Velocity and acceleration of a particle are `v=(2 hati) m/s` and `a = (4t hati+t^2 hatj) m /s^2` where, t is the time. Which type of motion is this ?

Explanation

To determine the type of motion of the particle given its velocity and acceleration, we can analyze the equations step by step.

### Step 1: Identify the given quantities
The velocity \( v \) of the particle is given as:
\[ v = 2 \hat{i} \, \text{m/s} \]
The acceleration \( a \) of the particle is given as:
\[ a = (4t \hat{i} + t^2 \hat{j}) \, \text{m/s}^2 \]

### Step 2: Analyze the velocity vector
The velocity vector \( v = 2 \hat{i} \) indicates that the particle is moving in the positive x-direction with a constant speed of 2 m/s. There is no component in the y-direction.

### Step 3: Analyze the acceleration vector
The acceleration vector \( a = 4t \hat{i} + t^2 \hat{j} \) shows that the acceleration has two components:
- In the x-direction: \( 4t \hat{i} \) which varies linearly with time \( t \).
- In the y-direction: \( t^2 \hat{j} \) which varies quadratically with time \( t \).

### Step 4: Determine the nature of motion
Since the acceleration is a function of time and has components in both the x and y directions, the motion of the particle is not uniform. The x-component of acceleration is increasing linearly with time, while the y-component is increasing quadratically with time.

### Step 5: Conclusion on the type of motion
Given that the acceleration is not constant and varies with time, the motion of the particle is classified as **non-uniform motion**. The displacement will not follow a simple parabolic path, and the velocity will not remain constant.

### Final Answer
The type of motion of the particle is **non-uniform motion**.