During a very quick stop, a car decelerates at 6.8 m/s^2. assume the forward motion of the car corresponds to a positive direction for the rotation of the tires (and that they do not slip on the ) what is the angular acceleration of its 0.280-m-radius tires, assuming they do not slip on the pavement? (b) how many revolutions do the tires make before coming to rest, given their initial angular velocity is 95.0 rad/s ? (c) how long does the car take to stop completely? (d) what distance does the car travel in this time?

-24.28571 rad/s²

29.57239 revolutions

3.91176 seconds

52.026478 m

Explanation:

= Tangential acceleration = -6.8 m/s²

r = Radius of wheel = 0.28

= Initial angular velocity = 95 rad/s

= Angle of rotation

= Final angular velocity

t = Time taken

Angular acceleration is given by

The angular acceleration is -24.28571 rad/s²

The number of revolutions is 29.57239

The time it takes for the car to stop is 3.91176 seconds

Linear distance

The distance the car travels is 52.026478 m

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