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

Options

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

Answer

To solve the problem, we need to find the acceleration of a particle moving in a circular path where its speed \( v \) varies with distance \( s \) according to the equation \( v = \alpha \sqrt{s} \), where \( \alpha \) is a positive constant.

### Step-by-Step Solution:

1. **Identify the Components of Acceleration**:
- In circular motion, the total acceleration \( a \) of a particle has two components:
- Tangential acceleration \( a_t \) (due to the change in speed).
- Centripetal acceleration \( a_c \) (due to the change in direction).

2. **Calculate Tangential Acceleration**:
- The tangential acceleration \( a_t \) can be expressed as:
\[ a_t = \frac{dv}{dt} \]
- We know \( v = \alpha \sqrt{s} \). To find \( \frac{dv}{dt} \), we can use the chain rule:
\[ \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} \]
- First, we find \( \frac{dv}{ds} \):
\[ \frac{dv}{ds} = \frac{d}{ds}(\alpha \sqrt{s}) = \frac{\alpha}{2\sqrt{s}} \]
- The speed \( v \) is also equal to \( \frac{ds}{dt} \):
\[ \frac{ds}{dt} = v = \alpha \sqrt{s} \]
- Now substituting back, we get:
\[ a_t = \frac{\alpha}{2\sqrt{s}} \cdot (\alpha \sqrt{s}) = \frac{\alpha^2}{2} \]

3. **Calculate Centripetal Acceleration**:
- The centripetal acceleration \( a_c \) is given by:
\[ a_c = \frac{v^2}{r} \]
- Substituting \( v = \alpha \sqrt{s} \):
\[ a_c = \frac{(\alpha \sqrt{s})^2}{r} = \frac{\alpha^2 s}{r} \]

4. **Combine the Accelerations**:
- The total acceleration \( a \) is the vector sum of \( a_t \) and \( a_c \). Since they are perpendicular, we can use the Pythagorean theorem:
\[ a = \sqrt{a_t^2 + a_c^2} \]
- Substituting the values:
\[ a = \sqrt{\left(\frac{\alpha^2}{2}\right)^2 + \left(\frac{\alpha^2 s}{r}\right)^2} \]
- Simplifying:
\[ a = \sqrt{\frac{\alpha^4}{4} + \frac{\alpha^4 s^2}{r^2}} = \alpha^2 \sqrt{\frac{1}{4} + \frac{s^2}{r^2}} \]

5. **Final Result**:
- The acceleration of the particle after traversing a distance \( s \) is:
\[ a = \alpha^2 \sqrt{\frac{1}{4} + \frac{s^2}{r^2}} \]