Posted by **nay** on Monday, February 23, 2009 at 8:41pm.

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**, Monday, February 23, 2009 at 9:01pm
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**, Tuesday, January 10, 2012 at 7:47am
119.37

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