---
title: "Stopping distance and brake energy"
subject: "Automotive Technology"
chapter: 7
key: "semester_5/automotive_technology:braking-distance-energy-numerical"
source: https://www.bctengineer.com/q/semester_5/automotive_technology/braking-distance-energy-numerical
site: BCT Engineering
---

# Stopping distance and brake energy

Automotive Technology · Chapter 7: Braking Systems

- Asked in: year not given, 8 marks
- Video: [L 32 Numerical Solution: To Find Stopping Distance | Vehicle Dynamics | Automobile](https://www.youtube.com/watch?v=4D0bLwr7JOI) — MECHANICAL & AUTOMOBILE ENGINEERING_LJIET
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/automotive_technology/braking-distance-energy-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/automotive_technology/chapter-7

## Question

A car of mass 1200 kg travels at 72 km/h on a level road with a tyre-road friction coefficient of 0.7. Assuming that all four wheels are on the verge of locking, calculate (i) the maximum deceleration, (ii) the braking force, (iii) the braking distance and time, (iv) the total stopping distance if the driver's reaction time is 1 s, and (v) the energy to be absorbed by the brakes, and the temperature rise of four discs of 6 kg each (specific heat 460 J/kg K) if all this energy is absorbed in them.

## Short answer

Maximum deceleration a = μg = 6.87 m/s², braking force F = ma = 8.24 kN. Braking distance s = u²/(2a) = 29.1 m, time t = u/a = 2.91 s; total stopping distance with 1 s reaction is 49.1 m. Kinetic energy to absorb is 240 kJ, causing a temperature rise of 21.7 K in four 6‑kg cast‑iron discs (c = 460 J/kg·K).

## Answer

**Given:** $m = 1200$ kg, $u = 72$ km/h $= 20$ m/s, $\mu = 0.7$, $g = 9.81$ m/s², $t_r = 1$ s.

#### (i) Maximum deceleration

$$
a = \mu g = 0.7 \times 9.81 = 6.87\ \text{m/s}^2
$$

#### (ii) Braking force

$$
F_b = m a = 1200 \times 6.867 = 8240\ \text{N} \quad (= \mu m g)
$$

#### (iii) Braking distance and time

$$
\begin{aligned} s_b &= \frac{u^2}{2a} = \frac{20^2}{2 \times 6.867} = 29.1\ \text{m} \\ t_b &= \frac{u}{a} = \frac{20}{6.867} = 2.91\ \text{s} \end{aligned}
$$

#### (iv) Total stopping distance with reaction time

Distance during reaction $= u\, t_r = 20 \times 1 = 20$ m.

$$
s_{total} = 20 + 29.1 = 49.1\ \text{m}
$$

#### (v) Energy and disc temperature rise

$$
E = \tfrac{1}{2} m u^2 = \tfrac{1}{2}(1200)(20)^2 = 240\,000\ \text{J} = 240\ \text{kJ}
$$

$$
\Delta T = \frac{E}{m_d c} = \frac{240\,000}{(4 \times 6) \times 460} = 21.7\ \text{K}
$$

In practice discs heat much more in repeated stops; the front brakes absorb 60 to 70 % of the energy.

**Answer:** $a = 6.87$ m/s²; $F_b = 8.24$ kN; $s_b = 29.1$ m in $2.91$ s; total stopping distance $= 49.1$ m; $E = 240$ kJ and $\Delta T \approx 21.7$ K.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With all wheels about to lock on a level road, which gives the maximum deceleration and braking force?
   A. a = mu/g = 0.07 m/s^2; F = 85 N
   B. a = g/mu = 14.0 m/s^2; F = 16.8 kN
   C. a = mu g = 6.87 m/s^2; F = mu m g = 8.24 kN
   D. a = mu g = 6.87 m/s^2; F = 1200 N
   Answer: C. Friction limits the force to mu m g = 0.7 x 1200 x 9.81 = 8240 N, giving a = mu g = 6.87 m/s^2. (Key point: Max deceleration and force)

2. What is the braking distance from 20 m/s?
   A. 58.2 m
   B. 49.1 m
   C. 29.1 m
   D. 20 m
   Answer: C. s = u^2/(2a) = 400/(2 x 6.867) = 29.1 m (braking time 2.91 s). (Key point: Braking distance)

3. How does the driver's 1 s reaction time change the stopping distance?
   A. Adds 10 m, giving 39.1 m
   B. Adds 2.91 m, giving 32 m
   C. No change
   D. Adds 20 m travelled at constant speed, giving 49.1 m
   Answer: D. During reaction the car moves u x t = 20 m at constant speed, so 20 + 29.1 = 49.1 m. (Key point: Total stopping distance)

4. How is the disc temperature rise found?
   A. Delta T = E/(m_disc c) with E = 1/2 m u^2 = 240 kJ and 24 kg of discs, giving 21.7 K
   B. Delta T = m u/(c), giving 52 K
   C. Delta T = E/(m c) using the 1200 kg car mass, giving 0.43 K
   D. Delta T = mu m g / c, giving 18 K
   Answer: A. All kinetic energy 240 kJ goes into 4 x 6 = 24 kg of disc: 240000/(24 x 460) = 21.7 K. (Key point: Energy and temperature rise)

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