c
(c) We can derive this equation from equations of motion so it is numerically correct.

\({S_t}\)= distance travelled in \(t^{th}\) second \(=\frac{{{\rm{Distance}}}}{{{\rm{time}}}} = [L{T^{ - 1}}]\)

\(u\) = velocity = \([L{T^{ - 1}}]\) and \(\frac{1}{2}a(2t - 1) = [L{T^{ - 1}}]\)

As dimensions of each term in the given equation are same, hence equation is dimensionally correct also.