Question

A particle is moving such that its position coordinates $(x,y)$ are $(2m,3m)$ at time $t=0,(6m,7m)$ at time $t=2s$ and $(13m,14m)$ at time $t=5s$.
Average velocity vector $\left({\vec{v}}_{av}\right)$ from $t=0$ to $t=5s$ 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 $t=5s$, 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,
${\vec{v}}_{av}=\frac{\Delta \vec{r}}{\Delta t}=\frac{11\hat{i}+11\hat{j}}{5-0}=\frac{11}{5}(\hat{i}+\hat{j})$