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 will analyze the motion of the ball in the circular slot and derive the expression for the angle between its velocity and acceleration.

### Step 1: Understand the given parameters
- The angular acceleration \( \alpha = 2 \, \text{rad/s}^2 \)
- At \( t = 0 \), the initial angular velocity \( \omega_0 = 0 \)

### Step 2: Calculate the angular velocity at time \( t \)
Using the equation of motion for angular velocity:
\[ \omega = \omega_0 + \alpha t \]
Substituting the known values:
\[ \omega = 0 + 2t = 2t \, \text{rad/s} \]

### Step 3: Calculate the tangential acceleration
The tangential acceleration \( a_t \) is given by:
\[ a_t = \alpha r \]
Substituting \( \alpha = 2 \):
\[ a_t = 2r \, \text{m/s}^2 \]

### Step 4: Calculate the centripetal acceleration
The centripetal acceleration \( a_c \) is given by:
\[ a_c = \omega^2 r \]
Substituting \( \omega = 2t \):
\[ a_c = (2t)^2 r = 4t^2 r \, \text{m/s}^2 \]

### Step 5: 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 \( a_t \):
\[ \tan \theta = \frac{4t^2 r}{2r} = 2t^2 \]

### Step 6: Solve for \( \theta \)
To find \( \theta \), we take the inverse tangent:
\[ \theta = \tan^{-1}(2t^2) \]

### Conclusion
The angle between the velocity and acceleration of the ball as a function of time \( t \) is:
\[ \theta = \tan^{-1}(2t^2) \]

### Final Answer
The correct option is \( \tan^{-1}(2t^2) \).

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