A particle is moving along x-y plane. Its x and y co-ordinates very with time as `x=2t^2 and y=t^3` Here, x abd y are in metres and t in seconds. Find average acceleration between a time interval from `t=0` to `t=2 s.`

Explanation

To find the average acceleration of the particle moving in the x-y plane, we need to follow these steps:

### Step 1: Determine the position vector
The position vector \( \mathbf{r} \) of the particle is given by:
\[ \mathbf{r}(t) = x(t) \hat{i} + y(t) \hat{j} = 2t^2 \hat{i} + t^3 \hat{j} \]

### Step 2: Calculate the velocity vector
The velocity vector \( \mathbf{v} \) is the derivative of the position vector with respect to time \( t \):
\[ \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(2t^2 \hat{i} + t^3 \hat{j}) = 4t \hat{i} + 3t^2 \hat{j} \]

### Step 3: Evaluate the velocity at the given times
Now, we need to find the velocity at \( t = 0 \) seconds and \( t = 2 \) seconds.

- For \( t = 0 \):
\[ \mathbf{v}(0) = 4(0) \hat{i} + 3(0)^2 \hat{j} = 0 \hat{i} + 0 \hat{j} = \mathbf{0} \]

- For \( t = 2 \):
\[ \mathbf{v}(2) = 4(2) \hat{i} + 3(2)^2 \hat{j} = 8 \hat{i} + 12 \hat{j} \]

### Step 4: Calculate the change in velocity
The change in velocity \( \Delta \mathbf{v} \) over the time interval from \( t = 0 \) to \( t = 2 \) seconds is:
\[ \Delta \mathbf{v} = \mathbf{v}(2) - \mathbf{v}(0) = (8 \hat{i} + 12 \hat{j}) - (0 \hat{i} + 0 \hat{j}) = 8 \hat{i} + 12 \hat{j} \]

### Step 5: Calculate the average acceleration
The average acceleration \( \mathbf{a}_{\text{avg}} \) is given by the change in velocity divided by the change in time:
\[ \mathbf{a}_{\text{avg}} = \frac{\Delta \mathbf{v}}{\Delta t} = \frac{8 \hat{i} + 12 \hat{j}}{2 - 0} = \frac{8 \hat{i} + 12 \hat{j}}{2} = 4 \hat{i} + 6 \hat{j} \]

### Final Answer
The average acceleration of the particle between \( t = 0 \) and \( t = 2 \) seconds is:
\[ \mathbf{a}_{\text{avg}} = 4 \hat{i} + 6 \hat{j} \, \text{m/s}^2 \]

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