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 analyze the motion of a particle moving in a circular path with constant tangential acceleration. We will derive the relationship between the tangential acceleration and the radial acceleration at a specific time.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The particle starts from rest and moves in a circle with constant tangential acceleration (\(a_t\)).
- At \(t = \frac{1}{2}\) seconds, the radial acceleration (\(a_r\)) is 25% of the tangential acceleration.

2. **Formulas**:
- The tangential acceleration is given by:
\[ a_t = \alpha \cdot r \]
where \(\alpha\) is the angular acceleration and \(r\) is the radius of the circular path.
- The radial (centripetal) acceleration is given by:
\[ a_r = \omega^2 \cdot r \]
where \(\omega\) is the angular velocity.

3. **Finding Angular Velocity**:
- Since the particle starts from rest, the initial angular velocity (\(\omega_0\)) is 0.
- The final angular velocity after time \(t\) can be expressed as:
\[ \omega = \omega_0 + \alpha t = 0 + \alpha t = \alpha t \]
- At \(t = \frac{1}{2}\) seconds, the angular velocity becomes:
\[ \omega = \alpha \cdot \frac{1}{2} = \frac{\alpha}{2} \]

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

5. **Setting Up the Equation**:
- According to the problem, at \(t = \frac{1}{2}\) seconds, the radial acceleration is 25% of the tangential acceleration:
\[ a_r = 0.25 \cdot a_t \]
- Substitute the expressions for \(a_r\) and \(a_t\):
\[ \frac{\alpha^2}{4} \cdot r = 0.25 \cdot (\alpha \cdot r) \]

6. **Simplifying the Equation**:
- Cancel \(r\) from both sides (assuming \(r \neq 0\)):
\[ \frac{\alpha^2}{4} = 0.25 \cdot \alpha \]
- Multiply both sides by 4:
\[ \alpha^2 = \alpha \]
- Rearranging gives:
\[ \alpha^2 - \alpha = 0 \]
- Factor out \(\alpha\):
\[ \alpha(\alpha - 1) = 0 \]

7. **Finding Values of Alpha**:
- This gives two solutions: \(\alpha = 0\) or \(\alpha = 1\).
- Since \(\alpha = 0\) would imply no acceleration, we discard this solution.

8. **Calculating Tangential Acceleration**:
- Thus, we have \(\alpha = 1 \, \text{rad/s}^2\).
- The tangential acceleration is then:
\[ a_t = \alpha \cdot r = 1 \cdot r = r \, \text{m/s}^2 \]

9. **Final Answer**:
- The value of tangential acceleration is correctly represented by:
\[ a_t = r \]