Answer
$v=\dfrac{-at^2+2s}{2t}$
Work Step by Step
Subtract $\frac{1}{2}at^2$ from both sides to find:
$s-\frac{1}{2}at^2=\frac{1}{2}at^2 +vt - \frac{1}{2}at^2
\\s-\frac{1}{2}at^2=vt$
Divide both sides by $t$ to find:
$\dfrac{s-\frac{1}{2}at^2}{t}=\dfrac{vt}{t}
\\\dfrac{s-\frac{1}{2}at^2}{t}=v
\\\dfrac{\frac{2s}{2} - \frac{at^2}{2}}{t}=v
\\\dfrac{\frac{2s-at^2}{2}}{t}=v
\\\dfrac{2s-at^2}{2t}=v
\\v=\dfrac{-at^2+2s}{2t}$
