Stopping distance (reaction and braking)

Problem

A car travels at $50\;\mathrm{km/h}$ in a built-up area. The driver has a reaction time of $0{,}8\;\mathrm{s}$ and the car subsequently decelerates with $a = 6{,}0\;\mathrm{m/s}^2$. What is the total stopping distance (reaction + braking)?

Solution

Converting the velocity. Recalling that $1\;\mathrm{km/h} = \dfrac{1}{3{,}6}\;\mathrm{m/s}$:

$v_0 = \frac{50}{3{,}6} \approx 13{,}9\;\mathrm{m/s}$

The stopping distance is the sum of two contributions: the stretch covered during the reaction time (at constant velocity) and the actual braking stretch.

Reaction stretch. During the reaction time the driver has not yet acted on the brake, so the car advances at constant velocity:

$d_\text{reaz} = v_0 \, t_\text{reaz} = 13{,}9 \cdot 0{,}8 \approx 11{,}1\;\mathrm{m}$

Braking stretch. During braking the motion is uniformly decelerated until it stops ($v = 0$). From the time-independent relation $v^2 = v_0^2 - 2a\,d_\text{fren}$ with $v = 0$:

$d_\text{fren} = \frac{v_0^2}{2a} = \frac{13{,}9^2}{2 \cdot 6{,}0} = \frac{192{,}9}{12} \approx 16{,}1\;\mathrm{m}$

Total distance.

$d = d_\text{reaz} + d_\text{fren} \approx 11 + 16 = 27\;\mathrm{m}$

Notice how much the reaction time weighs: even before starting to brake the car has already covered over $11\;\mathrm{m}$.

$\ev{d \approx 27\;\mathrm{m} \;(\text{reaction } 11\;\mathrm{m} + \text{braking } 16\;\mathrm{m})}$

Links

Topics: Cinematica Concepts: Moto uniformemente accelerato