A point moves along an arc of a circle of radius R. Its velocity depends on the distance covered s as `v=sqrts`, where a is a constant. Find the angle `alpha` between the vector of the total acceleration and the vector of velocity as a function of s.

Explanation

To solve the problem, we need to find the angle \( \alpha \) between the total acceleration vector and the velocity vector of a point moving along an arc of a circle with radius \( R \). The velocity \( v \) is given as \( v = a \sqrt{s} \), where \( a \) is a constant and \( s \) is the distance covered.

### Step-by-Step Solution

1. **Identify the Components of Motion**:
- The point moves along a circular path, which means it has both tangential and normal components of acceleration.
- The tangential acceleration \( a_t \) is associated with the change in speed along the path, while the normal acceleration \( a_n \) is due to the change in direction.

2. **Calculate the Tangential Acceleration**:
- The tangential acceleration \( a_t \) can be found using the relationship:
\[ a_t = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = \frac{dv}{ds} \cdot v \]
- Given \( v = a \sqrt{s} \), we find \( \frac{dv}{ds} \):
\[ \frac{dv}{ds} = \frac{a}{2\sqrt{s}} \]
- Thus, substituting \( v \):
\[ a_t = \left(\frac{a}{2\sqrt{s}}\right) \cdot (a\sqrt{s}) = \frac{a^2}{2} \]

3. **Calculate the Normal Acceleration**:
- The normal acceleration \( a_n \) is given by:
\[ a_n = \frac{v^2}{R} \]
- Substituting \( v = a \sqrt{s} \):
\[ a_n = \frac{(a\sqrt{s})^2}{R} = \frac{a^2 s}{R} \]

4. **Total Acceleration**:
- The total acceleration \( \mathbf{a} \) can be expressed as:
\[ \mathbf{a} = a_t \hat{t} + a_n \hat{n} \]
- In terms of magnitudes:
\[ |\mathbf{a}| = \sqrt{a_t^2 + a_n^2} = \sqrt{\left(\frac{a^2}{2}\right)^2 + \left(\frac{a^2 s}{R}\right)^2} \]
- Simplifying:
\[ |\mathbf{a}| = \sqrt{\frac{a^4}{4} + \frac{a^4 s^2}{R^2}} = \frac{a^2}{2} \sqrt{1 + \frac{4s^2}{R^2}} \]

5. **Finding the Angle \( \alpha \)**:
- The angle \( \alpha \) between the total acceleration vector and the velocity vector can be found using the cosine of the angle:
\[ \cos \alpha = \frac{a_t}{|\mathbf{a}|} \]
- Substituting \( a_t \) and \( |\mathbf{a}| \):
\[ \cos \alpha = \frac{\frac{a^2}{2}}{\frac{a^2}{2} \sqrt{1 + \frac{4s^2}{R^2}}} = \frac{1}{\sqrt{1 + \frac{4s^2}{R^2}}} \]
- Therefore, the angle \( \alpha \) can be expressed as:
\[ \alpha = \cos^{-1}\left(\frac{1}{\sqrt{1 + \frac{4s^2}{R^2}}}\right) \]

6. **Final Expression**:
- We can also express \( \tan \alpha \) using the relationship:
\[ \tan \alpha = \frac{2s}{R} \]
- Thus, the angle \( \alpha \) can also be written as:
\[ \alpha = \tan^{-1}\left(\frac{2s}{R}\right) \]

### Summary
The angle \( \alpha \) between the total acceleration vector and the velocity vector as a function of \( s \) is given by:
\[ \alpha = \tan^{-1}\left(\frac{2s}{R}\right) \]