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Laws of motion magnitude of average acceleration

  1. Sep 30, 2008 #1
    1. The problem statement, all variables and given/known data
    A rock is rolled in the sand. It starts at 5.0 m/s, moves in a straight line for a distance of 3.0m and then stops. What is the magnitude of the average acceleration?

    2. Relevant equations
    aavg = change in V/ change in time
    magnitude of acceleration = change of speed/time interval.

    3. The attempt at a solution
    i think i am thinking too hard about this problem. seems simple but cannot get the right answer
    found time = 1.67 s and tried to plug in for average acceleration. 5.0/1.67 = 2.99 m/s^2
    the correct answer is 4.2 m/s^2.
    ?? can't figure out what i am doing wrong.
     
  2. jcsd
  3. Sep 30, 2008 #2

    Hootenanny

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    Welcome to Physics Forums,

    Tell me, how did you find the time?
     
  4. Sep 30, 2008 #3
    thanks =)

    i took 5.0 m/s and divided by the distance of 3.0 m to get 1.67 seconds.
    so the ball traveled 3.0 m in 1.67 s at 5.0 m/s.
     
  5. Sep 30, 2008 #4

    Hootenanny

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    I'm afraid that's incorrect, it would be correct if 5.0m/s was the average velocity, but it isn't it's the initial velocity.

    Do you know any kinematics equations?
     
  6. Sep 30, 2008 #5
    ah, yes.. i overlooked that.

    well for average velocity its change in displacement over change in time, i think.
     
  7. Sep 30, 2008 #6

    Hootenanny

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    Whilst that is correct, I think that you may find some more useful formulae on this page.
     
  8. Sep 30, 2008 #7
    wow that was a lot easier than i made it out to be. thanks for the help.
     
  9. Sep 30, 2008 #8

    Hootenanny

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    My pleasure.
     
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