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. `7/3(hat(i)+hat(j))`
B. `2(hat(i)+hatj)`
C. `11/5(hat(i)+hat)j`
D. `1/5(13hat(i)+14hat(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 initial and final position vectors
At \(t = 0\) seconds, the position coordinates are:
\[ \vec{R_A} = (2 \, \text{m}, 3 \, \text{m}) = 2 \hat{i} + 3 \hat{j} \]
At \(t = 5\) seconds, the position coordinates are:
\[ \vec{R_C} = (13 \, \text{m}, 14 \, \text{m}) = 13 \hat{i} + 14 \hat{j} \]

### Step 2: Calculate the displacement vector
The displacement vector \(\vec{D}\) is given by:
\[ \vec{D} = \vec{R_C} - \vec{R_A} \]
Substituting the values:
\[ \vec{D} = (13 \hat{i} + 14 \hat{j}) - (2 \hat{i} + 3 \hat{j}) = (13 - 2) \hat{i} + (14 - 3) \hat{j} = 11 \hat{i} + 11 \hat{j} \]

### Step 3: Calculate the time interval
The time interval \(\Delta t\) is:
\[ \Delta t = t_f - t_i = 5 \, \text{s} - 0 \, \text{s} = 5 \, \text{s} \]

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

### 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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