A car moves at a constant speed v_0 = 12 m/s . When it…

A car moves at a constant speed $v_0 = 12\ \mathrm{m/s}$. When it brakes, it stops over a distance $s = 60\ \mathrm{m}$. What is the average acceleration of the car while braking?

Explanation of Method

Acceleration of a braking car.

Step 1: Choose the right equation

We know the initial speed, the final speed ($v = 0$) and the distance, but not the time. The equation that connects exactly these is

\[ v^2 = v_0^2 + 2 a s. \]

Step 2: Solve for $a$

\[ 0 = v_0^2 + 2 a s \;\Longrightarrow\; a = -\frac{v_0^2}{2s}, \]

\[ a = -\frac{(12\ \mathrm{m/s})^2}{2 \cdot 60\ \mathrm{m}} = -\frac{144}{120}\ \mathrm{m/s^2}. \]

\[a = -1.2\ \mathrm{m/s^2}\]

The minus sign means the acceleration points opposite to the motion: the car slows down by 1.2 m/s every second.

Check with the braking time

Time to stop: $ t = \dfrac{v_0}{|a|} = \dfrac{12}{1.2} = 10\ \mathrm{s} $

Average speed while braking is $ \tfrac{12 + 0}{2} = 6\ \mathrm{m/s} $, so $ s = 6 \cdot 10 = 60\ \mathrm{m}\ \checkmark $

$v\ (\mathrm{m/s})$ $t\ (\mathrm{s})$ $0$ $2$ $4$ $6$ $8$ $10$ $6$ $12$ $s = \tfrac{1}{2}\cdot 12 \cdot 10 = 60\ \mathrm{m}$ $\text{slope } a = -1.2\ \mathrm{m/s^2}$

Speed–time graph for uniform braking. The speed falls in a straight line from 12 m/s to 0; the area under the line is the braking distance, 60 m.

Result

\[\boxed{\,a = -1.2\ \mathrm{m/s^2}\,}\]