# A driver of a car traveling at 15.0 m/s applies the brakes, causing a uniform acceleration of -2.0 m/s^2. How long does it take the car to accelerate to a final speed of 10.0 m/s? How far has the car moved during the braking period?

Jul 20, 2015

Time needed: 2.5 s
Distance covered: 31.3 m

#### Explanation:

I'll start with the distance covered while decelerating. Since you know that the initial speed of the car is 15.0 m/s, and that its final speed must by 10.0 m/s, you can use the known acceleration to determine the distance covered by

${v}_{f}^{2} = {v}_{i}^{2} - 2 \cdot a \cdot d$

Isolate $d$ on one side of the equation and solve by plugging your values

$d = \frac{{v}_{i}^{2} - {v}_{f}^{2}}{2 a}$

d = ((15.0""^2 - 10.0""^2)"m"^cancel(2)cancel("s"^(-2)))/(2 * 2.0 cancel("m") cancel("s"^(-2))

$d = \textcolor{g r e e n}{\text{31.3 m}}$

To get the time needed to reach this speed, i.e. 10.0 m/s, you can use the following equation

${v}_{f} = {v}_{i} - a \cdot t$, which will get you

$t = \frac{{v}_{i} - {v}_{f}}{a}$

t = ((15.0 - 10.0)cancel("m")/cancel("s"))/(2.0cancel("m")/"s"^cancel(2)) = color(green)("2.5 s")

SIDE NOTE I left the value I got for the distance rounded to three sig figs.