The position x of a particle at time ‘t’ is given by $x=\frac{{\mathrm{v}}_{\mathrm{o}}}{\mathrm{a}}(1-{\mathrm{e}}^{-\mathrm{at}})$ where ${\mathrm{v}}_{\mathrm{o}}$ is a constant and  a > 0. The dimensions of ${\mathrm{v}}_{\mathrm{o}}$ and a are

$\mathrm{at}$ will be dimension less, so the dimensions of a will be T^-1. $\Rightarrow \left[\mathrm{a}\right]=\left[{\mathrm{T}}^{-1}\right]$

$\frac{{\mathrm{v}}_{\mathrm{o}}}{\mathrm{a}}$ will have the same dimensions are distance . $\left[\frac{{\mathrm{v}}_{\mathrm{o}}}{\mathrm{a}}\right]=\left[\mathrm{L}\right]\Rightarrow \left[{\mathrm{v}}_{\mathrm{o}}\right]=\left[{\mathrm{LT}}^{-1}\right]$

$\left[\mathrm{a}\right]=\left[{\mathrm{T}}^{-1}\right] \mathrm{\&} \left[{\mathrm{v}}_0\right]=\left[{\mathrm{M}}^0{\mathrm{LT}}^{-1}\right]$