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Average acceleration from average velocity?

  1. May 7, 2013 #1
    1. The problem statement, all variables and given/known data

    A particle travels from point A to point B following a semicircular path (a half-circle) of radius 5 m, and it travels at a constant speed of 1 m/s. Find the average acceleration of the particle.


    2. Relevant equations

    Average acceleration = change in velocity/ change in time

    Total time = distance / speed

    3. The attempt at a solution

    I already know the answer, and I understand why it's true:

    v0 = 1 m/s
    vf = -1 m/s

    total time = (5 pi m) / (1 m/s) = 5pi seconds

    Avg acceleration = (2 / (5pi) ) m/s^2

    My question is: Is the average acceleration ever equal to the (average velocity/ time) ? In this case, I know it's not because:

    Avg velocity = displacement / time = 10 m / 5pi sec = (2/ pi) m/s

    And, if I divided by the time, I would get a factor of pi^2 in the denominator, which is not in my original answer.

    I would really appreciate any help sorting this out! :smile:
     
  2. jcsd
  3. May 7, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Consider a 1-dimensional motion, ##v(t)=v_0 + at## with constant a, from t=0 to T. The average acceleration is a, the average velocity is v0+aT/2. As you are free to choose v0, you can get your equality.

    Particles at rest have zero average velocity and acceleration, they are a special case of my example for a=0.
     
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