A particle starts from the origin of coordinates at time `t = 0` and moves in the xy plane with a constant acceleration `alpha` in the y-direction. Its equation of motion is `y = beta x^(2)`. Its velocity component in the x-directon is

Options

A. Variable
B. `sqrt((2alpha)/(beta))`
C. `(alpha)/(2beta)`
D. `sqrt((alpha)/(2beta))`

Explanation

To solve the problem step by step, we need to analyze the motion of the particle given its trajectory and the acceleration.

### Step 1: Understand the given information
The particle starts from the origin (0,0) at time \( t = 0 \) and moves in the xy-plane with a constant acceleration \( \alpha \) in the y-direction. The equation of motion is given as:
\[ y = \beta x^2 \]

### Step 2: Differentiate the equation of motion
To find the velocity components, we first differentiate the equation of motion with respect to time \( t \):
\[ \frac{dy}{dt} = \frac{d}{dt}(\beta x^2) \]

Using the chain rule, we get:
\[ \frac{dy}{dt} = 2\beta x \frac{dx}{dt} \]
Let \( v_x = \frac{dx}{dt} \) be the velocity component in the x-direction. Thus, we can rewrite the equation as:
\[ v_y = 2\beta x v_x \]
where \( v_y = \frac{dy}{dt} \).

### Step 3: Differentiate again to find acceleration
Next, we differentiate \( v_y \) with respect to time \( t \) to find the acceleration in the y-direction:
\[ \frac{dv_y}{dt} = \frac{d}{dt}(2\beta x v_x) \]

Using the product rule, we have:
\[ \frac{dv_y}{dt} = 2\beta \left( v_x \frac{dx}{dt} + x \frac{dv_x}{dt} \right) \]
Since \( \frac{dx}{dt} = v_x \), we can substitute:
\[ \frac{dv_y}{dt} = 2\beta (v_x^2 + x \frac{dv_x}{dt}) \]

### Step 4: Relate acceleration to given information
The acceleration in the y-direction is given as \( \alpha \), so we can set:
\[ \frac{dv_y}{dt} = \alpha \]

Thus, we have:
\[ \alpha = 2\beta (v_x^2 + x \frac{dv_x}{dt}) \]

### Step 5: Solve for \( v_x \)
Since we need to find the expression for \( v_x \), we can rearrange the equation:
\[ \alpha = 2\beta v_x^2 + 2\beta x \frac{dv_x}{dt} \]

Assuming that \( x \) is a function of time, we can express \( x \) in terms of \( t \) and substitute it back into the equation. However, we can also find a simpler relationship by considering the motion in the x-direction.

### Step 6: Use the relationship between acceleration and velocity
From the equation of motion, we can express the acceleration in terms of the known quantities:
\[ v_x^2 = \frac{\alpha}{2\beta} \]

Taking the square root gives us:
\[ v_x = \sqrt{\frac{\alpha}{2\beta}} \]

### Conclusion
Thus, the velocity component in the x-direction is:
\[ v_x = \sqrt{\frac{\alpha}{2\beta}} \]