A particle of unit mass undergoes one dimensional motion such that its velocity varies according tov(x) = β x-2n, where β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by

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-2nβ2 x-2n-1

-2nβ2 x-4n-1

-2β2 x-2n+1

-2nβ2 x-4n+1

answer is B.

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Detailed Solution

Given, v = βx-2n a=dvdt=dxdt·dvdx⇒ a=vdvdx=βx-2n-2nβx-2n-1⇒ a=-2nβ2x-4n-1

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