A particle id moving in xy - plane with `y = x//2 and v_x = 4 - 2t`. Choose the correct options.

Options

A. Initial velocities in x and y directions are negative.
B. Initial velocities in x and y directions are positive.
C. Motion is first retarded, then accelerated.
D. Motion is first accelerated, then retarded.

Explanation

To solve the problem step-by-step, we will analyze the motion of the particle in the xy-plane given the equations provided.

### Step 1: Understand the given equations
We have two equations:
1. The relationship between y and x:
\[ y = \frac{x}{2} \]
2. The velocity in the x-direction:
\[ v_x = 4 - 2t \]

### Step 2: Find the relationship between velocities
From the equation \(y = \frac{x}{2}\), we can differentiate both sides with respect to time \(t\):
\[ \frac{dy}{dt} = \frac{1}{2} \frac{dx}{dt} \]
Here, \(\frac{dx}{dt}\) is the velocity in the x-direction (\(v_x\)) and \(\frac{dy}{dt}\) is the velocity in the y-direction (\(v_y\)). Thus, we can write:
\[ v_y = \frac{1}{2} v_x \]

### Step 3: Substitute the expression for \(v_x\)
Substituting \(v_x = 4 - 2t\) into the equation for \(v_y\):
\[ v_y = \frac{1}{2}(4 - 2t) = 2 - t \]

### Step 4: Analyze initial velocities
At time \(t = 0\):
- For \(v_x\):
\[ v_x = 4 - 2(0) = 4 \, \text{m/s} \]
- For \(v_y\):
\[ v_y = 2 - 0 = 2 \, \text{m/s} \]

Thus, the initial velocities are:
- \(u_x = 4 \, \text{m/s}\)
- \(u_y = 2 \, \text{m/s}\)

### Step 5: Determine accelerations
Next, we will find the acceleration in both x and y directions by differentiating the velocities with respect to time.

1. For \(v_x\):
\[ a_x = \frac{dv_x}{dt} = \frac{d(4 - 2t)}{dt} = -2 \, \text{m/s}^2 \]
2. For \(v_y\):
\[ a_y = \frac{dv_y}{dt} = \frac{d(2 - t)}{dt} = -1 \, \text{m/s}^2 \]

### Step 6: Analyze the motion
Both accelerations are negative, indicating that the motion is retarding in both x and y directions. Since both accelerations are constant and negative, the motion is uniformly retarded.

### Step 7: Conclusion
Based on the analysis, we can conclude:
- The initial velocities are positive.
- The motion is uniformly retarded in both directions.

Thus, the correct option is option 2.