A) \[0.5m/s\]
B) \[~1m/s\]
C) \[1.5m/s\]
D) \[2\text{ }m/s\]
Correct Answer: A
Solution :
[a] At, \[t=0,\,y=\frac{1}{1+{{x}^{2}}}\,\] or \[x=\sqrt{\frac{1-y}{y}}={{x}_{1}}\] At \[t=2s,\] \[y=\frac{1}{2+{{x}^{2}}-2x}=\frac{1}{1+{{(x-1)}^{2}}}\] or \[{{(x-1)}^{2}}=\frac{1-y}{y}\] or \[x=1+\sqrt{\frac{1-y}{y}}={{x}_{2}}\] \[\therefore \,\,\,\,Speed\text{ }of\text{ }the\text{ }wave\] \[v=\frac{\Delta x}{\Delta t}=\frac{{{x}_{2}}-{{x}_{1}}}{{{t}_{2}}-{{t}_{1}}}=\frac{1}{2-0}=0.5m/s\]You need to login to perform this action.
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