Dimensionally prove the 1st equation of motion.

• Dimension of velocity ($v_f$, $v_i$): $[\text{L}][\text{T}]^{-1}$ (Length per unit Time)
• Dimension of acceleration ($a$): $[\text{L}][\text{T}]^{-2}$ (Length per unit Time squared)
• Dimension of time ($t$): $[\text{T}]$ (Time)

The LHS is $v_f$.
Dimension of LHS = Dimension of $v_f = [\text{L}][\text{T}]^{-1}$

The RHS is $v_i + at$.
First, find the dimension of the term $v_i$:
Dimension of $v_i = [\text{L}][\text{T}]^{-1}$

Next, find the dimension of the term $at$:
Dimension of $at = (\text{Dimension of } a) \times (\text{Dimension of } t)$
Dimension of $at = ([\text{L}][\text{T}]^{-2}) \times ([\text{T}])$
Dimension of $at = [\text{L}][\text{T}]^{-2+1}$
Dimension of $at = [\text{L}][\text{T}]^{-1}$

For an equation to be dimensionally consistent, all terms added or subtracted must have the same dimensions. Here, both $v_i$ and $at$ have the dimension $[\text{L}][\text{T}]^{-1}$.
Therefore, the dimension of the RHS ($v_i + at$) is also $[\text{L}][\text{T}]^{-1}$.

Dimension of LHS = $[\text{L}][\text{T}]^{-1}$
Dimension of RHS = $[\text{L}][\text{T}]^{-1}$

Since the dimensions of the LHS and RHS are equal, the first equation of motion is dimensionally consistent.