Answer
See the work step by step.
Work Step by Step
(a) $v^2=v_0^2+2a\Delta x$
$\Delta x=x-x_0$. If $x=0$,
$\Delta x=-x_0$. Then:
$v^2=v_0^2+2a(-x_0)=v_0^2-2ax_0$
$v=±\sqrt {v_0^2-2ax_0}$
(b) $v=v_0+at$. Rearrange fot $t$:
$t=\frac{v-v_0}{a}$.
If $v=±\sqrt {v_0^2-2ax_0}$, then:
$t=\frac{±\sqrt {v_0^2-2ax_0}-v_0}{a}=\frac{-v_0±\sqrt {v_0^2-2ax_0}}{a}$
(c) From Math:
$ax^2+bx+c=0$. Then solving for $x$:
$x=\frac{-b±\sqrt {b^2-4ac}}{2a}$
Make: $x=t$, $a=\frac{1}{2}a$, $b=v_0$, $c=x_0$. Now we have:
$x_0+v_0t+\frac{1}{2}at^2=0$. Then solving for $t$:
$t=\frac{-v_0±\sqrt {v_0^2-4\frac{1}{2}ax_0}}{2\frac{1}{2}a}$
$t=\frac{-v_0±\sqrt {v_0^2-2ax_0}}{a}$
