The kinetic energy of a particle of mass m moving with speed v is given by `K=(1)/(2)mv^(2)`. If the kinetic energy of a particle moving along x-axis varies with x as `K(x)=9-x^(2)`, then The region in which particle lies is :

Options

A. `x ge 9`
B. `-3 le x le 3`
C. `0 le x le 9`
D. `-oo lt x lt oo`

Explanation

To solve the problem, we need to analyze the given kinetic energy function and determine the region in which the particle can move based on its kinetic energy being non-negative.

### Step-by-Step Solution:

1. **Understand the Kinetic Energy Function**:
The kinetic energy \( K \) of a particle is given by the equation:
\[ K = \frac{1}{2} mv^2 \]
We are also given that the kinetic energy varies with position \( x \) as:
\[ K(x) = 9 - x^2 \]

2. **Set Up the Condition for Kinetic Energy**:
Since kinetic energy cannot be negative, we set up the inequality:
\[ K(x) \geq 0 \]
Substituting the expression for \( K(x) \):
\[ 9 - x^2 \geq 0 \]

3. **Rearrange the Inequality**:
Rearranging the inequality gives:
\[ x^2 \leq 9 \]

4. **Solve for \( x \)**:
Taking the square root of both sides, we find:
\[ -3 \leq x \leq 3 \]
This means that \( x \) can take any value between -3 and 3, inclusive.

5. **Conclusion**:
Therefore, the region in which the particle lies is:
\[ x \in [-3, 3] \]

### Final Answer:
The particle lies in the region \( x \in [-3, 3] \).

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