Posted by **juana** on Thursday, May 26, 2011 at 11:41am.

The tires of a car make 58 revolutions as the car reduces its speed uniformly from 92 to 33 . The tires have a diameter of 0.86 .

What was the angular acceleration of the tires?

If the car continues to decelerate at this rate, how much more time is required for it to stop?

- physics -
**Henry**, Friday, May 27, 2011 at 9:36pm
1.C = pi*D = 3.14*0.86 = 2.7Ft.

d = 58rev * 2.7Ft/rev = 156.6Ft.

Vo = 92mi/h * 5280Ft/mi *(1h/3600s)=134.9Ft/s. = Inital velocity.

Vf = 33mi/h * 5280Ft/mi * (1h/3600s)=48.4Ft/s. = Final velocity.

To = d/Vo = 156.6Ft / 134.9Ft/s = 1.16s.

Tf = 156.6Ft / 48.4Ft/s = 3.23s.

a = (Vf-Vo)/(Tf-To)

a = (48.4-134.9) / (3.23-1.16) = -41.8Ft/s^2, Linear.

a=-41.8Ft/s^2 * 6.28rad/2.7Ft=

-97.2rad/s^2, Angular.

2. Vf = Vo + at = 0.

48.4 -97.2t = 0,

t = 0.50s.

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