A) \[\frac{1}{2}m{{s}^{-1}}\]
B) \[\frac{1}{3}m{{s}^{-1}}\]
C) \[\frac{1}{4}m{{s}^{-1}}\]
D) \[\frac{1}{5}m{{s}^{-1}}\]
Correct Answer: B
Solution :
At \[t=0,\,y=\frac{1}{2+{{x}^{2}}}\] or \[2y+y{{x}^{2}}=1\] or \[y{{x}^{2}}=+\,1-2y\] or \[x=\sqrt{\frac{1-2y}{y}}={{x}_{1}}\] (say) At \[t=3\,s,\] \[y=\frac{1}{[2+{{(x-1)}^{2}}]}\] or \[2+(x-1)=\frac{1}{y}\] or \[{{(x-1)}^{2}}=\frac{1}{y}-2=\frac{1-2y}{y}\] or \[x=1+\sqrt{\frac{1-2y}{y}}={{x}_{2}}\] (say) \[\therefore \]\[v=\frac{{{x}_{2}}-{{x}_{1}}}{{{t}_{2}}-{{t}_{1}}}=\frac{1+\sqrt{\frac{1-2y}{y}}-\sqrt{\frac{1-2y}{y}}}{3-1}=\frac{1}{3}m{{s}^{-1}}\]
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