A) \[{{v}_{0}}{{c}^{-kt}}\]
B) \[\frac{{{v}_{0}}}{\sqrt{2v_{0}^{2}kt+1}}\]
C) \[{{v}_{0}}\]
D) \[\frac{{{v}_{0}}}{2}\]
Correct Answer: B
Solution :
Here, \[\frac{dv}{dt}=-k{{v}^{3}}\] or \[\frac{dv}{{{v}^{3}}}=-kdt\] or \[\int_{{{v}_{0}}}^{v}{\frac{dv}{{{v}^{3}}}=\int_{0}^{t}{-kdt}}\] or \[\left[ \frac{-1}{2{{v}^{2}}} \right]_{{{v}_{0}}}^{v}=-kdt\] or \[\frac{1}{2v_{0}^{2}}-\frac{1}{2{{v}^{2}}}=-kt\] or \[\frac{1}{2{{v}^{2}}}=\frac{1}{2{{v}^{2}}}+kt\] or \[\frac{1}{2{{v}^{2}}}=\frac{1+2v_{0}^{2}kt}{2v_{0}^{2}}\] or \[{{v}^{2}}=\frac{v_{0}^{2}}{2v_{0}^{2}kt+1}\] or \[v=\frac{{{v}_{0}}}{\sqrt{(2v_{0}^{2}kt+1)}}\]
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