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 $t=5 \, s$.
Average velocity vector $\left(\vec{v}_{ av }\right)$ from $t=0$ to $t=5\, s$ is

• A. $ \frac{ 1}{ 5} (13 \hat{ i} + 14 \hat{j}) $
• B. $ \frac{ 7}{ 3} ( \hat{i} + \hat{j})$
• C. $2 ( \hat{i} + \hat{j})$
• D. $ \frac{11}{ 5} ( \hat{i} + \hat{j})$

Answer 1

Answer: (D)

At time $t=0$, the position vector of the particle is
$\vec{r}_{ 1 }=2 \hat{i}+3 \hat{j}$
At time $t=5 s$, the position vector of the particle is
$\vec{r}_{2}=13 \hat{i}+14 \hat{j}$
Displacement from $\vec{r}_{1}$ to $\vec{r}_{2}$ is
$\Delta \vec{r}=\vec{r}_{2}-\vec{r}_{1}=(13 \hat{i}+14 \hat{j})-(2 \hat{i}+3 \hat{j}) $
$= 11 \hat{i}+11 \hat{j} $
$\therefore$ Average velocity,
$ \vec{v}_{ av }=\frac{\Delta \vec{r}}{\Delta t}=\frac{11 \hat{i}+11 \hat{j}}{5-0}=\frac{11}{5}(\hat{i}+\hat{j})$