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 with a constant angular acceleration.

### Step-by-Step Solution:

1. **Identify the given values:**
- Angular acceleration, \( \alpha = 4 \, \text{rad/s}^2 \)
- Initial angular velocity, \( \omega_0 = 0 \, \text{rad/s} \) (since the particle is at rest at \( t = 0 \))

2. **Write the expressions for centripetal and tangential acceleration:**
- The centripetal acceleration \( a_c \) is given by:
\[ a_c = r \omega^2 \]
- The tangential acceleration \( a_t \) is given by:
\[ a_t = r \alpha \]

3. **Since we want to find the time when \( a_c = a_t \):**
\[ r \omega^2 = r \alpha \]
- We can simplify this equation by canceling \( r \) (assuming \( r \neq 0 \)):
\[ \omega^2 = \alpha \]

4. **Find the expression for angular velocity \( \omega \):**
- The angular velocity \( \omega \) at any time \( t \) under constant angular acceleration is given by:
\[ \omega = \omega_0 + \alpha t \]
- Since \( \omega_0 = 0 \):
\[ \omega = \alpha t \]

5. **Substituting \( \omega \) into the equation \( \omega^2 = \alpha \):**
\[ (\alpha t)^2 = \alpha \]
- This simplifies to:
\[ \alpha^2 t^2 = \alpha \]

6. **Dividing both sides by \( \alpha \) (assuming \( \alpha \neq 0 \)):**
\[ \alpha t^2 = 1 \]

7. **Solving for \( t^2 \):**
\[ t^2 = \frac{1}{\alpha} \]

8. **Taking the square root to find \( t \):**
\[ t = \frac{1}{\sqrt{\alpha}} = \frac{1}{\sqrt{4}} = \frac{1}{2} = 0.5 \, \text{seconds} \]

### Conclusion:
The time at which the magnitudes of centripetal acceleration and tangential acceleration are equal is \( t = 0.5 \, \text{seconds} \).