A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v ( x ) = βx − 2 n v(x) =βx-2n where, β 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

   Question:  From NCERT  | NEET 2015   || [Difficult level:Medium]  |Acceleration| 

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to $v\left(x\right)={\mathrm{\beta x}}^{-2n}$ where, $\beta$ 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-
1. $-2{\mathrm{n\beta }}^2{x}^{-2n-1}$
2. $-2{\mathrm{n\beta }}^2{x}^{-4n-1}$
3. $-2{\beta }^2{x}^{-2n+1}$
4. $-2{\mathrm{n\beta }}^2{x}^{-4n+1}$

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:2. 

$-2{\mathrm{n\beta }}^2{x}^{-4n-1}$

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NCERT Reference

XI Physics Page no. 48

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