Stopping Distance

This lesson covers thinking distance, braking distance, the factors that affect each, and the relationship between speed and braking distance, as required by the Edexcel GCSE Combined Science specification (1SC0). Understanding stopping distance is vital for road safety questions in the exam.

---

What Is Stopping Distance?

Stopping distance is the total distance a vehicle travels from the moment the driver sees a hazard to the moment the vehicle comes to a complete stop.

$\text{Stopping distance} = \text{Thinking distance} + \text{Braking distance}$Stopping distance=Thinking distance+Braking distance

graph LR
    A["Hazard seen"] -- "Thinking distance" --> B["Brakes applied"]
    B -- "Braking distance" --> C["Vehicle stops"]
    A -. "Stopping distance" .-> C

---

Thinking Distance

Thinking distance is the distance the vehicle travels during the driver's reaction time — the time between seeing the hazard and pressing the brake pedal.

$\text{Thinking distance} = \text{speed} \times \text{reaction time}$Thinking distance=speed×reaction time

Typical Reaction Time

A typical human reaction time is 0.2 s to 0.9 s, with about 0.7 s as a commonly used value.

Worked Example

A car is travelling at 20 m/s. The driver's reaction time is 0.5 s. What is the thinking distance?

$d_{thinking} = v \times t = 20 \times 0.5 = 10 \text{ m}$dthinking​=v×t=20×0.5=10 m

---

Factors Affecting Thinking Distance

Thinking distance increases if the reaction time is longer. Factors that increase reaction time include:

Factor	Why It Increases Reaction Time
Tiredness	Slower brain processing
Alcohol	Impaired judgement and slower reactions
Drugs / medication	Can affect concentration and response
Distractions	Using a phone, adjusting radio, etc.
Age	Reaction times tend to be longer in older drivers

Thinking distance also increases if the speed is higher, because the car covers more ground during the reaction time.

Exam Tip: Thinking distance is directly proportional to speed — double the speed, double the thinking distance. This is because thinking distance = speed × reaction time, and reaction time is constant.

---

Braking Distance

Braking distance is the distance the vehicle travels from the moment the brakes are applied to the moment it stops completely.

When the brakes are applied, a braking force acts to decelerate the vehicle. The kinetic energy of the vehicle is transferred to thermal energy in the brakes.

---

Factors Affecting Braking Distance

Factor	Effect
Speed	Higher speed → much greater braking distance
Condition of brakes	Worn brakes provide less braking force → longer braking distance
Condition of tyres	Worn tyres have less grip → longer braking distance
Road conditions	Wet, icy or oily roads reduce friction → longer braking distance
Mass of vehicle	Heavier vehicle has more kinetic energy to dissipate → longer braking distance

---

Braking Distance Is Proportional to Speed Squared

This is one of the most important relationships in this topic:

$\text{Braking distance} \propto v^2$Braking distance∝v2

This means:

• Double the speed → four times the braking distance
• Triple the speed → nine times the braking distance

Why?

The kinetic energy of a moving vehicle is:

$E_k = \frac{1}{2}mv^2$Ek​=21​mv2

When braking, all this kinetic energy must be transferred to thermal energy by the brakes. Since the braking force is roughly constant, the work done by the brakes equals:

$W = F \times d$W=F×d

Setting these equal: $Fd = \frac{1}{2}mv^2$Fd=21​mv2, so $d = \frac{mv^2}{2F}$d=2Fmv2​

Since m and F are approximately constant, $d \propto v^2$d∝v2.

Worked Example

At 20 m/s, a car has a braking distance of 14 m. Estimate the braking distance at 60 m/s.

Speed has tripled (60 ÷ 20 = 3), so braking distance increases by a factor of 3² = 9:

$d = 14 \times 9 = 126 \text{ m}$d=14×9=126 m

Exam Tip: The v² relationship is a favourite exam question. If speed doubles, braking distance quadruples. If speed triples, braking distance increases by 9 times. Always square the speed ratio.

---

Dangers of Large Decelerations

When a vehicle brakes very hard:

• A very large force is needed (from F = ma, large deceleration means large force).
• Brakes can overheat, reducing their effectiveness (brake fade).
• Tyres can lose grip and the vehicle may skid.
• Passengers experience large forces, increasing the risk of injury.

Worked Example

A 1200 kg car decelerates from 30 m/s to 0 in 3 s. What is the braking force?

$a = \frac{v - u}{t} = \frac{0 - 30}{3} = -10 \text{ m/s}^2$a=tv−u​=30−30​=−10 m/s2

$F = ma = 1200 \times 10 = 12\,000 \text{ N}$F=ma=1200×10=12000 N

---

Measuring Reaction Time

In the required practical, reaction time can be estimated using a ruler drop test:

1. A partner holds a ruler vertically.
2. The test subject places their fingers at the zero mark, ready to catch it.
3. The ruler is released without warning and the subject catches it as quickly as possible.
4. The distance the ruler falls is used to calculate reaction time using $s = \frac{1}{2}gt^2$s=21​gt2.

Rearranging: $t = \sqrt{\frac{2s}{g}}$t=g2s​​

Worked Example

The ruler falls 18 cm before being caught. Calculate the reaction time.

Convert: 18 cm = 0.18 m

$t = \sqrt{\frac{2 \times 0.18}{9.8}} = \sqrt{\frac{0.36}{9.8}} = \sqrt{0.0367} = 0.19 \text{ s}$t=9.82×0.18​​=9.80.36​​=0.0367​=0.19 s

Exam Tip: In the ruler drop test, repeat the experiment several times and calculate a mean to improve reliability. Also explain why an electronic timer is more accurate — it removes human error in measuring.

---

Summary

• Stopping distance = thinking distance + braking distance.
• Thinking distance = speed × reaction time. It is directly proportional to speed.
• Reaction time is affected by tiredness, alcohol, drugs, distractions and age.
• Braking distance depends on speed, brakes, tyres, road conditions and vehicle mass.
• Braking distance ∝ v²: double speed → 4× braking distance.
• Large decelerations are dangerous — risk of overheating brakes, skidding and injury.
• Reaction time can be measured with a ruler drop test and calculated using $t = \sqrt{2s/g}$t=2s/g​.

---