A particle of unit mass undergoes one dimensional motion such that its velocity varies according to
$v (x) = β x^{- 2 n}$ , 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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$- 2 {nβ}^{2} x^{- 2 n - 1}$

$- 2 β^{2} x^{- 2 n + 1}$

$- 2 {nβ}^{2} x^{- 4 n + 1}$

$- 2 {nβ}^{2} x^{- 4 n - 1}$

answer is B.

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Detailed Solution

Given, $v = {βx}^{- 2 n}$

$a = \frac{d v}{d t} = \frac{d x}{d t} \cdot \frac{d v}{d x}$

$\Rightarrow$ $a = v \frac{d v}{d x} = (β x^{- 2 n}) (- 2 n β x^{- 2 n - 1})$

$\Rightarrow$ $a = - 2 n β^{2} x^{- 4 n - 1}$

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