A) \[6:3\]
B) \[3:6\]
C) \[7:5\]
D) \[5:7\]
Correct Answer: C
Solution :
Here: initial velocity\[u=10\text{ }m/s\] First time limit\[{{t}_{1}}=3\text{ }sec\] Second time limit \[{{t}_{2}}=2\text{ }sec\] and \[g=10\text{ }m/{{s}^{2}}\] The distance covered in 3rd second \[{{s}_{n}}=u+\frac{g}{2}(2n-1)\] \[{{s}_{3}}=10+\frac{10}{2}(2\times 3-1)\] \[=35m\] Similarly, the distance covered in 2nd sec. \[{{s}_{2}}=10+\frac{10}{2}(2\times 2-1)\] \[=25m\] Hence the ratio of distances \[{{s}_{3}}:{{s}_{2}}=35:25\] \[=7:5\]
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