An ion’s position vector is initially $\overrightarrow{\mathbf{r}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\hat{\mathbf{i}}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$,and 10 slater

it is $\overrightarrow{\mathbf{r}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\hat{\mathbf{j}}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$, all in meters. In unit vector notation, what is its${\overrightarrow{\mathbf{v}}}_{\mathbf{a}\mathbf{v}\mathbf{g}}$during the 10 s ?

Initial position vector,${\overrightarrow{r}}_0=5.0\hat{i}-6.0\hat{j}+2.0\hat{k}$

Final position vector,$\overrightarrow{r}=-2.0\hat{i}+8.0\hat{j}-2.0\hat{k}$

Time interval,$∆\mathrm{t}=10\mathrm{s}$

The average velocity may be defined as the ratio of total displacement to the total time interval for the displacement to occur.It is a vector quantity, which has magnitude as well as direction.

The expression for the average velocity is given as:

${\overrightarrow{v}}_{avg}=\frac{∆\overrightarrow{r}}{∆t}$ ...(i)

Here, $∆\overrightarrow{r}$is the displacement and $∆t$is the time interval.

First find the displacement of the ion as:

$$\begin{gathered} ∆\overrightarrow{r}=\overrightarrow{r}-{\overrightarrow{r}}_0 \\ =\left(-2.0\hat{\mathrm{i}}+8.0\hat{\mathrm{j}}-2.0\hat{\mathrm{k}}\right)-\left(5.0\hat{\mathrm{i}}+6.0\hat{\mathrm{j}}-2.0\hat{\mathrm{k}}\right) \\ =-7.0\hat{\mathrm{i}}+14.0\hat{\mathrm{j}}-4.0\hat{\mathrm{k}} \end{gathered}$$

Using equation (i), the average velocity is calculated as:

$$\begin{gathered} {\overrightarrow{v}}_{avg}=\frac{\left(-7.0\hat{\mathrm{i}}+14.0\hat{\mathrm{j}}-4.0\hat{\mathrm{k}}\right)\mathrm{m}}{10\mathrm{s}} \\ =\left(-0.7\hat{\mathrm{i}}+1.4.\hat{\mathrm{j}}-0.4\hat{\mathrm{k}}\right)\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the average velocity of ion is $\left(-0.7\hat{\mathrm{i}}+1.4.\hat{\mathrm{j}}-0.4\hat{\mathrm{k}}\right)\mathrm{m}/\mathrm{s}$.