A particle is moving such that its position coordinates `(x, y)` are `(2m, 3m)` at time `t=0, (6m, 7m)` at time `t=2 s`, and `(13 m, 14m)` at time `t=5 s`.
Average velocity vector`(vec(V)_(av))` from `t=0` to `t=5 s` is

Options

A. `(1)/(5) (13 hati +14 hatj)`
B. `(7)/(3)(hati +hatj)`
C. `2(hati - hatj)`
D. `(11)/(5)(hati +hatj)`

Explanation

To find the average velocity vector \(\vec{V}_{av}\) of the particle from \(t = 0\) to \(t = 5\) seconds, we can follow these steps:

### Step 1: Identify the initial and final position vectors
The initial position vector \(\vec{r}_{initial}\) at \(t = 0\) seconds is given as:
\[ \vec{r}_{initial} = 2 \hat{i} + 3 \hat{j} \quad \text{(in meters)} \]
The final position vector \(\vec{r}_{final}\) at \(t = 5\) seconds is given as:
\[ \vec{r}_{final} = 13 \hat{i} + 14 \hat{j} \quad \text{(in meters)} \]

### Step 2: Calculate the net displacement
The net displacement \(\Delta \vec{r}\) can be calculated as:
\[ \Delta \vec{r} = \vec{r}_{final} - \vec{r}_{initial} \]
Substituting the values:
\[ \Delta \vec{r} = (13 \hat{i} + 14 \hat{j}) - (2 \hat{i} + 3 \hat{j}) \]
\[ \Delta \vec{r} = (13 - 2) \hat{i} + (14 - 3) \hat{j} = 11 \hat{i} + 11 \hat{j} \]

### Step 3: Calculate the time interval
The time interval \(\Delta t\) from \(t = 0\) to \(t = 5\) seconds is:
\[ \Delta t = t_{final} - t_{initial} = 5 \, \text{s} - 0 \, \text{s} = 5 \, \text{s} \]

### Step 4: Calculate the average velocity vector
The average velocity vector \(\vec{V}_{av}\) is given by the formula:
\[ \vec{V}_{av} = \frac{\Delta \vec{r}}{\Delta t} \]
Substituting the values:
\[ \vec{V}_{av} = \frac{11 \hat{i} + 11 \hat{j}}{5} \]
This can be simplified to:
\[ \vec{V}_{av} = \frac{11}{5} \hat{i} + \frac{11}{5} \hat{j} \]

### Final Answer
Thus, the average velocity vector from \(t = 0\) to \(t = 5\) seconds is:
\[ \vec{V}_{av} = \frac{11}{5} \hat{i} + \frac{11}{5} \hat{j} \]

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