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B Anyone explaining 'position equation'?

  1. Jul 23, 2016 #1
    Anyone explaining 'position equation'?
     
  2. jcsd
  3. Jul 23, 2016 #2

    phinds

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    What form are you talking about? Have you done any research to find the answer?
     
  4. Jul 23, 2016 #3

    jim mcnamara

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    IF this is from a textbook or homework please give us the whole problem or discussion. We have no way to give you a good answer the way the question was question.
     
  5. Jul 23, 2016 #4
    I have no clue as to the 'significance' of the word in question. I googled for it but no use. What in the world is the 'position equation',
    aka 'position function'? Please bear with my ignorance, and enlighten me about it. tia. text
     
  6. Jul 23, 2016 #5

    phinds

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    Please re-read post #3. You are not giving enough information.
     
  7. Jul 23, 2016 #6

    jtbell

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    If you saw 'position equation' on a web page, it would help us a lot if you link to the page so we can see how it is used. If it's in a book, you can copy the paragraph that it appears in.

    Words and phrases can have different meanings depending on the context. It would not help you at all if we make a guess and it turns out your source means something different.
     
  8. Jul 23, 2016 #7
    http://tinyurl.com/hbyfu9x
    Actually, I failed to locate the exact location where I've found the citation
    with regard to the question I've raised. Instead, I cite the link above
    which will lead to where 's=16t2 + 100...', the same equation I've misplaced.
    My question is where 16t2 came from. All the best, text
    PS: Specifically, you may find the question in question at page 105, which will open before you.
     
  9. Jul 23, 2016 #8

    Ssnow

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    ##-16t^2## corresponds to the term ##-\frac{1}{2}gt^{2}## in the law for accelerate motion, here the constant ##g## is not the same of the usual ##9,81 m/s^{2}## but is another ...
     
  10. Jul 23, 2016 #9
    Newtons second law: [itex] z'' = F/m[/itex]. This gives the acceleration. Integrate twice to get "position equation".

    This gives you: [itex] z(t) = \frac{1}{2} \frac{F}{m} t^2 + z'(0)t +z(0)[/itex]

    Edit: This gives the same result as Ssnow, assuming F=-mg which is the case for gravitation.

    [itex] g \approx 32 ft/s^2 [/itex], so you can see this means the first term is -16. No t term means no initial velocity, and the last term of 100 is the initial height
     
  11. Jul 23, 2016 #10
    I missed the secondary education, although I have a college degree in Japanese studies. I am moving forward albeit at a snail pace
    to make up for the lost time. Somewhere and somehow I remember seeing the explanation kindly given to me, so that I had an inkling
    of the question being related to physics, which has been confirmed.
     
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