A particle moves in a circle with constant tangential acceleration starting from rest. At t = 1/2 s radial acceleration has a value 25% of tangential acceleration. The value of tangential acceleration will be correctly represented by

Options

A. `a_t =R`
B. `a_t =R//2`
C. `a_t =2R`
D. none of these

Explanation

To solve the problem, we need to find the value of tangential acceleration \( a_t \) when the radial acceleration is 25% of the tangential acceleration at \( t = \frac{1}{2} \) seconds.

### Step-by-Step Solution:

1. **Understand the Definitions**:
- Tangential acceleration \( a_t \) is given by \( a_t = \alpha r \), where \( \alpha \) is the angular acceleration and \( r \) is the radius of the circular path.
- Radial (centripetal) acceleration \( a_r \) is given by \( a_r = \omega^2 r \), where \( \omega \) is the angular velocity.

2. **Initial Conditions**:
- The particle starts from rest, so the initial angular velocity \( \omega_i = 0 \).
- The angular velocity at time \( t \) can be calculated using the formula \( \omega_f = \omega_i + \alpha t \). Since \( \omega_i = 0 \), we have:
\[ \omega_f = \alpha t \]

3. **Calculate Angular Velocity at \( t = \frac{1}{2} \) seconds**:
- Substitute \( t = \frac{1}{2} \):
\[ \omega_f = \alpha \cdot \frac{1}{2} = \frac{\alpha}{2} \]

4. **Calculate Radial Acceleration**:
- Substitute \( \omega_f \) into the radial acceleration formula:
\[ a_r = \omega^2 r = \left(\frac{\alpha}{2}\right)^2 r = \frac{\alpha^2 r}{4} \]

5. **Set Up the Relationship**:
- According to the problem, at \( t = \frac{1}{2} \) seconds, the radial acceleration is 25% of the tangential acceleration:
\[ a_r = 0.25 a_t \]
- Substitute \( a_t = \alpha r \):
\[ a_r = 0.25 (\alpha r) \]

6. **Equate the Two Expressions for Radial Acceleration**:
- We have two expressions for \( a_r \):
\[ \frac{\alpha^2 r}{4} = 0.25 (\alpha r) \]
- Simplifying the right side:
\[ 0.25 (\alpha r) = \frac{1}{4} (\alpha r) \]

7. **Cancel Out Common Terms**:
- Since \( r \) is not zero, we can cancel \( r \) from both sides:
\[ \frac{\alpha^2}{4} = \frac{1}{4} \alpha \]

8. **Multiply Both Sides by 4**:
- This gives:
\[ \alpha^2 = \alpha \]

9. **Solve for \( \alpha \)**:
- Rearranging gives:
\[ \alpha^2 - \alpha = 0 \]
- Factor out \( \alpha \):
\[ \alpha(\alpha - 1) = 0 \]
- Thus, \( \alpha = 0 \) or \( \alpha = 1 \). Since the particle is moving with constant tangential acceleration, \( \alpha \) cannot be zero, so:
\[ \alpha = 1 \]

10. **Find Tangential Acceleration**:
- Now substitute \( \alpha \) back into the tangential acceleration formula:
\[ a_t = \alpha r = 1 \cdot r = r \]

### Final Answer:
The value of tangential acceleration \( a_t \) is correctly represented by \( r \).