A particle moves in a circular path such that its speed v varies with distance s as `v=alphasqrt(s)` where `alpha` is a positive constant. If the acceleration of the particle after traversing a distance s is `[alpha^2sqrt(x+s^2/R^2)]` find x.

Explanation

To solve the problem, we need to analyze the motion of a particle moving in a circular path with a varying speed. The speed \( v \) of the particle is given by the equation:

\[ v = \alpha \sqrt{s} \]

where \( \alpha \) is a positive constant and \( s \) is the distance traveled along the circular path. We are also given that the acceleration of the particle after traversing a distance \( s \) is:

\[ a = \alpha^2 \sqrt{x + \frac{s^2}{R^2}} \]

where \( R \) is the radius of the circular path. We need to find the value of \( x \).

### Step 1: Identify the components of acceleration
The total acceleration \( a \) of the particle can be broken down into two components:
1. **Radial acceleration** \( a_R \) (centripetal acceleration) which is directed towards the center of the circular path.
2. **Tangential acceleration** \( a_T \) which is due to the change in speed along the path.

The radial acceleration is given by:

\[ a_R = \frac{v^2}{R} \]

And the tangential acceleration can be found using the chain rule:

\[ a_T = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = \frac{dv}{ds} \cdot v \]

### Step 2: Calculate \( \frac{dv}{ds} \)
From the expression for \( v \):

\[ v = \alpha \sqrt{s} \]

Taking the derivative with respect to \( s \):

\[ \frac{dv}{ds} = \frac{\alpha}{2\sqrt{s}} \]

### Step 3: Calculate tangential acceleration \( a_T \)
Now substituting \( \frac{dv}{ds} \) and \( v \) into the expression for \( a_T \):

\[ a_T = \frac{\alpha}{2\sqrt{s}} \cdot (\alpha \sqrt{s}) = \frac{\alpha^2}{2} \]

### Step 4: Substitute into total acceleration
Now we can express the total acceleration \( a \) as the resultant of \( a_R \) and \( a_T \):

\[ a = \sqrt{a_R^2 + a_T^2} \]

Substituting \( a_R \) and \( a_T \):

\[ a_R = \frac{(\alpha \sqrt{s})^2}{R} = \frac{\alpha^2 s}{R} \]

Thus,

\[ a = \sqrt{\left(\frac{\alpha^2 s}{R}\right)^2 + \left(\frac{\alpha^2}{2}\right)^2} \]

### Step 5: Simplify the expression
Now we simplify the expression:

\[ a = \sqrt{\frac{\alpha^4 s^2}{R^2} + \frac{\alpha^4}{4}} = \alpha^2 \sqrt{\frac{s^2}{R^2} + \frac{1}{4}} \]

### Step 6: Compare with given acceleration
We know from the problem statement that:

\[ a = \alpha^2 \sqrt{x + \frac{s^2}{R^2}} \]

Equating the two expressions for acceleration:

\[ \alpha^2 \sqrt{\frac{s^2}{R^2} + \frac{1}{4}} = \alpha^2 \sqrt{x + \frac{s^2}{R^2}} \]

### Step 7: Square both sides
Squaring both sides gives:

\[ \frac{s^2}{R^2} + \frac{1}{4} = x + \frac{s^2}{R^2} \]

### Step 8: Solve for \( x \)
Subtracting \( \frac{s^2}{R^2} \) from both sides:

\[ \frac{1}{4} = x \]

Thus, we find:

\[ x = \frac{1}{4} \]

### Final Answer:
The value of \( x \) is:

\[ \boxed{\frac{1}{4}} \]