a
(a) Distance covered in \(5^{th}\) second,

\({S_{{5^{th}}}} = u + \frac{a}{2}(2n - 1) = 0 + \frac{a}{2}(2 \times 5 - 1) = \frac{{9a}}{2}\)

and distance covered in \(5\) second,

\({S_5} = ut + \frac{1}{2}a{t^2} = 0 + \frac{1}{2} \times a \times 25 = \frac{{25a}}{2}\)

\(\therefore \frac{{{S_{{5^{th}}}}}}{{{S_5}}} = \frac{9}{{25}}\)