A) \[-2n{{\beta }^{2}}{{x}^{-2n-1}}\]
B) \[-2n{{\beta }^{2}}{{x}^{-4n-1}}\]
C) \[-2n{{\beta }^{2}}{{x}^{-2n+1}}\]
D) \[-2n{{\beta }^{2}}{{e}^{-4n+1}}\]
Correct Answer: B
Solution :
We are given velocity of the particle \[\upsilon (x)=\beta {{x}^{-2n}}\] We know acceleration\[a=\upsilon \frac{d\upsilon }{dx}\] \[a=\beta {{x}^{-2n}}\frac{d}{dx}(\beta {{x}^{-2n}})\] \[={{\beta }^{2}}{{x}^{-2n}}(-2n){{x}^{-2n-1}}=-2n{{\beta }^{2}}{{x}^{-2n-1-2n}}\] \[a=-2n{{\beta }^{2}}{{x}^{-4n-1}}\]You need to login to perform this action.
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