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, we start with the given relationship between velocity and displacement:

\[ v = \sqrt{9 - x^2} \]

### Step 1: Express acceleration in terms of displacement
Acceleration \( a \) can be expressed in terms of velocity \( v \) and displacement \( x \) using the formula:

\[ a = v \frac{dv}{ds} \]

### Step 2: Differentiate velocity with respect to displacement
First, we need to find \( \frac{dv}{ds} \). We know that:

\[ v = \sqrt{9 - x^2} \]

To differentiate \( v \) with respect to \( x \), we use the chain rule:

\[ \frac{dv}{dx} = \frac{1}{2\sqrt{9 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{9 - x^2}} \]

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

\[ a = v \frac{dv}{dx} = \sqrt{9 - x^2} \cdot \left( \frac{-x}{\sqrt{9 - x^2}} \right) \]

This simplifies to:

\[ a = -x \]

### Step 4: Determine the range of \( x \)
From the original equation for velocity, we have:

\[ 9 - x^2 \geq 0 \]

This implies:

\[ x^2 \leq 9 \]
\[ -3 \leq x \leq 3 \]

### Step 5: Find the maximum acceleration
The acceleration \( a = -x \) will be maximum when \( |x| \) is maximum. The maximum value of \( |x| \) in the range \( -3 \leq x \leq 3 \) is 3. Therefore, we evaluate:

- If \( x = -3 \), then \( a = -(-3) = 3 \, \text{m/s}^2 \)
- If \( x = 3 \), then \( a = -3 \, \text{m/s}^2 \)

### Step 6: Conclusion
The magnitude of the maximum acceleration is:

\[ |a| = 3 \, \text{m/s}^2 \]

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

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