# Question #7609a

##### 1 Answer

Here's what I got.

#### Explanation:

The first thing to do here is to convert the two velocities of the car from *miles per hour* to *meters per second*

#80 color(red)(cancel(color(black)("mi")))/color(red)(cancel(color(black)("h"))) * (1color(red)(cancel(color(black)("h"))))/(60color(red)(cancel(color(black)("min")))) * (1color(red)(cancel(color(black)("min"))))/"60 s" * (1.61color(red)(cancel(color(black)("km"))))/(1color(red)(cancel(color(black)("mi")))) * (10^3"m")/(1color(red)(cancel(color(black)("km")))) = "35.8 m/s"#

#30 color(red)(cancel(color(black)("mi")))/color(red)(cancel(color(black)("h"))) * (1color(red)(cancel(color(black)("h"))))/(60color(red)(cancel(color(black)("min")))) * (1color(red)(cancel(color(black)("min"))))/"60 s" * (1.61color(red)(cancel(color(black)("km"))))/(1color(red)(cancel(color(black)("mi")))) * (10^3"m")/(1color(red)(cancel(color(black)("km")))) = "13.4 m/s"#

So, you know that it takes **decrease** from

As you know, **acceleration** is defined as the rate at which the velocity of an object changes with respect to time. In your case, it takes

#Deltav = |"13.4 m/s" - "35.8 m/s"| = "22.4 m/s"#

**SIDE NOTE** *Because you're looking for the magnitude of the acceleration, the change in velocity can be used without the minus sign that accompanies a decrease in velocity*.

This means that the magnitude of the acceleration will be

#a = (Deltav)/t = "22.4 m/s"/"3 s" = color(green)(bar(ul(|color(white)(a/a)color(black)("7.5 m/s"^2)color(white)(a/a)|)))#

I'll leave the answer rounded to two *sig figs*, but don't forget that you only have one sig fig for your values.

To calculate the distance covered by the car during its breaking time, use the equation

#color(purple)(bar(ul(|color(white)(a/a)color(black)(v^2 = v_0^2 - 2 * a * d)color(white)(a/a)|)))#

Here

**final velocity** of the car

**initial velocity** of the car

**acceleration**

**distance** covered

Rearrange the equation to solve for

#d = (v_0^2 - v^2)/(2 * a)#

Plug in your values to find

#d = (35.8^2"m"^color(red)(cancel(color(black)(2)))/color(red)(cancel(color(black)("s"^2))) - 13.4^2 "m"^color(red)(cancel(color(black)(2)))/color(red)(cancel(color(black)("s"^2))))/(2 * 7.5color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"^2)))) = color(green)(bar(ul(|color(white)(a/a)color(black)("73 m")color(white)(a/a)|)))#

Once again, I'll leave the answer rounded to two sig figs.