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)(13hat(i)+14hat(j))`
B. `(7)/(3)(hat(i)+hat(j))`
C. `(hat(i)+hat(j))`
D. `(11)/(5)(hat(i)+hat(j))`

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 position vectors at the given times
At \(t=0\):
\[ \vec{r_1} = 2 \hat{i} + 3 \hat{j} \quad \text{(in meters)} \]

At \(t=2\):
\[ \vec{r_2} = 6 \hat{i} + 7 \hat{j} \quad \text{(in meters)} \]

At \(t=5\):
\[ \vec{r_3} = 13 \hat{i} + 14 \hat{j} \quad \text{(in meters)} \]

### Step 2: Calculate the displacement vector \(\Delta \vec{r}\)
The displacement vector from \(t=0\) to \(t=5\) seconds is given by:
\[ \Delta \vec{r} = \vec{r_3} - \vec{r_1} \]
Substituting the values:
\[ \Delta \vec{r} = (13 \hat{i} + 14 \hat{j}) - (2 \hat{i} + 3 \hat{j}) \]
Calculating this gives:
\[ \Delta \vec{r} = (13 - 2) \hat{i} + (14 - 3) \hat{j} = 11 \hat{i} + 11 \hat{j} \]

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

### Step 4: Calculate the average velocity vector \(\vec{V}_{av}\)
The average velocity vector is given by:
\[ \vec{V}_{av} = \frac{\Delta \vec{r}}{\Delta t} \]
Substituting the values we found:
\[ \vec{V}_{av} = \frac{11 \hat{i} + 11 \hat{j}}{5} \]
This simplifies 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:
\[ \vec{V}_{av} = \frac{11}{5} \hat{i} + \frac{11}{5} \hat{j} \text{ m/s} \]

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