physics
posted by juana on .
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?

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.