physics

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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 -

    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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