A particle is moving such that its position coordinates $(x, y)$ are $(2 ext{ m}, 3 ext{ m})$ at time $t = 0

Question

A particle is moving such that its position coordinates $(x, y)$ are $(2	ext{ m}, 3	ext{ m})$ at time $t = 0$, $(6	ext{ m}, 7	ext{ m})$ at time $t = 2	ext{ s}$ and $(13	ext{ m}, 14	ext{ m})$ at time $t = 5	ext{ s}$. Average velocity vector ($\mathbf{v}_{av}$) from $t = 0$ to $t = 5	ext{ s}$ is:

Accepted Answer

1. **Definition:** Average velocity ($\mathbf{v}_{av}$) is defined as the ratio of the displacement ($\Delta \mathbf{r}$) to the corresponding time interval ($\Delta t$). It depends only on the initial and final positions, not the path taken or intermediate points .
   $$ \mathbf{v}_{av} = \frac{\mathbf{r}_2 - \mathbf{r}_1}{t_2 - t_1} $$
2. **Identify Coordinates:**
   - At $t_1 = 0$: Position vector $\mathbf{r}_1 = 2\hat{i} + 3\hat{j}$ meters.
   - At $t_2 = 5	ext{ s}$: Position vector $\mathbf{r}_2 = 13\hat{i} + 14\hat{j}$ meters.
   (Note: The position at $t=2	ext{ s}$ is irrelevant for the average velocity over the interval 0 to 5 s).
3. **Calculate Displacement ($\Delta \mathbf{r}$):**
   $$ \Delta \mathbf{r} = \mathbf{r}_2 - \mathbf{r}_1 = (13 - 2)\hat{i} + (14 - 3)\hat{j} = 11\hat{i} + 11\hat{j} 	ext{ m} $$
4. **Calculate Average Velocity:**
   $$ \mathbf{v}_{av} = \frac{11\hat{i} + 11\hat{j}}{5 - 0} = \frac{11}{5}(\hat{i} + \hat{j}) 	ext{ m/s} $$

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