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Solving for T in an acceleration equation

  1. Mar 3, 2013 #1
    http://imgur.com/HeqIYqR

    How do I solve for Δt when there's a Δt^2
     
  2. jcsd
  3. Mar 3, 2013 #2
    Just use the quadratic formula.
     
  4. Mar 3, 2013 #3

    So what would b, a and c be?
     
  5. Mar 3, 2013 #4
    when you set the equation equal to zero a is the t^2 value, b is value with t variable, and c is just the number without the variable with it.

    "ax^2 + bx + c"
     
  6. Mar 3, 2013 #5
     
    Last edited: Mar 3, 2013
  7. Mar 3, 2013 #6
    I thought you were trying to solve for the change in time? So you would set the equation equal to zero d = d∅ + v∅t + 1/2 vt^2
    and you just put d - d∅ = Δd
    Δd = v∅t + 1/2 vt^2

    where v∅=initial velocity

    so the v∅t would be the b value
    1/2 * v would be the a value
    you would subtract Δd from both sides to get the equation equal to 0 and Δd would be your c value

    it would look like
    0 = 1/2vt^ + v∅t - Δd
    0 = at^2 + bt + c

    so you'll get two answers one at the max height of trajectory when velocity = 0 and one at the end of the trajectory where velocity = 0 again
     
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