A particle is moving with a constant angular acceleration of `4rad//s^(2)` in a circular path. At time `t=0` , particle was at rest. Find the time at which the magnitudes of centripetal acceleration and tangential acceleration are equal.

Explanation

To solve the problem, we need to find the time at which the magnitudes of centripetal acceleration and tangential acceleration are equal for a particle moving in a circular path with a constant angular acceleration.

### Step-by-step Solution:

1. **Identify the given values:**
- Angular acceleration (α) = 4 rad/s²
- Initial angular velocity (ω₀) = 0 rad/s (since the particle is at rest at t=0)

2. **Formulas for centripetal and tangential acceleration:**
- Centripetal acceleration (a_c) is given by:
\[ a_c = r \omega^2 \]
- Tangential acceleration (a_t) is given by:
\[ a_t = r \alpha \]

3. **Express angular velocity (ω) in terms of time (t):**
- Since the particle starts from rest and has a constant angular acceleration, we can express ω as:
\[ \omega = \omega_0 + \alpha t = 0 + 4t = 4t \]

4. **Substitute ω into the centripetal acceleration formula:**
- Now substituting ω into the centripetal acceleration formula:
\[ a_c = r (4t)^2 = r \cdot 16t^2 \]

5. **Set centripetal acceleration equal to tangential acceleration:**
- We need to find the time when centripetal acceleration equals tangential acceleration:
\[ r \cdot 16t^2 = r \cdot 4 \]
- We can cancel r from both sides (assuming r ≠ 0):
\[ 16t^2 = 4 \]

6. **Solve for t:**
- Rearranging gives:
\[ t^2 = \frac{4}{16} = \frac{1}{4} \]
- Taking the square root:
\[ t = \frac{1}{2} \text{ seconds} \]

### Final Answer:
The time at which the magnitudes of centripetal acceleration and tangential acceleration are equal is **0.5 seconds**.