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Checking a relation in S and acceleration

  1. May 13, 2015 #1
    1. The problem statement, all variables and given/known data

    If S2 = at2+ 2bt+c, then the acceleration is
    A) Directly proportional to S
    B) Inversely proportional to S
    C) Directly proportional to S2
    D) inversely proportional to S3
    2. Relevant equations
    dS/dt = v
    dv/dt = A

    3. The attempt at a solution
    Differentiating both sides,
    2Sv = 2at + 2b
    Then again differentiating,
    2SA + 2v2 = 2a
    So,
    S = (a-v2)/A
    We can say that B option is correct or not?
    as there is also a velocity term which is not constant and can be written as differential of S.
     
  2. jcsd
  3. May 13, 2015 #2

    haruspex

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    As you suspect, the varying v is a problem. Your two stages of differentiation yield two equations with three variables, A, v and S. Use an equation to eliminate v.
     
  4. May 13, 2015 #3
    Which equation should I use?
    If I use the first equation which I have differentiated the term t will come which is also not constant.
     
  5. May 13, 2015 #4

    haruspex

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    You also have the undifferentiated initial equation.
     
  6. May 14, 2015 #5
    But from it I am only getting,
    ## S = \sqrt{at^2+2bt+c} ##
    What is the use of this?
     
  7. May 14, 2015 #6

    haruspex

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    You can use it to eliminate t, leaving an equation with only accn, S and constants. It may be messy.
     
  8. May 14, 2015 #7
    But the equation is not linear in terms of t, it's quadratic, we will not get a definite value for t.
     
  9. May 14, 2015 #8

    haruspex

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    I think you'll find there's some cancellation. It works out ok, just do it.
     
  10. May 14, 2015 #9

    ehild

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    You can choose the root which can be positive if you are worried about the two roots. And you can do the same with the other root.
     
    Last edited: May 14, 2015
  11. May 14, 2015 #10
    So getting, $$ t = \frac{-b+ \sqrt{b^2-ac+S^2a}}{a} $$
    Now what should I do?
    Substituting for v will get me a t term which is not constant.
     
  12. May 14, 2015 #11

    ehild

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    You got that v=(at+b)/S Substitute t here, to get v in terms of S.
     
  13. May 14, 2015 #12
    Hmm okay, there were lot of cancellations
    Finally getting
    ## S = \frac{ac-b^2}{A} ##
    So S is inversely proportional to A, means v is constant judging from the second differential of my attempt.
     
  14. May 14, 2015 #13

    ehild

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    I do not think it is correct. Check.
    And remember, you need the acceleration in terms of S.
     
  15. May 14, 2015 #14
    Oh yes, thanks
    I did not wrote the denominator S2
    So finally S3 is inversely proportional to A.
     
  16. May 14, 2015 #15

    ehild

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    So the acceleration is inverse
    The question is A, in terms of S. Formulate your answer this way.
     
  17. May 14, 2015 #16
    I know that both are same
    If acceleration is inversely proportional to S3,
    then S3 is also inversely proportional to acceleration.
     
  18. May 14, 2015 #17

    ehild

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    Yes, but you should answer the question. The two sentences are not the same. And you have to choose among the given answers. Is one of them identical with "S3 is inversely proportional to A"?
     
  19. May 14, 2015 #18
    Yes the 4th or D option is correct.
    Are you asking to me for a formula so others learn from this thread of A in terms of S?
     
  20. May 14, 2015 #19

    ehild

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    It would be nice :oldsmile:
     
  21. May 14, 2015 #20
    So should I write also how I differentiated in my attempt.
    What is product rule of differentiation, what is the meaning of differentiation,
    how I got the value of t, how we find quadratic roots etc.?
    In general if one wants to learn in this thread he/she should ask questions as we don't know what they are expecting.
    Also identical has identical or same meaning as same
    See
    https://www.google.co.in/#q=identical+meaning
    Would you say 3*1 is not same as 9/3 but they are identical?
     
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