A particle is moving such that its position coordinates (x, y) are (2 m, 3 m) at time t = 0, (6 m, 7 m) at tim

Question

A particle is moving such that its position coordinates (x, y) are (2 m, 3 m) at time t = 0, (6 m, 7 m) at time t = 2 s, and (13 m, 14 m) at time t = 5 s. The average velocity vector v_avg from t = 0 to t = 5 s is:

Accepted Answer

Average velocity is defined as the displacement divided by the corresponding time interval .
$$ \vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} = \frac{\vec{r}_f - \vec{r}_i}{t_f - t_i} $$

1.  **Identify Initial and Final Positions:**
    *   At $t_1 = 0$ s, position vector $\vec{r}_1 = 2\hat{i} + 3\hat{j}$ m.
    *   At $t_2 = 5$ s, position vector $\vec{r}_2 = 13\hat{i} + 14\hat{j}$ m.

2.  **Calculate Displacement ($ \Delta \vec{r} $):**
    $$ \Delta \vec{r} = \vec{r}_2 - \vec{r}_1 = (13\hat{i} + 14\hat{j}) - (2\hat{i} + 3\hat{j}) $$
    $$ \Delta \vec{r} = (13 - 2)\hat{i} + (14 - 3)\hat{j} = 11\hat{i} + 11\hat{j} 	ext{ m} $$

3.  **Calculate Time Interval ($ \Delta t $):**
    $$ \Delta t = 5 - 0 = 5 	ext{ s} $$

4.  **Calculate Average Velocity:**
    $$ \vec{v}_{avg} = \frac{11\hat{i} + 11\hat{j}}{5} = \frac{11}{5}(\hat{i} + \hat{j}) 	ext{ m/s} $$

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