Posted by nay on .
a car traveling at 30.0m/s undergoes a constant negative accerleration of magnitude of 2.00m/s^2 when the brakes are applied. how many revolutions does each tire make before the car comes to a complete stop,assuming thaat the car does not skid and that the tires have radii of 0.300m?

physics 
drwls,
First find out how far it travels while decelerating. Call that distance X. The divide X by 2 pi R to get the number of tire rotations.
The average speed while decelerating is 15.0 m/s. The time required to decelerate to zero speed is (30.0 m/s)/2 m/s^2 = 15 s
X = (average speed) * t = 225 m
Finally, divide that by 2 pi R. 
physics 
bob,
119.37