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What is the speed of the ball 10.0 sec later?

  1. Feb 17, 2007 #1
    INTRODUCTION:
    This is a problem from my Introduction to Physical Science class using "Conceptual Physics" 10th Ed.by Paul G. Hewitt

    EXACT PROBLEM:
    "A ball rolls down an inclined plane with a uniform acceleration of 1.00 m/sec^2. At a certain instant, its speed is 0.5 m/sec^2."

    PROBLEMS FACED:
    a) What is the speed of the ball 10.0 sec later?
    b) How far has the ball traveled in that 10.0 sec time interval?
    c) How far did the ball travel during the eighth second after it was released?

    MY THOUGHTS:
    I don't know what to do here really. I've had phyisics before, but that was for years ago (so much for saving that easy intro science class for senior year:rolleyes:). I am pretty much at a loss here. Any help for a lazy, "I just wanna graduate" senior would be of great help.
     
    Last edited: Feb 17, 2007
  2. jcsd
  3. Feb 17, 2007 #2

    hage567

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    Start by looking up the basic kinematic equations of linear motion in your book. Write out the things you know and the things you need to find. Find the equation that fits them together. Drawing a picture and labelling the points with the things you know can help.
     
  4. Feb 17, 2007 #3
    That doesn't help me at all. I take it I'm looking for something dealing with the equations for speed, average speed, acceleration. I honestly don't know.
     
    Last edited: Feb 17, 2007
  5. Feb 17, 2007 #4

    hage567

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  6. Feb 17, 2007 #5
    I know the equations, I don't know how to use them. I don't know the inputs.:frown:
     
  7. Feb 17, 2007 #6

    hage567

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    Does your book not talk about them? I'm not trying to be difficult, but I find it hard to believe there aren't any examples of how to use these equations in your textbook.

    To give you a hint for a):

    A ball is travelling down an incline, with an acceleration (a) of 1 m/s^2. At some time, it has a velocity ( call it Vo) of 0.5 m/s. At t = 10s after it had the velocity of 0.5 m/s, you want to know what the new velocity (V) is. So you know Vo, a, t, and want to find V. Which one of those equations listed in the formulary do you think you could use?
     
  8. Feb 17, 2007 #7
    ok, you answered my biggest problem. 0.5 is the starting point.
     
  9. Feb 17, 2007 #8
    OK, so here's my crack at part a:

    v = Vo + at
    v = .5 + (1.0)(10)
    v = .5 + 10
    v = 10.5 m/s

    part b:

    d = vt
    d = (10.5)(10)
    d = 105 m

    part c:

    v = Vo + at
    v = .5 + (1.0)(8)
    v = .5 + 8
    v = 8.5 m/s

    d = vt
    d = (8.5)(8)
    d = 68 m

    now was I correct in having to redo the velocity for part c?
     
    Last edited: Feb 17, 2007
  10. Feb 17, 2007 #9
    Since the ball is rolling wouldn't we have to use kinematic equations for uniformly accelerated rotational motion rather than linear motion (since the ball isn't sliding down the plane). So in terms of x, v, and a it's just a small gripe but use theta, omega and alpha.
     
    Last edited: Feb 17, 2007
  11. Feb 17, 2007 #10
    Another stab at part b:

    x = x0 + v0t + (1/2)at^2
    x = 0 + (.5 x 10) + .5(1)(10^2)
    x = 5 + 50
    x = 55m
     
  12. Feb 17, 2007 #11

    hage567

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    Here's some pointers on your answers:

     
  13. Feb 17, 2007 #12
    I fixed the units, and I solved part b again.

    a friend and I were wondering the same thing about whether it was AT 8 seconds or How Far it went from 8 to 9 seconds.
     
  14. Feb 17, 2007 #13

    hage567

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    Since it says during the eighth second, I would take it to be from t=7s to t=8s. If you count the interval between t=0s and t=1s as the first second. That's my opinion anyway.
     
  15. Feb 17, 2007 #14
    Everything redone. Final Answer, Regis:wink:

    part a:

    v = Vo + at
    v = .5 + (1.0)(10)
    v = .5 + 10
    v = 10.5 m/s

    part b:

    x = x0 + v0t + (1/2)at^2
    x = 0 + (.5 x 10) + .5(1)(10^2)
    x = 5 + 50
    x = 55m

    part c:

    v = Vo + at
    v = 0 + (1.0)(8)
    v = 0 + 8
    v = 8 m/s

    x = x0 + v0t + (1/2)at^2
    x = 0 + (8x1) + (1/2)(1.00m/s^2)(1s)^2
    x = 8 + .5
    x = 8.5m
     
    Last edited: Feb 17, 2007
  16. Feb 17, 2007 #15

    hage567

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    a) and b) look good. For c): If you are going to take the eighth second to be from t=8s to t=9s, then I agree with your answer is correct based on that.
     
  17. Feb 17, 2007 #16
    yeah, the problem is poorly worded if you ask me. I figure with the italics, it's stressing that point, so that's what I'm going with. I'll have to talk to my professor about it on Monday. Thanks for the help so far, as I have two more problems I have no idea about.
     
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