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Kinematic equations

  1. Sep 23, 2007 #1

    ~christina~

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    Can they be used if the acceleration is = to zero??

    I know that they can only be used if the acceleration is constant but is zero considered a constant??


    Thanks =):confused:
     
  2. jcsd
  3. Sep 23, 2007 #2
    What do you think? Try it and see what happens.
     
  4. Sep 23, 2007 #3

    ~christina~

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    What do you mean try it?? I don't have a particular question...if I plug it into...

    xf= xi + V*xi*t +(1/2)*ax*t^2 I get
    if a=0

    xf= xi +v*xi*t+ 0

    Does this mean it is okay? I'm not sure what it would mean to say it "can be used".
     
  5. Sep 23, 2007 #4
    Well, does the equation make sense when you made a=0?
     
  6. Sep 23, 2007 #5

    ~christina~

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    :cry: I don't know...seriously..if it is okay or doesn't make sense...

    is it yes?
    reasoning: it is for the position as a function of time but even though you cancel out the t^2 with the zero there is still another t in the equation but this wouldn't make sense for the other kinematic equation:

    vxf^2= vxi^2 +2ax(xf-xi)

    since if you put in 0 for the ax then the only thing left is vxf^2= vxi^2
    I'm not sure if that would make sense...but if it had 0 acceleration the object could still be moving so I dont' think the initial position would be the same as the final in all cases.

    Help...
     
  7. Sep 23, 2007 #6
    I dont know....*you're* going to tell me if it makes sense or not.

    You are on the right track. Think some more and then post. I want you to develop the skill of thinking for yourself. (That means no one else post an answer for her, PLEASE :grumpy:).
     
  8. Sep 24, 2007 #7

    ~christina~

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    that response doesn't help at all, it doesn't give me any clue as to if my thoughts are correct or incorrect...all you tell me is to think...I did that before and posted what I thought....can't you give me a hint or something???

    P.S telling me to think more isn't helping me...
     
  9. Sep 24, 2007 #8

    Astronuc

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    I think cyrus is hoping that one would realize if acceleration, a = 0, then the problem is one of constant velocity.

    Acceleration is the change in velocity as a function of time.
     
  10. Sep 24, 2007 #9

    ~christina~

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    Thank you...Astronuc
    That helped alot ..now I can keep it in mind that it does make sense.
     
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