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Let's analyze the problem step-by-step.

### Problem Statement Analysis
A particle is moving along the positive \( x \)-axis with a velocity given by:
\[
v = \alpha \sqrt{x}
\]
where \( \alpha \) is a constant, and \( x \) represents the position. We are given:
1. At \( t = 0 \), \( x = 0 \).
2. \( \alpha \) is a positive constant.

The problem has two parts:

#### Part (a)
Find the velocity \( v \) and acceleration \( a \) as functions of time \( t \).

#### Part (b)
Calculate the particle's average velocity over the time interval in which it travels a distance \( x = 2 \) from its starting point.

---

### Solution

#### Part (a): Finding Velocity and Acceleration as Functions of Time
Given:
\[
v = \frac{dx}{dt} = \alpha \sqrt{x}
\]
To find \( x(t) \), we separate variables and integrate.

1. Rewrite the equation as:
   \[
   \frac{dx}{\sqrt{x}} = \alpha \, dt
   \]
2. Integrate both sides:
   \[
   \int \frac{dx}{\sqrt{x}} = \int \alpha \, dt
   \]
   The left side integral becomes \( 2\sqrt{x} \), and the right side is \( \alpha t + C \):
   \[
   2\sqrt{x} = \alpha t + C
   \]
3. Apply the initial condition \( x = 0 \) when \( t = 0 \):
   \[
   2\sqrt{0} = \alpha \cdot 0 + C \Rightarrow C = 0
   \]
   Therefore:
   \[
   2\sqrt{x} = \alpha t
   \]
4. Solve for \( x \):
   \[
   x = \frac{\alpha^2 t^2}{4}
   \]

5. Now, substitute \( x(t) \) back to find \( v(t) \):
   \[
   v = \frac{dx}{dt} = \frac{d}{dt} \left( \frac{\alpha^2 t^2}{4} \right) = \frac{\alpha^2 t}{2}
   \]

6. To find acceleration \( a(t) \):
   \[
   a = \frac{dv}{dt} = \frac{d}{dt} \left( \frac{\alpha^2 t}{2} \right) = \frac{\alpha^2}{2}
   \]

So, the velocity and acceleration as functions of time are:
\[
v(t) = \frac{\alpha^2 t}{2} \quad \text{and} \quad a(t) = \frac{\alpha^2}{2}
\]

#### Part (b): Finding the Average Velocity
The average velocity over a distance of \( x = 2 \) is given by:
\[
\text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}}
\]
1. Set \( x = 2 \) in \( x = \frac{\alpha^2 t^2}{4} \):
   \[
   2 = \frac{\alpha^2 t^2}{4} \Rightarrow t^2 = \frac{8}{\alpha^2} \Rightarrow t = \frac{2\sqrt{2}}{\alpha}
   \]
2. The displacement is \( 2 \), so:
   \[
   \text{Average velocity} = \frac{2}{\frac{2\sqrt{2}}{\alpha}} = \frac{\alpha}{\sqrt{2}}
   \]

---

### Summary
- **(a)** \( v(t) = \frac{\alpha^2 t}{2} \) and \( a(t) = \frac{\alpha^2}{2} \).
- **(b)** Average velocity over the distance \( x = 2 \) is \( \frac{\alpha}{\sqrt{2}} \).

Would you like more details or have any specific questions?

---

### Further Questions
1. How can we generalize this approach for a different initial position?
2. What would change if \( \alpha \) were negative?
3. How would the solution differ for a time-dependent \( \alpha(t) \)?
4. What if the velocity were proportional to \( x^n \) instead of \( \sqrt{x} \)?
5. How can we verify the consistency of these results with dimensional analysis?

### Tip
When dealing with motion equations involving variable velocity, try to separate variables and integrate. It simplifies finding \( x(t) \) or \( v(t) \) effectively.

A particle moves along the positive x-axis with a velocity given by \( v = \alpha \sqrt{x} \), where \( \alpha \) is a positive constant. Given that at \( t = 0 \), \( x = 0 \), find the following:

(a) Express velocity and acceleration as functions of time.
(b) Determine the average velocity of the particle over the time interval in which it travels a distance of \( x = 2 \) from the starting point.

This problem explores a particle moving along the positive x-axis with a velocity \( v = \alpha \sqrt{x} \). Learn how to find the velocity and acceleration as functions of time and calculate the average velocity over a given distance. Suitable for students studying calculus and kinematics.

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