Answer:
[tex]\huge\boxed{t=\sqrt{\dfrac{2d}{a}}}[/tex]
Step-by-step explanation:
[tex]d=\dfrac{1}{2}at^2\\\\\dfrac{1}{2}at^2=d\qquad|\text{multiply both sides by 2}\\\\2\!\!\!\!\diagup\cdot\dfrac{1}{2\!\!\!\!\diagup}at^2=2\cdot d\\\\at^2=2d\qquad|\text{divide both sides by}\ a\neq0\\\\\dfrac{a\!\!\!\!\diagup t^2}{a\!\!\!\!\diagup}=\dfrac{2d}{a}\\\\t^2=\dfrac{2d}{a}\to t=\sqrt{\dfrac{2d}{a}}[/tex]
