if `theta` is angle between the velocity and acceleration of a particle moving on a circular path with decreasing speed, then

Options

A. `theta = 90^(@)`
B. `0^(@) lt theta lt 90^(@)`
C. `90^(@) lt theta lt 180^(@)`
D. `0^(@) le theta le 180^(@)`

Answer

To solve the problem, we need to analyze the motion of a particle moving in a circular path with decreasing speed. We will identify the directions of the velocity and acceleration vectors and determine the angle \( \theta \) between them.

### Step-by-Step Solution:

1. **Understanding Circular Motion**:
- A particle moving in a circular path experiences centripetal acceleration directed towards the center of the circle. This acceleration is responsible for changing the direction of the velocity vector but not its magnitude.

**Hint**: Remember that centripetal acceleration always points towards the center of the circular path.

2. **Identifying Velocity and Acceleration**:
- The velocity vector \( \vec{v} \) of the particle is tangent to the circular path at any point. Since the speed is decreasing, there is also a tangential acceleration \( \vec{a}_T \) acting in the direction opposite to the velocity vector.

**Hint**: The direction of tangential acceleration is opposite to the direction of the velocity when speed is decreasing.

3. **Components of Acceleration**:
- The total acceleration \( \vec{a} \) of the particle is the vector sum of the centripetal acceleration \( \vec{a}_C \) (towards the center) and the tangential acceleration \( \vec{a}_T \) (opposite to the velocity).

**Hint**: Visualize the two components of acceleration: one pointing inward (centripetal) and one pointing backward along the tangent (tangential).

4. **Analyzing the Angle \( \theta \)**:
- The angle \( \theta \) is formed between the velocity vector \( \vec{v} \) and the resultant acceleration vector \( \vec{a} \). Since \( \vec{a}_C \) is perpendicular to \( \vec{v} \) (90 degrees), and \( \vec{a}_T \) is directed opposite to \( \vec{v} \), the resultant acceleration \( \vec{a} \) will be directed at an angle greater than 90 degrees but less than 180 degrees.

**Hint**: Since \( \vec{a}_C \) is perpendicular to \( \vec{v} \), and \( \vec{a}_T \) pulls back, the resultant will tilt away from the velocity vector.

5. **Conclusion**:
- Therefore, the angle \( \theta \) between the velocity and the acceleration of the particle moving in a circular path with decreasing speed is greater than 90 degrees and less than 180 degrees.

**Final Answer**: \( 90^\circ < \theta < 180^\circ \)