## Precalculus (6th Edition)

$t=\pm\sqrt{\dfrac{2(r-r_{0})}{a}}$
$r=r_{0}+\dfrac{1}{2}at^{2}$, for $t$ Multiply the whole equation by $2$: $2\Big(r=r_{0}+\dfrac{1}{2}at^{2}\Big)$ $2r=2r_{0}+at^{2}$ Rearrange: $2r_{0}+at^{2}=2r$ Take $2r_{0}$ to subtract the right side: $at^{2}=2r-2r_{0}$ Take $a$ to divide the right side: $t^{2}=\dfrac{2r-2r_{0}}{a}$ Take the square root of both sides: $\sqrt{t^{2}}=\pm\sqrt{\dfrac{2r-2r_{0}}{a}}$ $t=\pm\sqrt{\dfrac{2r-2r_{0}}{a}}$ Take out common factor $2$ from the numerator inside the square root to show a better looking answer: $t=\pm\sqrt{\dfrac{2(r-r_{0})}{a}}$