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# Medical

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v=\beta x^{-2n} where b 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

Answers (1)
A avinash.dongre

@Vishal gond 

as v=\beta x^{-2n}

use a=\frac{dv}{dt}=\frac{dv}{dx}*\frac{dx}{dt}=\frac{dv}{dx}*v

and  \frac{dv}{dx}=-2n*\beta x^{-2n-1}

so a=-2n*\beta x^{-2n-1}*\beta x^{-2n}=-2n\beta ^2 x^{-4n-1}

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