A) \[\frac{2v_{0}^{3/2}}{3\alpha }\]
B) \[\frac{3v_{0}^{3/2}}{2\alpha }\]
C) \[\frac{v_{0}^{3/2}}{\alpha }\]
D) \[\frac{2{{\alpha }^{3}}}{3{{v}_{0}}}\]
Correct Answer: A
Solution :
| \[a=\frac{dv}{dt}=-\alpha \sqrt{v}\] |
| \[\Rightarrow \]\[\int_{{{v}_{0}}}^{v}{\frac{dv}{\sqrt{v}}}=-\alpha \int_{{{t}_{0}}=0}^{t}{dt}\] |
| (here, to is initial time) |
| \[\Rightarrow \]\[v={{\left( \sqrt{{{v}_{0}}}-\frac{\alpha t}{2} \right)}^{2}}={{v}_{0}}-\alpha \sqrt{{{v}_{0}}}.t+\frac{{{\alpha }^{2}}+{{t}^{2}}}{4}\] |
| \[\Rightarrow \]\[\frac{dx}{dt}={{v}_{0}}-\alpha \sqrt{{{v}_{0}}t}+\frac{{{\alpha }^{2}}{{t}^{2}}}{4}\] |
| Integrating |
| \[\Rightarrow \]\[x={{v}_{0}}t-\frac{\alpha \sqrt{{{v}_{0}}}{{t}^{2}}}{2}+\frac{{{\alpha }^{2}}{{t}^{3}}}{12}\] |
| Particle stops, when \[v=0\] |
| \[\Rightarrow \]\[\sqrt{{{v}_{0}}}-\frac{\alpha {{t}_{0}}}{2}=0\] |
| \[\Rightarrow \]\[{{t}_{0}}=\frac{2\sqrt{{{v}_{0}}}}{\alpha }\] |
| Substituting in Eq, (i), we get |
| \[x={{v}_{0}}\times \frac{2\sqrt{{{v}_{0}}}}{\alpha }-\frac{\alpha \sqrt{{{v}_{0}}}}{2}\times \frac{4{{v}_{0}}}{{{\alpha }^{2}}}+\frac{{{\alpha }^{2}}}{12}\times \frac{8v_{0}^{3/2}}{{{\alpha }^{3}}}\] |
| \[\Rightarrow \]\[x=\frac{2v_{0}^{3/2}}{3\alpha }\] |
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