A particle is moveint along the x-axis whose instantaneous speed is given by `v^(2)=108-9x^(2)`. The acceleration of the particle is.

Options

A. `-9x ms^(-2)`
B. `-18x ms^(-2)`
C. `(-9x)/(2) ms^(-2)`
D. None of there

Explanation

To find the acceleration of the particle whose instantaneous speed is given by the equation \( v^2 = 108 - 9x^2 \), we can follow these steps:

### Step 1: Differentiate the velocity equation
We start with the equation for velocity:
\[ v^2 = 108 - 9x^2 \]
To find acceleration, we need to differentiate both sides with respect to \( x \).

### Step 2: Apply the chain rule
Differentiating \( v^2 \) with respect to \( x \) using the chain rule gives:
\[ \frac{d(v^2)}{dx} = 2v \frac{dv}{dx} \]
On the right side, differentiating \( 108 - 9x^2 \) gives:
\[ \frac{d(108 - 9x^2)}{dx} = -18x \]
So, we have:
\[ 2v \frac{dv}{dx} = -18x \]

### Step 3: Solve for \( \frac{dv}{dx} \)
Now, we can isolate \( \frac{dv}{dx} \):
\[ v \frac{dv}{dx} = -9x \]
\[ \frac{dv}{dx} = \frac{-9x}{v} \]

### Step 4: Substitute for acceleration
We know that acceleration \( a \) can be expressed as:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v \]
Substituting \( \frac{dv}{dx} \) from the previous step:
\[ a = \left(\frac{-9x}{v}\right) \cdot v \]
This simplifies to:
\[ a = -9x \]

### Final Result
Thus, the acceleration of the particle is:
\[ a = -9x \, \text{m/s}^2 \]