According to the principle of dimensional homogeneity, the argument of an exponential function must be dimensionless. Therefore, the dimension of $\alpha t$ is $[M^0 L^0 T^0]$. $[\alpha][t] = [M^0 L^0 T^0] \implies [\alpha][T] = 1 \implies [\alpha] = [T^{-1}]$. Also, the equation must be dimensionally consistent, meaning the dimension of the left-hand side ($l$) must be equal to the dimension of the right-hand side ($\frac{v_0}{\alpha}$). $[l] = \frac{[v_0]}{[\alpha]}$ $[L] = \frac{[v_0]}{[T^{-1}]} \implies [v_0] = [L][T^{-1}] = [M^0 L^1 T^{-1}]$. Thus, the dimensions of $v_0$ and $\alpha$ are $[M^0 L^1 T^{-1}]$ and $[T^{-1}]$ respectively.