigger (1); braking distance is proportional to speed squared / four times the work needs to be done (1)
Model answer
80 m. Doubling the speed makes the kinetic energy four times larger, so the brakes have to do four times as much work, and the braking distance is four times bigger.
(Higher) A car travelling at 25 m/s stops in 0.20 s in a crash. A passenger has a mass of 60 kg. Estimate the force on the passenger.
Mark scheme
a = 25 ÷ 0.20 = 125 m/s² (1); F = m a = 60 × 125 (1); 7500 N (1)
Model answer
a = 25 / 0.20 = 125 m/s². F = m a = 60 × 125 = 7500 N.
Explain why large decelerations are dangerous, and why a driver should not brake as hard as possible on a wet road.
Mark scheme
a large deceleration needs a large force (F = m a) (1); this can damage the body of the passenger / cause injury (1); braking too hard on a wet road makes the tyres skid (1); the friction decreases and the car loses control / takes longer to stop (1)
Model answer
A large deceleration needs a large force, which can injure the passengers. On a wet road, braking too hard makes the wheels skid, so friction reduces and the car can lose control.
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