A) 1 : 1
B) 2 : 1
C) 1:2
D) 3:1
Correct Answer: C
Solution :
[c] \[{{V}^{2}}={{2}^{2}}+\frac{2as}{2}\] \[\Rightarrow {{v}^{2}}-4=as\] \[{{14}^{2}}-{{v}^{2}}=\frac{2as}{2}\] \[\Rightarrow 196-{{v}^{2}}=as\] From (i) and (ii) \[{{v}^{2}}-4=196-{{v}^{2}}\Rightarrow v=10m/s\] Now \[{{t}_{1}}=\frac{v-2}{a}=\frac{10-2}{a}=\frac{8}{a}\] \[{{t}_{2}}=\frac{14-v}{a}=\frac{14-10}{a}=\frac{4}{a}\Rightarrow \frac{{{t}_{2}}}{{{t}^{1}}}=\frac{4/a}{8/a}=\frac{1}{2}\]You need to login to perform this action.
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