Question
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?
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
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.
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.