Braking distance at 90 km/h

Problem

Car braking. A car at $90\;\text{km/h}$ brakes with a constant deceleration of $5\;\text{m/s}^2$. What distance does it cover before stopping?

Solution

Let us convert the initial speed into metres per second, recalling that $1\;\text{km/h} = \tfrac{1}{3{,}6}\;\text{m/s}$:

$v_0 = \frac{90}{3{,}6} = 25\;\mathrm{m/s}$

Braking is a uniformly decelerated motion: the acceleration is $a = -5\;\mathrm{m/s^2}$ (opposite to the motion) and the final speed is $v = 0$. Let us use the relation linking speed and distance without time:

$v^2 = v_0^2 + 2\,a\,s \quad\Longrightarrow\quad 0 = v_0^2 - 2 \cdot 5 \cdot s$

Solving for the stopping distance:

$s = \frac{v_0^2}{2 \cdot 5} = \frac{25^2}{10} = \frac{625}{10} = 62{,}5\;\mathrm{m}$

Even on good tarmac, from $90\;\text{km/h}$ it takes over $60$ metres to stop: seatbelts and safe following distances exist for exactly this reason.

$\ev{s = 62{,}5\;\mathrm{m}}$

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Topics: Cinematica Concepts: Moto uniformemente accelerato