The co-ordinate of the particle in x-y plane are given as `x=2+2t+4t^(2)` and `y=4t+8t^(2)` :-
The motion of the particle is :-

Options

A. along a straight line
B. uniformly accelerated
C. along a parabolic path
D. non-uniformly accelerated

Answer

To analyze the motion of the particle given the coordinates \( x = 2 + 2t + 4t^2 \) and \( y = 4t + 8t^2 \), we will follow these steps:

### Step 1: Differentiate the x-coordinate with respect to time
The x-coordinate is given by:
\[ x = 2 + 2t + 4t^2 \]
To find the velocity in the x-direction, we differentiate \( x \) with respect to \( t \):
\[ \frac{dx}{dt} = \frac{d}{dt}(2 + 2t + 4t^2) = 0 + 2 + 8t = 2 + 8t \]

### Step 2: Differentiate the y-coordinate with respect to time
The y-coordinate is given by:
\[ y = 4t + 8t^2 \]
To find the velocity in the y-direction, we differentiate \( y \) with respect to \( t \):
\[ \frac{dy}{dt} = \frac{d}{dt}(4t + 8t^2) = 4 + 16t \]

### Step 3: Differentiate the velocities to find acceleration
Now, we differentiate the velocity in the x-direction to find the acceleration:
\[ \frac{d^2x}{dt^2} = \frac{d}{dt}(2 + 8t) = 0 + 8 = 8 \, \text{m/s}^2 \]

Next, we differentiate the velocity in the y-direction to find the acceleration:
\[ \frac{d^2y}{dt^2} = \frac{d}{dt}(4 + 16t) = 0 + 16 = 16 \, \text{m/s}^2 \]

### Step 4: Analyze the motion
Since both the x-acceleration and y-acceleration are constant (independent of time), we conclude that the motion of the particle is uniformly accelerated.

### Conclusion
The correct option regarding the motion of the particle is:
**Uniformly accelerated motion.**

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