A) \[\frac{{{v}_{0}}}{2ktv_{0}^{2}}\]
B) \[\frac{{{v}_{0}}}{1+2ktv_{0}^{2}}\]
C) \[\frac{{{v}_{0}}}{\sqrt{1-2kv_{0}^{2}}}\]
D) \[\frac{{{v}_{0}}}{\sqrt{1+2kt\,v_{0}^{2}}}\]
Correct Answer: D
Solution :
Given, acceleration \[a=-k{{v}^{3}}\] Initial velocity at cut-off, \[{{v}_{1}}={{v}_{0}}\] Initial time of cut-off, \[t=0\] and final time after cut off,\[{{t}_{2}}=t\] Again, \[a=\frac{dv}{dt}=-k{{v}^{3}}\] or \[\frac{dv}{{{v}^{3}}}=-kdt\] Integrating both sides, with in the condition ofmotion \[\int_{{{v}_{0}}}^{v}{\frac{dv}{{{v}^{3}}}=-\int_{0}^{t}{k\,dt}}\] or \[\left[ -\frac{1}{2{{v}^{2}}} \right]_{{{v}_{0}}}^{v}=-[kt]_{0}^{t}\] or \[\frac{1}{2{{v}^{2}}}-\frac{1}{2v_{0}^{2}}=kt\] or \[v=\frac{{{v}_{0}}}{\sqrt{1+2kt\,v_{0}^{2}}}\]
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