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

Explanation

To find the acceleration of a particle whose velocity varies according to the equation \( v(x) = \beta x^{-2n} \), we can follow these steps:

### Step 1: Understand the relationship between acceleration, velocity, and position.
Acceleration \( a \) is defined as the rate of change of velocity with respect to time, which can be expressed mathematically as:
\[ a = \frac{dv}{dt} \]
However, since the velocity \( v \) is a function of position \( x \), we can use the chain rule to express acceleration in terms of \( x \):
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} \]
Here, \( \frac{dx}{dt} = v \), so we can rewrite the equation as:
\[ a = v \cdot \frac{dv}{dx} \]

### Step 2: Differentiate the velocity function with respect to position.
Given the velocity function:
\[ v(x) = \beta x^{-2n} \]
we need to find \( \frac{dv}{dx} \). We differentiate \( v \) with respect to \( x \):
\[ \frac{dv}{dx} = \frac{d}{dx} (\beta x^{-2n}) = \beta \cdot (-2n) x^{-2n-1} \]
Thus,
\[ \frac{dv}{dx} = -2n \beta x^{-2n-1} \]

### Step 3: Substitute \( v \) and \( \frac{dv}{dx} \) back into the acceleration equation.
Now, substituting \( v \) and \( \frac{dv}{dx} \) into the acceleration formula:
\[ a = v \cdot \frac{dv}{dx} = \left(\beta x^{-2n}\right) \cdot \left(-2n \beta x^{-2n-1}\right) \]
This simplifies to:
\[ a = -2n \beta^2 x^{-2n} x^{-2n-1} = -2n \beta^2 x^{-4n-1} \]

### Final Result:
Thus, the acceleration of the particle as a function of \( x \) is:
\[ a(x) = -2n \beta^2 x^{-4n-1} \]

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