Braking distance at 50 km/h

Problem

Estimate the braking distance of a car travelling at $50\;\text{km/h}$ in the city, assuming a typical deceleration on dry asphalt ($a \approx 7\;\text{m/s}^2$). Add also the distance covered during the driver’s reaction time ($\sim 1\;\text{s}$). Is it consistent with what is taught in driving school?

Solution

Velocity conversion. Recalling that $1\;\text{km/h} = 1/3{.}6\;\text{m/s}$:

$v = \frac{50}{3{.}6} \approx 13{.}9\;\text{m/s}$

Actual braking distance. During braking the motion is uniformly decelerated; using $v^2 = v_0^2 - 2 a\, d_{\text{fren}}$ with zero final velocity:

$d_{\text{fren}} = \frac{v^2}{2a} = \frac{(13{.}9)^2}{2\cdot 7} = \frac{193}{14} \approx 13{.}8\;\text{m}$

$\ev{d_{\text{fren}} \approx 14\;\mathrm{m}}$

fully consistent with the estimate $d \sim 10$–$15\;\text{m}$.

Distance during the reaction time. During the reaction time $t_r \approx 1\;\text{s}$ the driver has not yet braked: the car proceeds at constant velocity $v$, covering

$d_{\text{reaz}} = v\, t_r = 13{.}9\cdot 1 \approx 13{.}9\;\text{m}$

Total stopping distance. Adding reaction and braking:

$\ev{d_{\text{arresto}} = d_{\text{reaz}} + d_{\text{fren}} \approx 14 + 14 = 28\;\mathrm{m}}$

Consistency with driving school. Yes. The stopping-distance tables teach that at $50\;\text{km/h}$ the total is about $25$–$30\;\text{m}$, roughly half due to the reaction time and half to braking: exactly the order of magnitude obtained here. Note that the braking distance grows with the square of the velocity, while the reaction distance grows only linearly.

Collegamenti

Argomenti: Kinematics Concetti: Uniformly accelerated motion Competenze: Fermi estimates