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Homework Help: Determine the acceleration after 6 seconds.

  1. Jul 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Determine the acceleration after 6 seconds.

    2. Relevant equations

    s(t) = sqrt12t - 8 where s is in metres and t is in seconds.

    3. The attempt at a solution


    this is where i don't understand...they got a positive value for 36 in the book and i got a negative value...here is my work below that got me to a negative value...what did i do wrong?

    a=v'(t)=0*sqrt12t-8 - 6/sqrt12t-8*6/(sqrt12t-8)^2
    = - 36/sqrt12t-8/(sqrt12t-8)^2
    = - 36/(12t-8)^3
  2. jcsd
  3. Jul 18, 2011 #2
    Hard to tell what your original equation is..



    It's kind of hard for me to see what any of your work is too.
    Last edited: Jul 18, 2011
  4. Jul 18, 2011 #3

    Ray Vickson

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    Do you mean s = sqrt(12t - 8) or s = sqrt(12t) - 8? Use brackets, or else typeset your material in TeX.

    If you mean s = sqrt(12t - 8), then v = ds/dt = 6/sqrt(12t - 8). If you mean s = sqrt(12t) - 8,
    then v = (1/2)sqrt(12/t).

  5. Jul 18, 2011 #4
    From the work done already, it looks to be the first option.

    Could you explicitly state the entire question, also? Often times with acceleration/velocity problems the manner in which the problem is set up/interpreted can account for the sign differences that using strictly mathematics doesn't show.

    And as mentioned, your work is very confusing. But, from what I can see, are you trying to use the quotient rule in getting the acceleration? What you need to use is the chain rule (which I am assuming you did for the correct velocity function) after moving the square root expression in the denominator up. Remember your power/exponent rules.
  6. Jul 18, 2011 #5
    Sorry, the full question is: Given the position function s(t) = sqrt12t-8 where s is in metres and t is in seconds. Determine the acceleration after 6 seconds.
  7. Jul 18, 2011 #6
    First i did this:

  8. Jul 18, 2011 #7
    therefore v(t)=6/sqrt12t-8
  9. Jul 18, 2011 #8
    a = v'(t)=d(6)/dt*sqrt12t-8 - d(sqrt12t-8)/dt*6/(sqrt12t - 8)^2
  10. Jul 18, 2011 #9
    Hrm, I worked the problem and I also get a negative value. I also checked the graph and by eye it looks like it should be just barely negative.

    Are you sure your initial function isn't a velocity function (and thus you'only derive once, that gives a positive value.)
  11. Jul 18, 2011 #10
    Ya i am sure, but it could be that the answer in this book is wrong? The initial function is(according to the book) a position function.
  12. Jul 18, 2011 #11
    I could have made an error, but the graph points to a slightly falling acceleration due to it's concavity.

    Was the book's answer. 75?
  13. Jul 18, 2011 #12
    the books answer was:
    a(6) = 0.07m/s^2
  14. Jul 18, 2011 #13
    Then in my opinion, the book either has a sign error or your function was copied incorrectly.
  15. Jul 18, 2011 #14

    Ray Vickson

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    No. A correct statement would be: therefore, v(t) = 6/sqrt(12t - 8). I don't understand what you have against writing expressions clearly, using brackets. It really does not require extra work.

  16. Jul 18, 2011 #15
    ya it does on my computer...nothing "against" writing expressions clearly...that is silly.
  17. Jul 18, 2011 #16

    Ray Vickson

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    I don't believe you. Before, you wrote v=s'(t)=1/2(12t-8)^-1/2(12) and that has lots of brackets.

  18. Jul 18, 2011 #17
    i don't care buddy...you need a holiday.
  19. Jul 18, 2011 #18
    Right... about the mathematics, can anyone else confirm that the book's solution does not match the problem we have been given?
  20. Jul 18, 2011 #19
    The answer in the book is wrong. It showing a positive acceleration when it should be a negative value. The book made a sign error.

    a(t) = v'(t) = \frac{-36}{(12t-8)^{\frac{3}{2}}}

    For t = 6 sec

    a(6) = v'(6) \, \approx \, -0.07 m/sec^{2}
  21. Jul 18, 2011 #20
    Yes, the book must be wrong if we're all getting matching answers.
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