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Textbook made mistake in algebra, help

  1. Sep 4, 2014 #1
    1. The problem statement, all variables and given/known data
    There is an image but I will reiterate my problem.
    The writer is solving for t. Okay, easy enough.

    2. Relevant equations

    y-y0 = v0 t + 1/2 a t^2

    Textbook says

    t = sqrt( (2(y-y0) - 2 v0) / a )

    3. The attempt at a solution

    Shouldn't it be

    t = sqrt( (2(y-y0) - 2 v0 t) / a )

    Let's take it slow.

    Multiply both sides by two.

    2(y-y0) = 2 v0 t + a t^2

    Bring (2 v0 t) to left side.

    2(y-y0) - 2 v0 t = a t^2

    Divide both sides by a.

    (2(y-y0) - 2 v0 t) / a = t^2

    Sqrt both sides.

    t = sqrt( (2(y-y0) - 2 v0 t) / a )

    This really bugs me.
     

    Attached Files:

  2. jcsd
  3. Sep 4, 2014 #2

    andrevdh

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    dimensionally you can see theirs is wrong
     
  4. Sep 4, 2014 #3

    SteamKing

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    Writing in standard form:

    y-y0 = v0 t + 1/2 a t^2 becomes

    (1/2) a t^2 + v0 t - (y-y0) = 0

    so that for the general quadratic in t, A t^2 + B t + C = 0,

    A = a/2

    B = v0

    C = -(y - y0) = (y0 - y)

    and the roots are

    t = [itex]\frac{-v_{0} \pm \sqrt{v^{2}_{0}-4(a/2)(y_{0}-y)}}{a}[/itex]

    With your algebra:

    t = sqrt( (2(y-y0) - 2 v0 t) / a )

    you wind up with 't' on both sides of the equation, which isn't conducive to obtaining a solution for t without iterating.
     
  5. Sep 4, 2014 #4

    BvU

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    My 'guess' is v0 = 0 somewhere in the problem statement.

    The textbook (or solution manual) write is completely wrong with his t = ...
    He/she should have written y-y0 = 1/2 a t^2 first and then proceeded to t = ... (without the -2v0).
     
  6. Sep 4, 2014 #5
    I hope this isn't illegal but I would like to copy the webpage the page of the textbook to show you guys the problem. Will this be illegal?
     
  7. Sep 4, 2014 #6

    berkeman

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    Staff: Mentor

    Just the one page for this purpose should be fine.
     
  8. Sep 4, 2014 #7
    The page is attached as an image.
     

    Attached Files:

  9. Sep 4, 2014 #8

    AlephZero

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    The book is wrong. They already said v0 = 0 so they should have removed it from the equation.

    Your equation
    is "right", but it's only useful in this problem because v0 = 0 and therefore to v0t = 0.
    If v0 is not 0, you have to solve the quadratic equation as Steamking said.
     
  10. Sep 4, 2014 #9
    Stupid wrong books. Paid good money...

    Alright I really like SteamKing's answer. Clearly states how the t went missing. Thanks. Really helps
     
  11. Sep 6, 2014 #10
    Yeah the book doubly messed up. First, as you show the answer is flat out wrong and second it doesn't even make sense to try to do what they did. Doing what they did would trap you with a t on both sides which is useless unless you lucked out to have v0=0. They forget to bring the t over though which makes it look like they actually used a method that would be some general form to solve things. It really makes no sense and it doesn't even give the right dimensions that even work out.

    Either they should have presented it full out in general with the whole -b+/srt( )/blah etc. formula or just plugged in 0 for v0 to start and then just did d=1/2 * a * t^2 and then get t= sqrt(2d/a)

    Man that is really messed up. It's hard to believe that is published in a book like that.
    Is that in the published book or just in some supplementary web material for the book?
     
  12. Sep 6, 2014 #11
    It came with the webassign package. So it's hard to tell it's from the book or supplements because even the chapters look like supplements (i.e. each chapter and section is a weblink to a different one-page webpage)
     
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