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Acceleration around a running track

  1. Apr 26, 2016 #1
    1. The problem statement, all variables and given/known data
    He is running around a running track training for a 400m race. His velocity is 5.11m/s around the circular end of the track, which has a radius of 31.8m.
    I'm trying to find his acceleration which I know is 0.821ms-2 but I can't find out how to get it.

    2. Relevant equations
    v=rω
    d=rθ
    a=rα
    ω=θ/t
    α=ω/t
    ω=2πf
    ƒ=1/T

    3. The attempt at a solution
    I first calculated his angular velocity of 0.160rad/s. Then re-arranged d=rθ to θ=d/r and got 12.57rad. Then I re-arranged ω=θ/t to get t=ω/θ and got 0.0127s which I know must be wrong... anyways I then used α=ω/t and got 12.59rads-2 and finally a=rα and I got 400ms-2 and the answer is 0.821ms-2.
     
    Last edited by a moderator: Apr 26, 2016
  2. jcsd
  3. Apr 26, 2016 #2

    haruspex

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    The statement of the question is not quite right. The 5.11m/s is a speed, not a velocity. Velocity is a vector so also has a direction. The speed is just the magnitude of the velocity.
    Since the speed is constant, the acceleration consists of its changes in direction. What kind of acceleration do you associate with an object going around in circles? What equations do you know that relate to that?
     
  4. Apr 26, 2016 #3
    If it is asking to find the acceleration then I assume I will be using a=rα to convert the angular acceleration from α=ω/t to 'linear' (not sure if that's that right word) acceleration. The problem is I can't find the time or if the displacement of 400m is relevant. And yes I think it is meant to say speed but it says velocity.
     
  5. Apr 27, 2016 #4

    haruspex

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    No, that formula is for a as the tangential acceleration. Since speed is constant, there is no tangential acceleration, and so no angular acceleration.
    When speed is constant, the only possible acceleration is at right angles to the direction of travel. Have you heard of centripetal acceleration?
     
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