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?
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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.