1. Oct 26, 2008

### xregina12

The driver of a car traveling at 12.0 m/s applies the brakes and undergoes a constnant decelertion of 1.90m/s^2.
How many revolutions does each tire make before the car comes to a stop assuming that the car does not skid and that the tires of radii of 0.40 m? answer in units of rev.

0=144+2(-1.90)(d)
d=37.89meters
d=r(θ)
revolutions=94.725/(2pi)=15.076 revolutions.
However, I did not get a correct answer. Can anyone help? Did I do this the right way?

2. Oct 26, 2008

### R.Harmon

You've worked out the distance it takes to stop correctly, but I don't follow your working after that. You know the radius of the tyre is 0.4m, so it's circumference = 2Pi*0.4=2.5137. The car stops in a distance of 37.89 meters, so the number of revolutions is just the distance it takes to stop divided by the circumference of the tyre. Hope that helps.
EDIT: Actually looking at that method I get the same answer as you, so I'm not sure why that's not the correct answer. If this is from an online homework site are you entering the correct number of decimals?

Last edited: Oct 26, 2008