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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a
-2nβ2 x-2n-1
b
-2nβ2 x-4n-1
c
-2β2 x-2n+1
d
-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