A) \[9:5\]
B) \[5\,\,:\,\,7\]
C) \[5\,\,:\,\,9\]
D) \[7\,\,:\,\,9\]
Correct Answer: C
Solution :
In fifth second, the distance travelled by body A in fifth second is equal to distance travelled by body B in the third second from their start. The distance covered by the body in the nth second of motion is \[{{S}_{n}}=u+\frac{a}{2}\,(2n-1)\] Where u is initial velocity and a in acceleration. Distance covered by the body A in 5th second after its start, with acceleration\[{{a}_{1}}\], is \[{{({{S}_{5}})}_{A}}=0+\frac{{{a}_{1}}}{2}(2\times 5-1)=\frac{9{{a}_{1}}}{2}\] Time taken by second body \[= 5 - 2 = 3 s\] \[{{({{S}_{3}})}_{B}}=0+\frac{{{a}_{2}}}{2}(2\times 3-1)=\frac{5{{a}_{2}}}{2}\] \[~\left( {{S}_{5}} \right)A={{\left( {{S}_{3}} \right)}_{B}}\] \[\therefore \,\,\,\,\,\,\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{5}{9}\]
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