Worked exercise — Collision avoidance

Problem

Two cars travel along the same straight road in opposite directions: the first at $80\;\text{km/h}$, the second at $100\;\text{km/h}$. The initial distance between them is $200\;\text{m}$. Both drivers react after $1{,}0\;\text{s}$ and then brake with a deceleration of $6\;\text{m/s}^2$. Is the collision avoided?

Solution

Speed conversion. With $1\;\text{km/h} = 1/3{,}6\;\text{m/s}$:

$v_1 = \frac{80}{3{,}6} \approx 22{,}2\;\text{m/s}, \qquad v_2 = \frac{100}{3{,}6} \approx 27{,}8\;\text{m/s}$

Since they move towards each other, the relative closing speed is the sum: $v_\text{rel} = 22{,}2 + 27{,}8 = 50\;\text{m/s}$.

Reaction phase ($1\;\text{s}$). During this second no one is braking yet: the two cars keep a constant speed and close the gap by

$\Delta s_\text{reaz} = v_\text{rel}\cdot t = 50\cdot 1 = 50\;\text{m}$

This leaves $200 - 50 = 150\;\text{m}$ of margin.

Braking phase. Each car, braking with deceleration $a = 6\;\text{m/s}^2$, uses up a braking distance $d = v^2/(2a)$:

$d_1 = \frac{22{,}2^2}{12} \approx 41{,}1\;\text{m}, \qquad d_2 = \frac{27{,}8^2}{12} \approx 64{,}4\;\text{m}$

Total distance used up by the two braking manoeuvres: $d_1 + d_2 \approx 105{,}5\;\text{m}$.

Comparison. Since $105{,}5\;\text{m} < 150\;\text{m}$, the cars stop before colliding:

$\ev{\text{urto evitato, margine residuo} \approx 45\;\mathrm{m}}$

The $50\;\text{m}$ lost during the second of reaction time are the price paid for human response time: with instantaneous braking the margin would have been even greater.

Links

Topics: Cinematica Concepts: Moto uniformemente accelerato Skills: Scelta degli assi