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= betax^2`. Its velocity component in the x-direction is

Options

A. (a)variable
B. `(b)sqrt((2alpha)/beta)`
C. `(c)alpha/(2beta)`
D. `(d)sqrt(alpha/(2beta))`

Explanation

To solve the problem step by step, we will analyze the motion of the particle and derive the required velocity component in the x-direction.

### 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 relationship between the velocities, we will differentiate the equation of motion with respect to time \( t \):
\[ \frac{dy}{dt} = \frac{d}{dt}(\beta x^2) \]
Using the chain rule, we have:
\[ \frac{dy}{dt} = 2\beta x \frac{dx}{dt} \]
Here, \( \frac{dx}{dt} \) is the velocity in the x-direction, denoted as \( v_x \).

### Step 3: Differentiate Again to Find Acceleration
Next, we differentiate the equation \( \frac{dy}{dt} = 2\beta x v_x \) with respect to time \( t \) again:
\[ \frac{d^2y}{dt^2} = 2\beta \left( \frac{dx}{dt} v_x + x \frac{d^2x}{dt^2} \right) \]
Where \( \frac{d^2x}{dt^2} \) is the acceleration in the x-direction, denoted as \( a_x \).

### Step 4: Simplify the Expression
Since the problem states that the acceleration in the x-direction is zero (as the acceleration is only in the y-direction), we have:
\[ a_x = 0 \]
Thus, the equation simplifies to:
\[ \frac{d^2y}{dt^2} = 2\beta v_x^2 \]

### Step 5: Relate to the Given Acceleration in the y-Direction
The acceleration in the y-direction is given as \( \alpha \). Therefore, we can set:
\[ \alpha = 2\beta v_x^2 \]

### Step 6: Solve for the Velocity Component in the x-Direction
Rearranging the equation gives us:
\[ v_x^2 = \frac{\alpha}{2\beta} \]
Taking the square root to find \( v_x \):
\[ v_x = \sqrt{\frac{\alpha}{2\beta}} \]

### Final Answer
The velocity component in the x-direction is:
\[ v_x = \sqrt{\frac{\alpha}{2\beta}} \]

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