Asked by juana

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?

Answers

Answered by Henry
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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