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. `-2nbeta^(2)x^(-2n-1)`
B. `-2nbeta^(2)x^(-4n-1)`
C. `-2beta^(2)x^(-2n+1)`
D. `-2nbeta^(2)e^(-4n+1)`

Answer

b

Explanation

`v(x)=betax^(-2n)`
`(dv(x))/(dt)=beta(-2n)x^(-2n-1)(dx)/(dt)`
`beta(-2n)x^(-2n-1)(betax^(-2n)) [v(x)=(dx)/(dt)]`
`a(x)=-beta^(2)(2n)x^(-4n-1)`