Thinking, braking and stopping distance · Reaction time · Factors affecting each · Braking deceleration
AQA GCSE Physics 4.5 | Year 11 Higher Tier
🧠 Define thinking distance, braking distance and stopping distance
⏱️ Understand what reaction time is and what factors affect it
📏 Explain what factors affect braking distance
⚡ Use the equation: stopping distance = thinking distance + braking distance
🔢 Calculate deceleration during braking using v² = u² + 2as
🚗 Evaluate the dangers of large decelerations and long stopping distances
🛑 The Three Key Distances
When a driver needs to stop a vehicle in an emergency, the total distance covered before the car comes to rest is called the stopping distance. It is made up of two separate parts:
Thinking Distance: The distance travelled by the vehicle from the moment the driver perceives a hazard to the moment they apply the brakes. During this time the vehicle travels at constant speed.
Braking Distance: The distance travelled by the vehicle from the moment the brakes are applied to the moment the vehicle comes to rest. The vehicle decelerates during this phase.
Both thinking distance and braking distance increase as the initial speed of the vehicle increases. However, they are affected by different factors, so it is important to treat them separately.
At 30 mph (≈ 13 m/s), the Highway Code gives a typical stopping distance of about 23 m. At 70 mph (≈ 31 m/s) this rises to approximately 96 m — more than four times as far! This is because braking distance depends on v², not v.
⏱️ Reaction Time and Thinking Distance
Reaction time is the time interval between a driver perceiving a hazard and beginning to apply the brakes. A typical human reaction time is 0.2 s to 0.9 s, though values vary enormously between individuals and circumstances.
Thinking distance (m) = Speed (m/s) × Reaction time (s)
dthink = v × tr
Because this is a simple proportional relationship, thinking distance increases linearly with speed — double the speed, double the thinking distance.
Factors that increase reaction time:
Alcohol: Slows nerve signal transmission; reaction times can double or triple.
Tiredness / fatigue: Reduces mental alertness and processing speed.
Distraction: Using a mobile phone, adjusting the radio, or talking to passengers diverts attention.
Drugs (prescription or recreational): Many medications warn against driving; recreational drugs severely impair reaction speed.
Age: Reaction time tends to increase with age on average.
Reaction time can be measured experimentally using a ruler-drop test: a ruler is dropped without warning and the distance it falls before being caught is converted to a time using s = ½gt².
Speed (m/s)
Speed (mph)
Thinking dist. at t = 0.7 s
13
30
9.1 m
22
50
15.4 m
31
70
21.7 m
🔧 Braking Distance and Deceleration
When the brakes are applied, friction between the brake pads and wheels (and between tyres and road) converts the vehicle's kinetic energy into thermal energy (heat). This decelerates the vehicle. The braking distance is the distance over which this occurs.
Using the equation of motion with final velocity v = 0:
v² = u² + 2as
0 = u² + 2as
Braking distance: s = −u² ÷ (2a)
(where a is negative = deceleration)
Because braking distance depends on u², doubling the speed quadruples the braking distance. This is a crucial safety implication — at motorway speeds, stopping distances are enormous.
Factors that increase braking distance:
Higher initial speed: More kinetic energy (KE = ½mv²) must be removed by braking.
Wet or icy roads: Reduce friction between tyres and road surface, requiring more distance to dissipate the same KE.
Worn tyres: Reduced tread depth means less grip; aquaplaning risk increases in wet conditions.
Worn or faulty brakes: Reduced braking force means smaller deceleration, so longer distance required.
Greater vehicle mass: More KE to dissipate at any given speed (KE = ½mv²).
Downhill gradient: Component of gravity acts in the direction of motion, opposing braking.
Large decelerations are dangerous because the braking force on passengers can cause serious injury. Seatbelts and crumple zones extend the time over which deceleration occurs, reducing the force on occupants (F = ma → smaller a, smaller F).
📐 Energy and Braking Force
We can also analyse braking using the work-energy theorem. The brakes must do work against the vehicle's kinetic energy to bring it to rest:
Work done by brakes = Kinetic energy of vehicle
F × d = ½ × m × v²
Braking distance: d = mv² ÷ (2F)
where F is the braking force, m is the mass of the vehicle, v is the initial speed, and d is the braking distance.
This equation shows clearly why:
A larger braking force F → shorter braking distance (but larger deceleration)
Greater mass m → longer braking distance for the same force
Larger speed v → much longer braking distance (v² dependence)
Emergency braking can produce decelerations of 6–10 m/s². At these values, an unrestrained passenger experiences a force several times their body weight — explaining why seatbelts are legally required.
Quantity
Symbol
SI Unit
Stopping distance
dstop
metres (m)
Thinking distance
dthink
metres (m)
Braking distance
dbrake
metres (m)
Speed
v or u
metres per second (m/s)
Reaction time
tr
seconds (s)
Deceleration (magnitude)
a
metres per second² (m/s²)
Braking force
F
newtons (N)
Mass
m
kilograms (kg)
📊 Speed and Stopping Distance: The v² Relationship
It is vital to understand the non-linear relationship between speed and stopping distance at GCSE and beyond. While thinking distance scales linearly with speed, braking distance scales with v².
For a vehicle decelerating uniformly from speed u to rest with deceleration magnitude a:
Total stopping distance = (u × tr) + u² ÷ (2a)
As u increases, the second term dominates — meaning at higher speeds, braking distance becomes far more important than thinking distance in determining the total stopping distance.
Example comparison (a = 6 m/s², tr = 0.7 s):
Speed u (m/s)
Thinking dist. (m)
Braking dist. (m)
Total (m)
10
7
8.3
15.3
20
14
33.3
47.3
30
21
75.0
96.0
Trebling the speed from 10 m/s to 30 m/s increases braking distance by a factor of 9 (3²) but only triples the thinking distance. Total stopping distance increases by a factor of about 6.3 — much more than most drivers appreciate.
Example 1: A car travels at 20 m/s. The driver has a reaction time of 0.6 s. After braking, the car decelerates uniformly at 8 m/s². Calculate: (a) the thinking distance, (b) the braking distance, (c) the total stopping distance.
1Identify the quantities: u = 20 m/s, tr = 0.6 s, a = −8 m/s² (deceleration), v = 0 m/s
2Thinking distance: dthink = u × tr = 20 × 0.6 = 12 m
3Braking distance using v² = u² + 2as: 0 = (20)² + 2 × (−8) × s 0 = 400 − 16s 16s = 400 s = 400 ÷ 16 = 25 m
4Total stopping distance: dstop = 12 + 25 = 37 m
Thinking distance = 12 m | Braking distance = 25 m | Stopping distance = 37 m
Example 2: A lorry of mass 12 000 kg is travelling at 25 m/s. The braking force applied is 45 000 N. Calculate (a) the deceleration of the lorry, (b) the braking distance.
1Identify the quantities: m = 12 000 kg, u = 25 m/s, F = 45 000 N, v = 0 m/s
2Deceleration using Newton's second law: F = ma a = F ÷ m = 45 000 ÷ 12 000 = 3.75 m/s² (deceleration)
3Braking distance using v² = u² + 2as: 0 = (25)² + 2 × (−3.75) × s 0 = 625 − 7.5s s = 625 ÷ 7.5 = 83.3 m
4Alternative check using energy: F × d = ½mv² d = (½ × 12000 × 625) ÷ 45000 = 3 750 000 ÷ 45 000 = 83.3 m ✓
Deceleration = 3.75 m/s² | Braking distance = 83.3 m
Example 3: A car has a stopping distance of 56 m when travelling at 18 m/s. The driver's reaction time is 0.8 s. Calculate the deceleration during braking.
1Find thinking distance: dthink = u × tr = 18 × 0.8 = 14.4 m
3Find deceleration using v² = u² + 2as (v = 0, u = 18 m/s, s = 41.6 m): 0 = (18)² + 2 × a × 41.6 0 = 324 + 83.2a 83.2a = −324 a = −324 ÷ 83.2 = −3.90 m/s²
4State the answer with correct sign: The deceleration magnitude is 3.90 m/s²
Deceleration = 3.9 m/s² (to 2 s.f.)
Example 4: A driver is travelling at 30 m/s on a dry road (braking deceleration 9 m/s²). The road becomes icy, reducing the braking deceleration to 1.5 m/s². How much further does the car travel before stopping if the reaction time is 0.5 s in both cases? (Assume same speed at start of each journey.)
1Thinking distance is the same in both cases (same speed, same reaction time): dthink = 30 × 0.5 = 15 m
2Braking distance on dry road (a = 9 m/s²): sdry = u² ÷ (2a) = (30)² ÷ (2 × 9) = 900 ÷ 18 = 50 m Total dry = 15 + 50 = 65 m
3Braking distance on icy road (a = 1.5 m/s²): sice = (30)² ÷ (2 × 1.5) = 900 ÷ 3 = 300 m Total icy = 15 + 300 = 315 m
4Difference: 315 − 65 = 250 m further on ice
On ice the car travels 250 m further before stopping — nearly the length of 2.5 football pitches!
Q1. A vehicle travels at constant speed during which part of the stopping distance?
Q2. A car is travelling at 15 m/s. The driver's reaction time is 0.5 s. What is the thinking distance?
Q3. Which of the following factors affects braking distance but NOT thinking distance?
Q4. A car brakes from 24 m/s to rest with a deceleration of 6 m/s². What is the braking distance?
Q5. If a driver doubles their speed, what happens to the braking distance (assuming constant deceleration)?
Challenge 1 (6 marks): A car of mass 1200 kg is travelling at 28 m/s on a dry road. The driver has a reaction time of 0.75 s. The braking force applied by the brakes is 9600 N.
(a) Calculate the thinking distance. [1]
(b) Calculate the deceleration during braking. [2]
(c) Calculate the total stopping distance. [3]
(a) dthink = 28 × 0.75 = 21 m
(b) a = F ÷ m = 9600 ÷ 1200 = 8 m/s²
(c) v² = u² + 2as → 0 = 28² − 2 × 8 × s → s = 784 ÷ 16 = 49 m
Total stopping distance = 21 + 49 = 70 m
Challenge 2 (5 marks): Two cars, A and B, both travel at 20 m/s. Car A has new tyres on a dry road (deceleration 8 m/s²). Car B has worn tyres on a wet road (deceleration 3.2 m/s²). Both drivers have a reaction time of 0.6 s.
(a) Show that the braking distance of Car A is 25 m. [2]
(b) Calculate the total stopping distance of Car B. [3]
(c) Compare the two total stopping distances and explain the difference in terms of friction and energy. [2]
(a) s = u² ÷ (2a) = 400 ÷ 16 = 25 m ✓
(b) Thinking distance (both) = 20 × 0.6 = 12 m
Braking distance B: s = 400 ÷ (2 × 3.2) = 400 ÷ 6.4 = 62.5 m
Total stopping distance B = 12 + 62.5 = 74.5 m
(c) Car A total = 12 + 25 = 37 m. Car B total = 74.5 m — over twice as far.
Worn tyres on a wet road provide significantly less friction force between tyre and road. Less friction → smaller braking force → smaller deceleration. The same kinetic energy (½mv² = same for both) must be dissipated, but with less force, requiring a greater distance (F × d = KE, so smaller F → larger d).
Challenge 3 (6 marks): In an investigation into reaction times, a student drops a ruler from rest without warning and another student catches it. The second student catches the ruler after it falls 18 cm.
(a) Use s = ½gt² to calculate the reaction time. Use g = 10 m/s². [3]
(b) If this student were driving at 20 m/s, calculate their thinking distance. [1]
(c) A different student, who has consumed alcohol, has a reaction time of 1.2 s. Calculate the additional thinking distance compared to the first student if both drive at 20 m/s. [2]
(a) s = 0.18 m, g = 10 m/s²
0.18 = ½ × 10 × t²
t² = 0.18 ÷ 5 = 0.036
t = √0.036 = 0.19 s (to 2 s.f.)
(b) dthink = 20 × 0.19 = 3.8 m
(c) Sober thinking distance = 3.8 m
Alcohol thinking distance = 20 × 1.2 = 24 m
Additional distance = 24 − 3.8 = 20.2 m further
This illustrates the extreme danger of drink-driving.
Challenge 4 — Extended Response (6 marks): Explain why stopping distances increase significantly at higher speeds. In your answer, refer to thinking distance, braking distance, kinetic energy and the relationship between these quantities and speed.
Model Answer:
Thinking distance increases linearly with speed (d = v × tr). Doubling speed doubles the thinking distance because the car covers more ground in the same reaction time.
Braking distance increases with the square of speed. This is because kinetic energy KE = ½mv² depends on v². To stop, the brakes must remove all this kinetic energy. Using the work-energy theorem, F × d = ½mv², so d = mv² ÷ (2F). Doubling speed quadruples the kinetic energy and therefore quadruples the braking distance (for constant braking force).
Combined effect: At low speeds, thinking distance is the larger component. At high speeds, the v² dependence of braking distance dominates, causing total stopping distance to grow much faster than speed increases. Drivers travelling at motorway speeds (≈ 31 m/s) require stopping distances in excess of 90 m — the length of a football pitch. This is why speed limits exist and why tailgating at high speeds is so dangerous.