Question #7609a

1 Answer
Sep 19, 2016

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 #"3 s"# for the velocity of the car to decrease from #"35.8 m/s"# to #"13.4 m/s"#.

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 #"3 s"# for the velocity of the car to decrease by

#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

#v# - the final velocity of the car
#v_0# - the initial velocity of the car
#a# - the acceleration
#d# - the distance covered

Rearrange the equation to solve for #d#

#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.