Stopping distance and brake energy

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.