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How do you calculate stopping distances and the forces involved in motoring?

Calculating overall stopping distance from thinking and braking distances, the typical Highway Code figures, and the science of motoring - force, momentum and kinetic energy.

A CCEA GCSE Motor Vehicle and Road User Studies answer on calculating overall stopping distance from thinking and braking distances, the Highway Code figures, and the science of motoring including force, momentum and kinetic energy.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

CCEA wants you to calculate the overall stopping distance from its parts, recall the typical Highway Code figures, and handle the science of motoring - force (F = ma), momentum (p = mv) and kinetic energy (Ek=12mv2E_k = \tfrac12 m v^2). This is the most mathematical part of Unit 1 and links the road-safety topic to the physics behind it.

The answer

Overall stopping distance

To find the overall distance, add the thinking and braking distances. To find one part, subtract: braking distance == overall −- thinking.

Typical Highway Code figures

The Highway Code gives typical dry-road stopping distances. Useful ones to know:

  • At 30 mph: thinking 9 m + braking 14 m == 23 m.
  • At 60 mph: thinking 18 m + braking 55 m == 73 m.
  • At 70 mph: thinking 21 m + braking 75 m == 96 m.

Notice the braking distance grows much faster than the thinking distance, because braking distance depends on the square of the speed.

The science of motoring

Three physics relationships explain why speed and mass matter.

  • A heavier or faster vehicle has more momentum and needs a bigger force (or longer time) to stop.
  • Kinetic energy depends on the square of the speed, so doubling the speed gives four times the energy - which the brakes must turn into heat, and which makes a crash far more severe.

Worked example: the braking force

Worked example: kinetic energy and crash severity

Examples in context

Example 1. Why braking distance leaps. From 30 to 60 mph the speed doubles, so the kinetic energy and braking distance roughly quadruple - the 14 m braking distance at 30 mph becomes about 55 m at 60 mph.

Example 2. Heavy vehicle, more momentum. A loaded lorry has far more momentum than a car at the same speed, so it needs a much greater force and distance to stop.

Try this

Q1. Thinking distance is 12 m and braking distance is 24 m. What is the overall stopping distance? [1 mark]

  • Cue. 12+24=3612 + 24 = 36 m.

Q2. Calculate the momentum of a 1500 kg car moving at 10 m/s. [2 marks]

  • Cue. p=mv=1500×10=15 000p = mv = 1500 \times 10 = 15\,000 kg m/s.

Q3. Calculate the kinetic energy of a 1000 kg car moving at 10 m/s. [2 marks]

  • Cue. Ek=12×1000×102=50 000E_k = \tfrac{1}{2} \times 1000 \times 10^2 = 50\,000 J.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA style3 marksAt a certain speed the thinking distance is 15 m and the braking distance is 38 m. Calculate the overall stopping distance, and state what could increase the thinking distance.
Show worked answer →

Overall stopping distance == thinking distance ++ braking distance.

=15+38=53 m.= 15 + 38 = 53\ \text{m}.

The thinking distance could be increased by anything that lengthens the reaction time: alcohol, drugs, tiredness, distraction (such as a phone), or being a less experienced driver. (Higher speed also increases it.)

Markers reward adding the two distances to get 53 m, and a valid cause of a longer thinking distance.

CCEA style4 marksA car of mass 1200 kg is travelling at 20 m/s. (a) Calculate its momentum. (b) Calculate its kinetic energy. State the units.
Show worked answer →

(a) Momentum =mass×velocity= \text{mass} \times \text{velocity}:

p=mv=1200×20=24 000 kg m/s.p = m v = 1200 \times 20 = 24\,000\ \text{kg m/s}.

(b) Kinetic energy =12mv2= \tfrac{1}{2} m v^2:

Ek=12×1200×202=12×1200×400=240 000 J.E_k = \tfrac{1}{2} \times 1200 \times 20^2 = \tfrac{1}{2} \times 1200 \times 400 = 240\,000\ \text{J}.

Markers reward p = mv = 24 000 kg m/s, and KE = ½mv² = 240 000 J, with correct units.

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