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Motion in 1 Dimension

  1. May 5, 2009 #1
    1. The problem statement, all variables and given/known data
    A box slides along a surface with a positive initial velocity. It the experiences an acceleration of -0.25m/s^2. After traveling 4.00 meters, its velocity is +0.50m/s. How long did it take for the boxto travel the 4.00 meters?


    2. Relevant equations
    [tex]\Delta[/tex]x=Vot + 1/2 a t^2 ??


    3. The attempt at a solution
    4.00m=(.50m/s)t+1/2(.25m/s^2)(t^2)

    (.125m/s^2)(t^2)+(.50m/s)(t)-(4.00m)

    -(.50) + - [tex]\sqrt{(.50)^2-4(.125)(-4)/2(.125)}[/tex]

    [tex]\sqrt{2}[/tex]-.50 = .9seconds?
     
  2. jcsd
  3. May 5, 2009 #2
    Not from my calculations.

    From you problematic, [tex]v_0[/tex] is the initial velocity of the box, which is not equal to [tex]0.5m/s[/tex]

    Cheers
     
  4. May 5, 2009 #3
    You got the quadratic formula wrong. Everything is over 2a, not just 4ac.

    I had to double-check your initial set-up, because you had acceleration as positive, but after a bit of algebra it's correct. I'm not sure if that was by accident or intentional. You didn't show that step, if it was intentional. V-initial isn't given, so you have to solve for V-initial in terms of V-final, a, and t, which happens to be what you got.
     
  5. May 5, 2009 #4
    i set it over 2a...im just wondering if i set the previous equation up properly
     
  6. May 5, 2009 #5
    You did, but I suspect it was due to shear luck, because you didn't show how you went from Vo to Vf, and you didn't show how you went from 'a' being negative to 'a' being positive. According to my early-morning algebra, the set-up is right, however.

    And no, you didn't set it all over 2a. If you had, you would have gotten the correct answer.
     
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