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Constant acceleration

  1. Sep 16, 2012 #1
    1. A boulder initially at rest rolls down a hill with an acceleration of 3.69m/s^2. If it accelerates for 7.72 seconds, how far will it move?


    2. d=v0(initial velocity)t+.5at^2



    3. d=(0)(7.72)+.5(-9.8)(3.27)^2
    -39.24+-52.4
    -91.64m
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 16, 2012 #2

    HallsofIvy

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    Why is t= 7.72 in the first term and 3.27 in the second?

    Where did the "-39.24" come from? And why the "+-" for the second term?

     
  4. Sep 16, 2012 #3
    You are using the correct equation, but your values for 'a' and 't' are wrong.
    The acceleration of the ball is given to you in the question, so you don't need to use 9.81m/s². And also the time stated in the question is 7.72s.
     
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