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Homework Help: Constant slope acceleration problem

  1. Sep 29, 2005 #1
    I got this problem:

    A ball is rolled up a constant slope with an initial velocity of 11.0 m/s and after 9.3 s the ball is rolling down the slope with a velocity of 7.3 m/s. What is the acceleration of the ball on the slope.

    Here's what I tried: I divided the problem in 2 for the ball going up, and the ball going down.

    1st Givens: Vi: 11 m/s Vf: 0 t:9.3
    2nd Givens: Vi: 7.3 m/s Vf :0

    I then got the displacement of half of the slope from the first set of givens, which gave me 51.15. That's where I got stuck. I tried getting a acceleration for both sides of the slope and then averaging, but that wasn't right.

    Please help
     
  2. jcsd
  3. Sep 29, 2005 #2

    kreil

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    Gold Member

    [tex]a=\frac{{\Delta}v}{t} \implies a=\frac{v_f-v_i}{t}[/tex]

    **hint** velocity has direction, so when the ball is rolling down the slope, it has a negative velocity.

    .
     
  4. Sep 29, 2005 #3
    Ah yes, but there is still a problem. This is what changes then:
    2nd Givens: Vi: -7.3 m/s D:-51.15 Vf: 0(I'm not sure if this is right)
    So I can get a time for when it's going down of 14 sec.
    So I added the inicial velocities of the ball going up and going down and then divided it by 23.3(9.3 sec + 14 sec). But it gave a large decimal.

    The teacher gave us the answer of -2.0 m/s^2.

    Damn, I'm still stuck. Please advise.
     
  5. Sep 29, 2005 #4

    kreil

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    Gold Member

    The problem tells you that it starts at 11m/s, and 9.3s later it is going down at -7.3 m/s. just plug this information in to the equation:

    [tex]a=\frac{-7.3m/s-11m/s}{9.3s}=\frac{-18.3}{9.3}=-1.96m/{s^2}[/tex]
     
  6. Sep 29, 2005 #5
    Wow, I just complicating myself with all my equations. Thank you so much for your help
     
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