A ball starts moving in a circular slot with angular acceleration `alpha = 2 rad //s^2 ` at t=0 The angle between its velocity and acceleration varies with time as

Options

A. `90^@`
B. `tan ^(-1) (2t^2)`
C. ` tan ^(-1) (t//2)`
D. `45^@`

Explanation

To solve the problem step by step, we need to analyze the motion of the ball in a circular slot with the given angular acceleration.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Angular acceleration, \( \alpha = 2 \, \text{rad/s}^2 \)
- At \( t = 0 \), the initial angular velocity \( \omega_0 = 0 \).

2. **Determine Angular Velocity**:
- The angular velocity \( \omega \) at any time \( t \) can be calculated using the formula:
\[ \omega = \omega_0 + \alpha t \]
- Since \( \omega_0 = 0 \), we have:
\[ \omega = 0 + 2t = 2t \, \text{rad/s} \]

3. **Calculate Tangential and Centripetal Acceleration**:
- **Tangential Acceleration (\( A_t \))**:
- It is given by:
\[ A_t = \alpha R = 2R \, \text{m/s}^2 \]
- **Centripetal Acceleration (\( A_c \))**:
- It is given by:
\[ A_c = \frac{V^2}{R} = \frac{(\omega R)^2}{R} = \frac{(2t R)^2}{R} = \frac{4t^2 R^2}{R} = 4t^2 R \, \text{m/s}^2 \]

4. **Determine the Angle Between Velocity and Acceleration**:
- The angle \( \theta \) between the velocity vector and the resultant acceleration vector can be found using the tangent of the angle:
\[ \tan \theta = \frac{A_c}{A_t} \]
- Substituting the values of \( A_c \) and \( A_t \):
\[ \tan \theta = \frac{4t^2 R}{2R} = \frac{4t^2}{2} = 2t^2 \]

5. **Find \( \theta \)**:
- Thus, we can express \( \theta \) as:
\[ \theta = \tan^{-1}(2t^2) \]

### Final Answer:
The angle between the velocity and acceleration of the ball varies with time as:
\[ \theta = \tan^{-1}(2t^2) \]