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Average Acceleration Confusion

  1. Sep 6, 2010 #1
    1. The problem statement, all variables and given/known data

    From t = 0 to t = 5.00 min, a man stands still, and from t = 5.00 min to t = 10.0 min, he walks briskly in a straight line at a constant speed of 2.5 m/s.

    What is his average acceleration aavg in the interval 2.00 min to 8.00 min?

    2. Relevant equations

    I know that in the formula the A_avg is ∆V/∆t... but I can't make it work for me. It says the answer is .00694 m/s^2, but I keep getting a different answer.

    3. The attempt at a solution

    I tried to calculate the final velocity (V_f): ∆d/∆t: [2.5 m/s x 180 s]/480s. (or 2.5 m/s x 3 minutes)/8 minutes). This give me 0.975 m/s^2.
    The initial velocity (V_i) seems to me that it would be 0 at t=2 minutes.
    So A = [.975-0]/∆t = [.975 m/s]/360s = .0027 m/s^2.

    but this is wrong.

    Help please. where did I mess up?
     
  2. jcsd
  3. Sep 6, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The final velocity is given--no need to calculate it! (You're calculating the average velocity, which is not needed.)
     
  4. Sep 6, 2010 #3
    !!!
    Thank you so much. I knew it was something silly I was messing up.
    :blush:
     
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