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

Explanation

To find the acceleration of the particle as a function of its position \( x \), we start with the given velocity function:

\[ v(x) = \beta x^{-2n} \]

### Step 1: Understand the relationship between acceleration, velocity, and position
The acceleration \( a \) can be expressed as the derivative of velocity with respect to time:

\[ a = \frac{dv}{dt} \]

However, since we have velocity as a function of position \( x \), we can use the chain rule to express acceleration in terms of \( x \):

\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v \]

### Step 2: Differentiate the velocity function with respect to \( x \)
Now we need to find \( \frac{dv}{dx} \). We differentiate \( v(x) \):

\[ v(x) = \beta x^{-2n} \]

Using the power rule, we get:

\[ \frac{dv}{dx} = -2n \beta x^{-2n - 1} \]

### Step 3: Substitute \( v \) and \( \frac{dv}{dx} \) into the acceleration formula
Now we substitute \( v \) and \( \frac{dv}{dx} \) into the equation for acceleration:

\[ a = v \cdot \frac{dv}{dx} \]

Substituting the expressions we found:

\[ a = \left( \beta x^{-2n} \right) \left( -2n \beta x^{-2n - 1} \right) \]

### Step 4: Simplify the expression
Now we simplify the expression for acceleration:

\[ a = -2n \beta^2 x^{-2n} x^{-2n - 1} = -2n \beta^2 x^{-4n - 1} \]

### Final Result
Thus, the acceleration of the particle as a function of \( x \) is given by:

\[ a(x) = -2n \beta^2 x^{-4n - 1} \]