How to Solve for Time in Kinematics Equations

[Firestrider]

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 . 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.

d = v_{i}t + \frac{1}{2}at^{2}

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

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

\frac{1}{2}at^{2} + v_{i}t = d

t^{2} + \frac{2v_{i}t}{a} = \frac{2d}{a}

t^{2} + \frac{2v_{i}t}{a} + \frac{v_{i}^{2}}{a^{2}} = \frac{2d}{a} + \frac{v_{i}^{2}}{a^{2}}

(t + \frac{v_{i}}{a})^{2} = \frac{2d}{a} + \frac{v_{i}^{2}}{a^{2}}

t + \frac{v_{i}}{a} = \sqrt{\frac{2d}{a} + \frac{v_{i}}{a^{2}}}

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

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

t = - \frac{v_{i}}{a}\frac{\pm\sqrt{v_{i} + 2ad}}{a}

t = \frac{-v_{i}\pm\sqrt{v_{i} + 2ad}}{a}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?

 

[xman]

are you referring the the kinematic eq.
x=x_{0}+v_{0} t+\frac{a}{2}t^{2}
if so why do you not just use the quadratic formula if you're trying to solve for t.
t = \frac{-v_{0} \pm \sqrt{v_{0}^{2}-2ax_{0}}}{a}
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.

 

[Firestrider]

Ya that's what I did... if only the latex from the first post would show up

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

And this is the final equation I got:

t = \frac{-v_{i} \pm \sqrt{v_{i} + 2ad}}{a}

 

[Firestrider]

Is the final equation right? I don't think it is.

 

[robphy]

Is the final equation right? I don't think it is.

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.

 

[Firestrider]

Tried that, didn't work! Plugged in 5's and got a 3m as displacement.

 

[xman]

Ya that's what I did... if only the latex from the first post would show up

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

And this is the final equation I got:

t = \frac{-v_{i} \pm \sqrt{v_{i} + 2ad}}{a}

You should get with your equation
t = \frac{-v_{i} \pm \sqrt{v_{i}^{2}-2a (\pm d)}}{a}
remember
d=d_{f}-d_{i}
so if
d_{f}<d_{i}
your result wouldn't be correct.