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A motorcyclist is traveling along a road and accelerates for 4.84 s to pass another cyclist. The angular acceleration of each wheel is 6.76 rad/s2, and, just after passing, the angular velocity of each is 77.6 rad/s, where the plus signs indicate counterclockwise directions. What is the angular displacement of each wheel during this time?

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Answer:

296 rad

Explanation:

time (t) = 4.84 s

angular acceleration (a) = 6.76 rad/s^{2}

angular velocity (ω) = 77.6 rad/s

what is the angular displacement (θ) of the wheel during this time?

from the equation of angular kinematics angular displacement (θ) = ω₀t + [tex]\frac{1}{2}at^{2}[/tex]

where

  • a is angular acceleration = 6.76
  • t is time = 4.84
  • ω₀ is initial angular acceleration and can be gotten from the formula below

        ω = ω₀ + at

        77.6 = ω₀ + (6.76 x 4.84)

        ω₀ = 77.6 - (6.76 x 4.84) = 44.88 rad/s

  • now we can substitute all required values into the formula for angular displacement

angular displacement (θ) =ω₀t +  [tex]\frac{1}{2}at^{2}[/tex]

angular displacement (θ) =[tex] (44.88 x 4.84) + \frac{1}{2} x 6.76 x 4.84^{2}[/tex]

angular displacement (θ) = 296 rad

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