A particle moves in a circular path such that its speed `1v` varies with distance `s` as `v = sqrt(s)`, where `prop` is a positive constant. Find the acceleration of the particle after traversing a distance `s`.

Explanation

To find the acceleration of a particle moving in a circular path where its speed varies with distance as \( v = \sqrt{s} \), we need to consider both the centripetal and tangential components of acceleration.

### Step-by-Step Solution:

1. **Identify the given relationship**:
The speed \( v \) of the particle is given by:
\[
v = \alpha \sqrt{s}
...