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 = (VfVo)/(TfTo)
a = (48.4134.9) / (3.231.16) = 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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