A particle is moving such that its position coordinates $(x, y)$ are $(2 \mathrm{~m}, 3 \mathrm{~m})

Question

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

Professor PC Thomas Classes · Answer

(d) : At time $t=0$, the position vector of the particle is $\vec{r}_{1}=2 \hat{i}+3 \hat{j}$.\ At time $t=5 \mathrm{~s}$, 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\ $\Delta \vec{r}=\vec{r}_{2}-\vec{r}_{1}=(13 \hat{i}+14 \hat{j})-(2 \hat{i}+3 \hat{j})=11 \hat{i}+11 \hat{j}$\ $	herefore \quad$ Average velocity, $\vec{v}_{\mathrm{av}}=\frac{\Delta \vec{r}}{\Delta t}=\frac{11 \hat{i}+11 \hat{j}}{5-0}=\frac{11}{5}(\hat{i}+\hat{j})$