The velocity of a particle varies with its displacement as `v=(sqrt(9-x^2))ms^-1`.
Find the magnitude of the maximum acceleration of the particle.

Options

A. `3 ms^(-2)`
B. `4 ms^(-2)`
C. `3.5 ms^(-2)`
D. `5 ms^(-2)`

Answer

To find the magnitude of the maximum acceleration of the particle given the velocity as \( v = \sqrt{9 - x^2} \, \text{m/s} \), we can follow these steps:

### Step 1: Express acceleration in terms of displacement
The acceleration \( a \) can be expressed as:
\[ a = \frac{dv}{dt} \]
Using the chain rule, we can write:
\[ a = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \]

### Step 2: Differentiate the velocity function
Given \( v = \sqrt{9 - x^2} \), we first differentiate \( v \) with respect to \( x \):
\[ v^2 = 9 - x^2 \]
Differentiating both sides with respect to \( x \):
\[ 2v \frac{dv}{dx} = -2x \]
From this, we can isolate \( \frac{dv}{dx} \):
\[ \frac{dv}{dx} = \frac{-x}{v} \]

### Step 3: Substitute back to find acceleration
Now substituting \( \frac{dv}{dx} \) back into the expression for acceleration:
\[ a = v \cdot \frac{dv}{dx} = v \cdot \left(\frac{-x}{v}\right) = -x \]
Thus, the acceleration is:
\[ a = -x \]

### Step 4: Find the maximum magnitude of acceleration
The magnitude of acceleration is given by:
\[ |a| = |x| \]
From the velocity equation \( v = \sqrt{9 - x^2} \), we know that \( 9 - x^2 \geq 0 \), which implies:
\[ x^2 \leq 9 \quad \Rightarrow \quad |x| \leq 3 \]
Thus, the maximum value of \( |x| \) is 3.

### Step 5: Conclusion
Therefore, the maximum magnitude of acceleration is:
\[ |a|_{\text{max}} = 3 \, \text{m/s}^2 \]

### Final Answer
The magnitude of the maximum acceleration of the particle is \( 3 \, \text{m/s}^2 \).

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