A particle starts from the origin of co-ordinates 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

Options

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

Answer

To solve the problem step by step, we need to analyze the motion of the particle given its equation of motion and the conditions provided.

### 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 component in the x-direction, we need to differentiate the equation of motion with respect to time \( t \):
\[ y = \beta x^2 \]
Differentiating both sides with respect to \( t \):
\[ \frac{dy}{dt} = 2\beta x \frac{dx}{dt} \]
Let \( v_x = \frac{dx}{dt} \) (the velocity in the x-direction) and \( v_y = \frac{dy}{dt} \) (the velocity in the y-direction). Thus, we have:
\[ v_y = 2\beta x v_x \]

### Step 3: Find the acceleration in the y-direction
The acceleration in the y-direction is given as \( \alpha \). We can express the acceleration as the second derivative of \( y \) with respect to time:
\[ \frac{d^2y}{dt^2} = \alpha \]
Differentiating \( v_y = 2\beta x v_x \) with respect to time \( t \):
\[ \frac{d^2y}{dt^2} = 2\beta \left( \frac{dx}{dt} \cdot \frac{dx}{dt} + x \cdot \frac{d^2x}{dt^2} \right) \]
This can be rewritten as:
\[ \frac{d^2y}{dt^2} = 2\beta v_x^2 + 2\beta x \frac{d^2x}{dt^2} \]

### Step 4: Set the acceleration in the y-direction equal to \( \alpha \)
Since the acceleration in the x-direction is zero (as given), we have:
\[ \frac{d^2x}{dt^2} = 0 \]
Thus, the equation simplifies to:
\[ \alpha = 2\beta v_x^2 \]

### Step 5: Solve for \( v_x \)
Rearranging the equation gives:
\[ v_x^2 = \frac{\alpha}{2\beta} \]
Taking the square root, we find:
\[ v_x = \sqrt{\frac{\alpha}{2\beta}} \]

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