Explanation

We can derive this equation from equations of motion so it is numerically correct $S_{1} =$distance trvelled in $t^{th}$ second = $\frac{Distance}{time} = \lbrack LT^{- 1}\rbrack$ $u =$ velocity =$\lbrack LT^{- 1}\rbrack$ and $\frac{1}{2}a(2t - 1) = \lbrack LT^{- 1}\rbrack$ As dimensions of each term in the given equation are same, hence equation is dimensionally correct also