# Thread: Solve for time

1. ## Solve for time

Can somebody help me with this? How the hell do I solve for t:-

x=vt+1/2at^2

2. Use the quadratic formula:
The solutions to $\displaystyle ax^2+ bx+ c= 0$ are given by
$\displaystyle \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$

Replace the "a", "b", and "c" with the coefficients from your equation"
$\displaystyle \frac{1}{2}a t^2+ vt- x= 0$:
So "a" is $\displaystyle \frac{1}{2}a$, "b" is $\displaystyle v$, and "c" is $\displaystyle -x$.

3. You could rearrange it (e.g. $\displaystyle at^2+2vt-2x=0$) and complete the square, or sub in the appropriate values into the quadratic formula.
You ok with doing this?

4. Well, here I'd use the quadratic formula, or you can complete the square.

Rearrange;
$\displaystyle \frac{1}{2}at^2 + vt - x =0$

$\displaystyle t = \frac{-v \+- \sqrt{v^2 - 4(\frac{a}{2})(-x)}}{2(\frac{a}{2})}$

$\displaystyle t = \frac{-v \+- \sqrt{v^2 + 2ax}}{a}$

But since time is positive, you take only the positive sign;

$\displaystyle t = \frac{-v + \sqrt{v^2 + 2ax}}{a}$

Completing the square;

$\displaystyle \frac{1}{2}at^2 + vt - x =0$

$\displaystyle \frac{a}{2}(t^2 + \frac{2v}{a}t) - x= 0$

$\displaystyle \frac{a}{2}((t + \frac{v}{a})^2 - \frac{v^2}{a^2}) - x= 0$

Rearrange;

$\displaystyle (t + \frac{v}{a})^2 = \frac{2x}{a} + \frac{v^2}{a^2}$

$\displaystyle t = \sqrt{\frac{2x}{a} + \frac{v^2}{a^2}} - \frac{v}{a}$

EDIT: Oops, a little too late =S

5. Thanks for that halls.

6. I know about the quadratic formula and have used it before. For some reason I was trying to solve using other methods which were inapropriate in this case.