A particle moves in a circular path such that its speed v varies with distance as `v=alphasqrts` where `alpha` is a positive constant. Find the acceleration of particle after traversing a distance S?

Options

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

Answer

To find the acceleration of a particle moving in a circular path with a speed that varies with distance, we start with the given relationship between speed and distance:

1. **Given Relationship**:
\[ v = \alpha \sqrt{s} \]
where \( \alpha \) is a positive constant.

2. **Finding the Tangential Acceleration**:
The tangential acceleration \( a_t \) can be found using the relationship:
\[ a_t = \frac{dv}{dt} \]
To find \( \frac{dv}{dt} \), we can use the chain rule:
\[ \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} \]
We first find \( \frac{dv}{ds} \):
\[ \frac{dv}{ds} = \frac{d}{ds}(\alpha \sqrt{s}) = \frac{\alpha}{2\sqrt{s}} \]
Now, since \( \frac{ds}{dt} = v = \alpha \sqrt{s} \), we can substitute:
\[ a_t = \frac{\alpha}{2\sqrt{s}} \cdot \alpha \sqrt{s} = \frac{\alpha^2}{2} \]

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

4. **Total Acceleration**:
The total acceleration \( a \) is the vector sum of the tangential and centripetal accelerations:
\[ a = \sqrt{a_t^2 + a_r^2} \]
Substituting \( a_t \) and \( a_r \):
\[ a = \sqrt{\left(\frac{\alpha^2}{2}\right)^2 + \left(\frac{\alpha^2 s}{r}\right)^2} \]
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**:
Therefore, the acceleration of the particle after traversing a distance \( s \) is:
\[ a = \alpha^2 \sqrt{\frac{1}{4} + \frac{s^2}{r^2}} \]