A particle is moving in `x-y` plane with `y=x/2` and `V_(x)=4-2t`. The displacement versus time graph of the particle would be

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Explanation

To solve the problem, we need to find the displacement versus time graph for a particle moving in the x-y plane with the given conditions. Let's break it down step by step.

### Step 1: Understand the Motion in the x-y Plane
The particle's motion is described by the equation \( y = \frac{x}{2} \). This indicates that the y-coordinate is always half of the x-coordinate.

### Step 2: Analyze the Velocity in the x-direction
We are given the velocity in the x-direction as:
\[ V_x = 4 - 2t \]
This represents how the x-component of the velocity changes with time.

### Step 3: Relate Velocity to Displacement
The velocity in the x-direction can be expressed as:
\[ V_x = \frac{dx}{dt} \]
Thus, we can write:
\[ \frac{dx}{dt} = 4 - 2t \]

### Step 4: Integrate to Find Displacement in the x-direction
To find the displacement \( x(t) \), we integrate the equation:
\[ dx = (4 - 2t) dt \]
Integrating both sides from \( t = 0 \) to \( t \) and \( x = 0 \) to \( x \):
\[ \int dx = \int (4 - 2t) dt \]
This gives:
\[ x = 4t - t^2 + C \]
Assuming the initial condition \( x(0) = 0 \), we find \( C = 0 \). Thus:
\[ x(t) = 4t - t^2 \]

### Step 5: Find the Corresponding y-coordinate
Using the relationship \( y = \frac{x}{2} \):
\[ y(t) = \frac{1}{2}(4t - t^2) = 2t - \frac{t^2}{2} \]

### Step 6: Find the Displacement
The total displacement \( D \) from the origin can be calculated using the Pythagorean theorem:
\[ D = \sqrt{x^2 + y^2} \]
Substituting \( x(t) \) and \( y(t) \):
\[ D = \sqrt{(4t - t^2)^2 + \left(2t - \frac{t^2}{2}\right)^2} \]

### Step 7: Simplify the Expression
Calculating \( D \):
1. Calculate \( (4t - t^2)^2 = 16t^2 - 8t^3 + t^4 \)
2. Calculate \( \left(2t - \frac{t^2}{2}\right)^2 = 4t^2 - 2t^3 + \frac{t^4}{4} \)

Now combine:
\[ D^2 = (16t^2 - 8t^3 + t^4) + (4t^2 - 2t^3 + \frac{t^4}{4}) \]
\[ D^2 = 20t^2 - 10t^3 + \frac{5t^4}{4} \]
Thus, the displacement \( D \) is:
\[ D = \sqrt{20t^2 - 10t^3 + \frac{5t^4}{4}} \]

### Step 8: Analyze the Displacement vs Time Graph
To understand the displacement versus time graph, we observe how \( D \) changes with \( t \). The leading term is \( 20t^2 \), which indicates that the displacement will initially increase quadratically with time, but the presence of the cubic and quartic terms will affect the curvature of the graph as \( t \) increases.

### Summary of the Solution
The displacement versus time graph will start at the origin and initially rise, following a quadratic trend, but will eventually curve due to the influence of the cubic and quartic terms.