c
The distance covered in fifth second for first body is equal to the distance travelled in \(3^{\text {rd }}\) second for second body, \(S _{5}= S _{3}\)

\(S_{t}= u +\frac{(2 t -1) a}{2}\)

\(S_{5}=0+\frac{(2 \times 5-1) a}{2}\)

\(S_{5}=\frac{9 a_{1}}{2}\)

\(S_{3}=0+\frac{5}{2} a_{2}=\frac{5}{2} a_{2}\)

\(\frac{9 a_{1}}{2}=\frac{5}{2} a_{2}\)

\(\frac{a_{1}}{a_{2}}=\frac{5}{9}\)