A) 9 : 5
B) 5 : 7
C) 5 : 9
D) 7: 9
Correct Answer: C
Solution :
Key Idea: In fifth second, the distance travelled by body A in fifth second is equal to distance travelled by body B in 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 is acceleration. Distance covered by the body A in 5th second after its start, with acceleration \[{{a}_{1}}\] is \[({{S}_{5}})=0+\frac{{{a}_{1}}}{2}(2\times 5-1)=\frac{9{{a}_{1}}}{2}\] Time taken by second body \[=5-2=3\text{ }s\] \[{{({{S}_{3}})}_{B}}=0+\frac{{{a}_{2}}}{2}(2\times 3-1)=\frac{5{{a}_{2}}}{2}\] Given, \[{{({{S}_{5}})}_{A}}={{({{S}_{3}})}_{B}}\] \[\therefore \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{5}{9}\]
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