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

Answer

A

Explanation

According to question,
`v(x) = bx^(-2n)` So, `(dv)/(dx) = -2 n b x^(-2n-1)`
Acceleration of the particle as function of x,
`a= v (dv)/(dx) = bx^(-2n) {b(-2n)x^(-2n-1)}= - 2nb^(2)x^(-4n-1)`