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Finding Time

  1. Jan 28, 2009 #1
    k so i used
    x-xo= Vo t + 1/2 at^2

    plugged in all my values
    then moved everything to one side and used the quadratic formula

    here is my question.

    when i solve for it i get
    t=40.28 and t= 40.58

    So which t is my time?
  2. jcsd
  3. Jan 28, 2009 #2
    Er.. well, you seem to not have included the actual question?
  4. Jan 28, 2009 #3
    lets just use imaginary numbers

    if u solve for t and get one + and one - answer u know time can't be negative if you are traveling forward.. so the answer is the + one

    but what if your t comes out to be both + and you are going foward with -acceleration
  5. Jan 28, 2009 #4
    This isn't something that has a general rule - sometimes both of them can be the correct time and other times there's some condition that rules the other one out.

    Looking at your equation now it's the one used for 2D motion.. Taking that as an example, if you're considering a projectile and get two times it should be because the object is actually at the same horizontal position twice - once when going up and once when going down. It depends on what you want to calculate, in that case.

    Probably this is no help, but I honestly can't do much more. Sorry ;/
  6. Jan 28, 2009 #5
    I actually got it... had to think about it for a while..
    my guess is that its the lower time because thats when the object gets there first...
    the second time is when v=0 and the -acc. is making it go backwards so it goes back past the same position

  7. Jan 28, 2009 #6
    Yeah, if you're considering something that's thrown straight up into the air that would be the case, I'd say.
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