Question

Approximately, the stopping distance of a car travelling at v km/h is given by the expression 5/24(v +(v^2)/32)m (v is measured in km/h), the 1st term representing the `reaction' distance of the driver and the second the distance through which the brakes are actually applied. Two cars are 35 m apart
when the 1st car, travelling at 45 km/h, makes an emergency stop. The driver of the second car has no prior warning of this hazard. Show that a collision must occur if the speed of a second car exceeds 71 km/h.

Answer 1

Approximately, the stopping distance of a car travelling at $\mathrm{v} \, \mathrm{km/h}$ is given by the expression $5/24(\mathrm{v} + (\mathrm{v}^2)/32) \, \mathrm{m}$ (v is measured in $\mathrm{km/h}$), the 1st term representing the 'reaction' distance of the driver and the second the distance through which the brakes are actually applied. Two cars are $35 \, \mathrm{m}$ apart when the 1st car, travelling at $45 \, \mathrm{km/h}$, makes an emergency stop. The driver of the second car has no prior warning of this hazard. Show that a collision must occur if the speed of a second car exceeds $71 \, \mathrm{km/h}$.

Solution

the stopping distance of the 1st car

$$S = \frac{5}{24} \left(45 + \frac{45^2}{32}\right) = 22,5 \, \mathrm{m}$$

on the other side

$$S = vt - \frac{at^2}{2}; \, v = at \gg S = vt - \frac{vt}{2} = \frac{vt}{2} \gg t = \frac{2S}{v}$$

time of stopping

$$t = \frac{2 \cdot 22,5 \cdot 3.6}{45} = 3.6 \, \mathrm{s}$$

So if the speed of a second car exceeds $71 \, \mathrm{km/h}$

$$S_2 = Vt = \frac{71}{3.6} \cdot 3.6 = 71 \, \mathrm{m}$$

total distance of two cars for $3.6 \, \mathrm{s}$

$$S = S + S_2 = 22,5 + 71 = 93,5 \, \mathrm{m} > 35 \, \mathrm{m}$$

So a collision must occur if the speed of a second car exceeds $71 \, \mathrm{km/h}$.