Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kinematics: Solve for Time

  1. Mar 6, 2006 #1
    Ok well I'm making a program to solve all my physics formulas quick and easy. But for the equation D = ViT + .5AT^2 I can't solve for T. I forgot some of my algebra 2 skills, which would come in handy here :rolleyes:. This is what I have so far:

    Since this is somewhat similar to deriving the quadratic equation I wrote that down to try to mirror it.

    [tex]d = v_{i}t + \frac{1}{2}at^{2}[/tex]

    [tex]0 = v_{i}t + \frac{1}{2}at^{2} - d[/tex]

    [tex]\frac{1}{2}at^{2} + v_{i}t - d = 0[/tex]

    [tex]\frac{1}{2}at^{2} + v_{i}t = d[/tex]

    [tex]t^{2} + \frac{2v_{i}t}{a} = \frac{2d}{a}[/tex]

    [tex]t^{2} + \frac{2v_{i}t}{a} + \frac{v_{i}^{2}}{a^{2}} = \frac{2d}{a} + \frac{v_{i}^{2}}{a^{2}}[/tex]

    [tex](t + \frac{v_{i}}{a})^{2} = \frac{2d}{a} + \frac{v_{i}^{2}}{a^{2}}[/tex]

    [tex]t + \frac{v_{i}}{a} = \sqrt{\frac{2d}{a} + \frac{v_{i}}{a^{2}}}[/tex]

    [tex]t = - \frac{v_{i}}{a}\pm\sqrt{\frac{2d}{a} + \frac{v_{i}}{a^{2}}}[/tex]

    [tex]t = - \frac{v_{i}}{a}\pm\sqrt{\frac{2da}{a^{2}} + \frac{v_{i}}{a^{2}}}[/tex]

    [tex]t = - \frac{v_{i}}{a}\frac{\pm\sqrt{v_{i} + 2ad}}{a}[/tex]

    [tex]t = \frac{-v_{i}\pm\sqrt{v_{i} + 2ad}}{a}[/tex]

    Is there any way of simlifing this more? Any help is appreciated. Well I can't seem to get my LaTeX image to show up, can anyone edit it so it will?
    Last edited: Mar 6, 2006
  2. jcsd
  3. Mar 6, 2006 #2
    are you referring the the kinematic eq.
    [tex] x=x_{0}+v_{0} t+\frac{a}{2}t^{2} [/tex]
    if so why do you not just use the quadratic formula if you're trying to solve for t.
    [tex] t = \frac{-v_{0} \pm \sqrt{v_{0}^{2}-2ax_{0}}}{a} [/tex]
    if you want to derive the quad. equation then, start with the first formula, complete the square in terms of t, and solve remember in completing the square you have to make it such that the coefficient in front of the squared term is 1, otherwise it's a little messier.
    Last edited: Mar 6, 2006
  4. Mar 6, 2006 #3
    Ya thats what I did... if only the latex from the first post would show up :grumpy:

    This is the first equation:
    [tex] d = v_{i}t + \frac{1}{2}at^2 [/tex]

    And this is the final equation I got:

    [tex] t = \frac{-v_{i} \pm \sqrt{v_{i} + 2ad}}{a} [/tex]
    Last edited: Mar 6, 2006
  5. Mar 7, 2006 #4
    Is the final equation right? I don't think it is.
  6. Mar 7, 2006 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The usual response to this is: "plug your solution back into your initial equation!"

    However, before carrying that out, it's a good idea to check that the units of your expression are consistent.
  7. Mar 7, 2006 #6
    Tried that, didn't work! Plugged in 5's and got a 3m as displacement.
  8. Mar 7, 2006 #7
    You should get with your equation
    [tex] t = \frac{-v_{i} \pm \sqrt{v_{i}^{2}-2a (\pm d)}}{a} [/tex]
    [tex] d=d_{f}-d_{i}[/tex]
    so if
    [tex] d_{f}<d_{i} [/tex]
    your result wouldn't be correct.
    Last edited: Mar 7, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Kinematics: Solve for Time
  1. Solve For It (Replies: 2)

  2. Solve It! (Replies: 2)