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 step by step, we need to analyze the motion of the particle moving in a circular path with a varying speed. The speed \( v \) is given by the equation \( v = \alpha \sqrt{s} \), where \( \alpha \) is a positive constant. We also know the expression for the acceleration of the particle after traversing a distance \( s \) is given as \( a = \alpha^2 \sqrt{x + \frac{s^2}{R^2}} \). Our goal is to find the value of \( x \).

### Step 1: Understanding the Components of Acceleration
The total acceleration \( a \) of a particle moving in a circular path can be expressed as the vector sum of tangential acceleration \( a_t \) and radial (centripetal) acceleration \( a_r \):
\[ a = \sqrt{a_t^2 + a_r^2} \]

### Step 2: Finding Tangential Acceleration
The tangential acceleration \( a_t \) can be defined as the rate of change of velocity with respect to time:
\[ a_t = \frac{dv}{dt} \]
Given \( v = \alpha \sqrt{s} \), we differentiate \( v \) with respect to \( t \):
\[ \frac{dv}{dt} = \frac{d}{dt}(\alpha \sqrt{s}) = \alpha \cdot \frac{1}{2\sqrt{s}} \cdot \frac{ds}{dt} \]
Since \( \frac{ds}{dt} = v = \alpha \sqrt{s} \), we substitute this into the equation:
\[ \frac{dv}{dt} = \alpha \cdot \frac{1}{2\sqrt{s}} \cdot (\alpha \sqrt{s}) = \frac{\alpha^2}{2} \]

### Step 3: Finding Radial Acceleration
The radial (centripetal) acceleration \( a_r \) is given by:
\[ a_r = \frac{v^2}{R} \]
Substituting \( v = \alpha \sqrt{s} \):
\[ a_r = \frac{(\alpha \sqrt{s})^2}{R} = \frac{\alpha^2 s}{R} \]

### Step 4: Total Acceleration
Now, we can express the total acceleration \( a \) as:
\[ a = \sqrt{a_t^2 + a_r^2} = \sqrt{\left(\frac{\alpha^2}{2}\right)^2 + \left(\frac{\alpha^2 s}{R}\right)^2} \]
This simplifies to:
\[ a = \sqrt{\frac{\alpha^4}{4} + \frac{\alpha^4 s^2}{R^2}} = \alpha^2 \sqrt{\frac{1}{4} + \frac{s^2}{R^2}} \]

### Step 5: Equating to Given Acceleration
We know from the problem statement that:
\[ a = \alpha^2 \sqrt{x + \frac{s^2}{R^2}} \]
Setting the two expressions for acceleration equal to each other:
\[ \alpha^2 \sqrt{\frac{1}{4} + \frac{s^2}{R^2}} = \alpha^2 \sqrt{x + \frac{s^2}{R^2}} \]
Dividing both sides by \( \alpha^2 \) (since \( \alpha \) is positive):
\[ \sqrt{\frac{1}{4} + \frac{s^2}{R^2}} = \sqrt{x + \frac{s^2}{R^2}} \]

### Step 6: Squaring Both Sides
Squaring both sides gives:
\[ \frac{1}{4} + \frac{s^2}{R^2} = x + \frac{s^2}{R^2} \]
Subtracting \( \frac{s^2}{R^2} \) from both sides results in:
\[ \frac{1}{4} = x \]

### Conclusion
Thus, the value of \( x \) is:
\[ \boxed{\frac{1}{4}} \]