If some function say x varies linearly with time and we want to find its average value in a given time interval we can directly find it by `(x_(i)+x_(f))/(2)`. Here, `x_(i)` is the initial value of x and `x_(f)` its final value y co-ordinates of a particle moving in x-y plane at some instant are : `x=2t^(2)` and `y=3//2 t^(2)`. The average velocity of particle in at time interval from `t=1 s` to `t=2s` is :-

Options

A. `(8hat(i)+5hat(j)) m//s`
B. `(12 hat(i)+9hat(j)) m//s`
C. `(6hat(i)+4.5 hat(j)) m//s`
D. `(10 hat(i)+6hat(j)) m//s`

Explanation

To find the average velocity of a particle moving in the x-y plane given the equations \( x = 2t^2 \) and \( y = \frac{3}{2}t^2 \) over the time interval from \( t = 1 \) s to \( t = 2 \) s, we can follow these steps:

### Step 1: Find the initial and final positions
We need to calculate the initial and final positions of the particle at \( t = 1 \) s and \( t = 2 \) s.

- **At \( t = 1 \) s:**
\[ x_i = 2(1)^2 = 2 \quad \text{and} \quad y_i = \frac{3}{2}(1)^2 = \frac{3}{2} \]
So, the initial position \( \mathbf{r_i} = (2, \frac{3}{2}) \).

- **At \( t = 2 \) s:**
\[ x_f = 2(2)^2 = 8 \quad \text{and} \quad y_f = \frac{3}{2}(2)^2 = 6 \]
So, the final position \( \mathbf{r_f} = (8, 6) \).

### Step 2: Calculate the displacement
The displacement \( \Delta \mathbf{r} \) is given by:
\[ \Delta \mathbf{r} = \mathbf{r_f} - \mathbf{r_i} = (8 - 2, 6 - \frac{3}{2}) = (6, 4.5) \]

### Step 3: Calculate the time interval
The time interval \( \Delta t \) is:
\[ \Delta t = t_f - t_i = 2 - 1 = 1 \text{ s} \]

### Step 4: Calculate the average velocity
The average velocity \( \mathbf{V_{avg}} \) is given by:
\[ \mathbf{V_{avg}} = \frac{\Delta \mathbf{r}}{\Delta t} = \frac{(6, 4.5)}{1} = (6, 4.5) \]

### Final Answer
Thus, the average velocity of the particle from \( t = 1 \) s to \( t = 2 \) s is:
\[ \mathbf{V_{avg}} = 6 \mathbf{i} + 4.5 \mathbf{j} \]