A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to
`v(x) = beta x^(-2 n)`
where `beta` and `n` are constant and `x` is the position of the particle. The acceleration of the particle as a function of `x` is given by.

Options

A. (a) ` ` -2 n beta^2 x^(-4n-1)`
B. (b) ` `-2 beta^(2)x^(-2n+1)`
C. (c ) ` - 2 n beta^2 x^(-4) n+ 1)`
D. (d) ` -2 n beta^2 x^(-2n-1)`

Answer

A

Explanation

Here `v = beta x^(-2n) `
` :. ` (dv)/(dx) =- 2n -1 `, As ` a = (dv)/(dt) = (dv)/(dx) xx (dx)/(dt)`
` or ` a = ( v dv) /(dx) = beta x^(-2n) xx -2 n beta x ^(-2n-1)`
` a=- 2 n beta^2 x^(-4n-1)` .