91Ó°ÊÓ

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

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

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:

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

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

First find the displacement of the ion as:

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

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

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

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