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Accelerating Ball

  1. Mar 13, 2009 #1
    1. The problem statement, all variables and given/known data
    A ball initially at rest and constantly accelerating covers a distance of 10 metres during its third second of motion. What is its acceleration?


    2. Relevant equations
    No idea which equations I need. I'm guessing I need
    a = (v-u)/t

    3. The attempt at a solution
    I've got no idea how to even apporach this. I'm thinking:
    s = 3 s
    u = 0 m/s
    v =
    a =
    t =
     
  2. jcsd
  3. Mar 13, 2009 #2

    tiny-tim

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    Homework Helper

    Welcome to PF!

    Hi ride4life! Welcome to PF! :smile:

    Hint: call the acceleration a, and find the distances after 2 seconds and after 3 seconds, and find the value of a that makes their difference equal 10 metres :wink:
     
  4. Mar 13, 2009 #3
    What formula should I use to figure the distanes covered in each second?

    0 - 1 = ?
    1 - 2 = ?
    2 - 3 = 10m = 10m/s/s

    I was thinking in the 1st second it travelled 2.5m and in the 2nd second it travelled 5m. I think that turns out to be 2.5m/s/s. That's just a guess. That's all I can figure out. :(
     
  5. Mar 13, 2009 #4

    rl.bhat

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    You can use, displacement s = ut +1/2*a*t^2.
    Find s for 2s and 3s. Then find the difference which is equal to 10m.
     
  6. Mar 13, 2009 #5
    So for
    2 - 3 = 10m

    s = 10m
    u = ?
    a = ?
    t = 1s
     
  7. Mar 13, 2009 #6

    rl.bhat

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    Ball starts from rest. So u = 0.
    S2 = 1/2*a*(2)^2........(1)
    S3 = 1/2*a*(3)^2........(2)
    Find S3 - S2, equate it to 10m, and solve for a.
     
  8. Mar 13, 2009 #7
    Thanks for the help, I got 4m/s/s which is the right answer. :D
     
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