Answer
The equations $v=\frac{1}{2}at$ and $v^{2}=ax$ are dimensionally consistent.
Work Step by Step
Dimension of a (acceleration): $\frac{[L]}{[T]^{2}}$.
Dimension of t (time): [T].
Dimension of x (distance): [L].
Dimension of v (velocity): $\frac{[L]}{[T]}$.
The numbers 2 and $\frac{1}{2}$ have no dimensions.
(a) $v=\frac{1}{2}at$
$\frac{[L]}{[T]}=\frac{[L]}{[T]^{2}}\times[T]$
$\frac{[L]}{[T]}=\frac{[L]}{[T]}$.
This equation is dimensionally consistent.
(b) $v=\frac{1}{2}at^{2}$
$\frac{[L]}{[T]}=\frac{[L]}{[T]^{2}}\times[T]^{2}$
$\frac{[L]}{[T]}=[L]$. But, $\frac{[L]}{[T]}\ne[L]$.
This equation is not dimensionally consistent.
(c) $t=\frac{a}{v}$
$[T]=\frac{\frac{[L]}{[T]^{2}}}{\frac{[L]}{[T]}}$
$[T]=\frac{[L]}{[T]^{2}}\frac{[T]}{[L]}$
$[T]=\frac{1}{[T]}=[T]^{-1}$. But, $[T]\ne[T]^{-1}$.
This equation is not dimensionally consistent.
(d) $v^{2}=ax$
$(\frac{[L]}{[T]})^{2}=\frac{[L]}{[T]^{2}}[L]$
$\frac{[L]^{2}}{[T]^{2}}=\frac{[L]^{2}}{[T]^{2}}$.
This equation is dimensionally consistent.
