A) 5 : 9
B) 5 : 7
C) 9 : 5
D) 9 : 7
Correct Answer: A
Solution :
Distance travelled in 5th second \[{{S}_{n}}=u+\frac{a}{2}(2n-1)\] \[\therefore \] Distance travelled by car A in 5th second \[{{S}_{A}}=0+\frac{{{a}_{1}}}{2}(2\times 5-1)=\frac{9{{a}_{1}}}{2}\] Distance travelled by car B in 5th second \[{{S}_{B}}=0+\frac{{{a}_{2}}}{2}(2\times 3-1)=\frac{5{{a}_{1}}}{2}\] Given, \[{{S}_{A}}={{S}_{B}}\] \[\therefore \] \[\frac{9{{a}_{1}}}{2}=\frac{5{{a}_{2}}}{2}\] \[\therefore \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{5}{9}\]
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