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23 February, 16:20

A car accelerates uniformly from rest and reaches a speed of 21.5 m/s in 11.4 s. The diameter of a tire is 66.5 cm. Find the number of revolutions the tire makes during this motion, assuming no slipping. Answer in units of rev.

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Answers (2)
  1. 23 February, 17:23
    0
    57.39 rev

    Explanation:

    From circular motion,

    s = rθ ... Equation 1

    Where s = distance, r = radius, θ = angular distance.

    make θ the subject of the equation

    θ = s/r ... Equation 2

    Where can look for s using any of the equation of motion

    s = (v+u) t/2 ... Equation 3

    Where v and u = Final and initial velocity respectively, t = time.

    Given: v = 21.5 m/s, u = 0 m/s (at rest), t = 11.4 s

    Substitute into equation 3

    s = (21.5+0) 11.4/2

    s = 122.55 m.

    given: r = 66.5/2 = 33.25 cm = 0.3325 m

    Substitute into equation 2

    θ = 122.55/0.3325

    θ = 368.57 rad

    θ = (360.57*0.159155) rev

    θ = 57.39 rev
  2. 23 February, 17:34
    0
    58.6886 revolutions

    Explanation:

    First we need to know the total distance travelled by the car, and we can do that using Torricelli formula:

    V2 = Vo2 + 2aDS

    V = 21.5

    Vo = 0

    a = 21.5/11.4 = 1.886

    (21.5) ^2 = 2*1.886*DS

    DS = 462.25/3.772 = 122.5477 m

    For each revolution of the tire, the car moves the circunference of the tire, which is pi*d = 3.14*66.5 = 208.81 cm = 2.0881 m

    So, to know the number of revolutions, we divide the total travel distance by the circunference of the tire:

    122.5477/2.0881 = 58.6886
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