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11 August, 21:09

A car accelerates uniformly from rest and reaches a speed of 22.0 m/s in 9.00 s. Assuming the diameter of a tire is 58.0 cm, (a) find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. (b) What is the final angular speed of a tire in revolutions per second?

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  1. 12 August, 00:16
    0
    Given:

    Initial velocity, u = 0 m/s (at rest)

    Final velocity, v = 22 m/s

    Time, t = 9 s

    Diameter, d = 58 cm

    Radius, r = 0.29 m

    Using equation of motion,

    v = u + at

    a = (22 - 0) / 9

    = 2.44 m/s^2

    v^2 = u^2 + 2a * S

    S = (22^2 - 0^2) / 2 * 2.44

    = 99.02 m

    S = r * theta

    Theta = 99.02/0.29

    = 341.44 °

    1 rev = 360°

    341.44°,

    = 341.44/360

    = 0.948 rev

    = 0.95 rev

    B.

    Final angular speed, wf = v/r

    = 22/0.29

    = 75.86 rad/s
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