A) \[-2n{{\beta }^{2}}\,{{x}^{-2n-1}}\]
B) \[-2n{{\beta }^{2}}\,{{x}^{-4n-1}}\]
C) \[-2{{\beta }^{2}}\,\,{{x}^{-2n+1}}\]
D) \[-2n{{\beta }^{2}}\,\,\,{{e}^{-4n+1}}\]
Correct Answer: B
Solution :
Given, \[v=\beta {{x}^{-2n}}\] |
\[a=\frac{dv}{dt}=\frac{dx}{dt}.\frac{dv}{dx}\] |
\[\Rightarrow \] \[\,a=v\frac{dv}{dx}=(\beta {{x}^{-2n}})(-2n\beta {{x}^{-2n-1}})\] |
\[\Rightarrow \] \[a=-2n{{\beta }^{2}}{{x}^{-4n}}^{-1}\] |
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