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

## Answer this Question

## Related Questions

- physics - A car initially traveling at 31.0 m/s undergoes a constant negative ...
- PHYSICS - The driver of a car traveling at 25.0 m/s applies the brakes and ...
- Physics - A car initially traveling at 33.4 m/s undergoes a constant negative ...
- Physics - A car initially traveling at 33.4 m/s undergoes a constant negative ...
- physic - Hi!Need help please. Question:A car initially traveling at 24.0 m/s ...
- calc - a) A car is traveling at 50 mi/h when the brakes are fully applied, ...
- college Physics - A car is traveling at 7.0 m/s when the driver applies the ...
- physics - A car initially traveling at 26 m/s hits the brakes, generating a ...
- physics - The brakes are applied to a car traveling on a dry, level highway. A ...
- physics - a)Two cars are traveling along a straight line in the same direction, ...