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 \[2+{{(x-1)}^{2}}=\frac{1}{y}\] or \[{{(x-1)}^{2}}=\frac{1}{y}-2=\frac{1-2y}{y}\] or \[x=1+\sqrt{\frac{1-2y}{y}}={{x}_{2}}\] \[\therefore \]\[v=\frac{{{x}_{2}}-{{x}_{1}}}{{{t}_{2}}-{{t}_{1}}}=\frac{1+\sqrt{\frac{1-2y}{y}}-\sqrt{\frac{1-2y}{y}}}{3-1}=\frac{1}{3}m{{s}^{-1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
