A) \[y=\frac{{{k}_{1}}}{2{{k}_{2}}}\,{{x}^{2}}\]
B) \[y=\frac{{{k}_{2}}}{2{{k}_{1}}}\,{{x}^{2}}\]
C) \[y=\frac{2{{k}_{1}}}{{{k}_{2}}}\,{{x}^{2}}\]
D) \[y=\frac{2{{k}_{2}}}{{{k}_{1}}}\,{{x}^{2}}\]
Correct Answer: B
Solution :
\[v={{k}_{1}}\,i\,\,+\,\,{{k}_{2}}xj\] \[\therefore \,\,\,\text{ }x={{k}_{1}}\,\,t~~~~~~~~\,\,\,\,\,\,\,\,\,.\,..\left( 1 \right)\] \[\And \,\,{{v}_{y}}={{k}_{2}}\,x\,j\] \[\Rightarrow \,\,\frac{dy}{dt}\,\,=\,\,{{k}_{2}}\,{{k}_{1}}\,t\] \[\int{dy\,\,=\,\,\int{{{k}_{2}}\,{{k}_{1}}t\,dt}\,}\] \[\Rightarrow \,\,y\,\,=\,\,{{k}_{1}}{{k}_{2}}\frac{{{t}^{2}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....(2)\] By (1) & (2) \[y=\frac{{{k}_{2}}}{2{{k}_{1}}}\,{{x}^{2}}\]You need to login to perform this action.
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