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

Velocity of the body is given by

$V\left(x\right)=\beta x^{-2n}\dots \left(1\right)$

When $\beta$ and n are constant, x is the position of the particle.
We need to find out acceleration in terms of position (x).

We know, acceleration , $a=\frac{dv}{dt}$

Differentiating equation (1),

$a=\frac{dv\left(x\right)}{dt}=\beta (-2n)x^{-2n-1}\frac{dx}{dt}$

$=-2nv\beta x^{(-2n-1)}$

Substituting the value of v from (1)

$\begin{array}{rl} & a=-2n\beta x^{-(2n-1)}\beta x^{-2n}\\ & a=-2n{\beta }^2{x}^{-4n-1}\end{array}$