A particle moves in a circular path such that its speed `v` varies with distance `s` as `v = alpha sqrt(s)`, where `alpha` is a positive constant. Find the acceleration of the particle after traversing a distance `s`.

Options

A. ` alpha sqrt((1)/(4) + (s^2)/(R^2))`
B. ` alpha^2 sqrt((1)/(4) + (s)/(R))`
C. ` alpha sqrt((1)/(4) + (s)/(R))`
D. ` alpha^2 sqrt((1)/(4) + (s^2)/(R^2))`

Explanation

To find the acceleration of a particle moving in a circular path with a speed that varies with distance, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between speed and distance**: We are given that the speed \( v \) of the particle varies with distance \( s \) as:
\[ v = \alpha \sqrt{s} \]
where \( \alpha \) is a positive constant.

2. **Differentiate the speed with respect to time**: To find the acceleration, we need to differentiate \( v \) with respect to time \( t \). Using the chain rule, we have:
\[ \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} \]

3. **Calculate \( \frac{dv}{ds} \)**: First, we differentiate \( v \) with respect to \( s \):
\[ \frac{dv}{ds} = \frac{d}{ds}(\alpha \sqrt{s}) = \alpha \cdot \frac{1}{2\sqrt{s}} = \frac{\alpha}{2\sqrt{s}} \]

4. **Substitute \( \frac{ds}{dt} \)**: We know that \( \frac{ds}{dt} = v = \alpha \sqrt{s} \). Now substituting this into the equation for \( \frac{dv}{dt} \):
\[ \frac{dv}{dt} = \frac{\alpha}{2\sqrt{s}} \cdot (\alpha \sqrt{s}) = \frac{\alpha^2}{2} \]

5. **Calculate the radial acceleration \( A_r \)**: The radial acceleration is given by:
\[ A_r = \frac{v^2}{R} \]
Substituting \( v = \alpha \sqrt{s} \):
\[ A_r = \frac{(\alpha \sqrt{s})^2}{R} = \frac{\alpha^2 s}{R} \]

6. **Calculate the total acceleration \( A \)**: The total acceleration \( A \) is the vector sum of the tangential acceleration \( A_t \) and the radial acceleration \( A_r \):
\[ A = \sqrt{A_t^2 + A_r^2} \]
Substituting \( A_t = \frac{\alpha^2}{2} \) and \( A_r = \frac{\alpha^2 s}{R} \):
\[ A = \sqrt{\left(\frac{\alpha^2}{2}\right)^2 + \left(\frac{\alpha^2 s}{R}\right)^2} \]

7. **Factor out common terms**: We can factor out \( \alpha^2 \):
\[ A = \alpha \sqrt{\frac{\alpha^4}{4} + \frac{\alpha^4 s^2}{R^2}} = \alpha^2 \sqrt{\frac{1}{4} + \frac{s^2}{R^2}} \]

### Final Result:
Thus, the acceleration of the particle after traversing a distance \( s \) is:
\[ A = \alpha^2 \sqrt{\frac{1}{4} + \frac{s^2}{R^2}} \]