# On a dry road, a car with good tires may be able to brake with a constant deceleration of 5.74 m/s^2. (a) How long does such a car, initially traveling at 29.3 m/s, take to stop? (b) How far does it travel in this time ?

Sep 9, 2015

Time needed to stop: $\text{5.10 s}$
Distance covered: $\text{74.8 m}$

#### Explanation:

This time, you know that at $t = 0$ the car has a speed of $\text{29.3 m/s}$.

At this moment, it starts slowing down with a constant deceleration of ${\text{5.74 m/s}}^{2}$. After covering a distance $d$, the car comes to a complete stop.

This means that you can write

${\underbrace{{v}^{2}}}_{\textcolor{b l u e}{= 0}} = {v}_{0}^{2} - 2 \cdot a \cdot d$

The negative sign is there to show that the acceleration is oriented in the opposite direction to the direction of movement, which is what you would expect for a car that's slowing down.

So, you have

$d = {v}_{0}^{2} / \left(2 \cdot a\right) = \left(29.3 \text{^2"m"^color(red)(cancel(color(black)(2)))/color(red)(cancel(color(black)("s"^2))))/(2 * 5.74color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"^2)))) = color(green)("74.8 m}\right)$

The time needed for the car to come to a complete stop is

${\underbrace{v}}_{\textcolor{b l u e}{= 0}} = {v}_{0} - a \cdot t$

t = v_0/a = (29.3color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"))))/(5.74color(red)(cancel(color(black)("m")))/"s"^color(red)(cancel(color(black)(2)))) = color(green)("5.10 m/s")