Can somebody help me with this? How the hell do I solve for t:-
x=vt+1/2at^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 $\displaystyle -x$.
Well, here I'd use the quadratic formula, or you can complete the square.
Quadratic formula:
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