Figure $\mathbf{4}\mathbf{-}\mathbf{30}$gives the path of a squirrel moving about on level ground, from point A (at time$\mathit{t}\mathbf{=}\mathbf{0}$), to points B (at $\mathit{t}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{min}$), C (at $\mathit{t}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{min}$), and finally D (at $\mathit{t}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{min}$). Consider the average velocities of the squirrel from point A to each of the other three points. Of them, what are the (a) magnitude and (b) angle of the one with the least magnitude and the (c) magnitude and (d) angle of the one with the greatest magnitude?

From the figure,

The position vector of point A, $\stackrel{\rightharpoonup }{{\mathrm{r}}_{\mathrm{A}}}=\left(15\mathrm{m}\right)\hat{\mathrm{i}}+(-15\mathrm{m})\overbrace{\mathrm{j}}$

The position vector of point B, $\stackrel{\rightharpoonup }{{\mathrm{r}}_B}=\left(30\mathrm{m}\right)\hat{\mathrm{i}}+(-45\mathrm{m})\overbrace{\mathrm{j}}$

The position vector of point C, $\stackrel{\rightharpoonup }{{\mathrm{r}}_C}=\left(20\mathrm{m}\right)\hat{\mathrm{i}}+(-15\mathrm{m})\overbrace{\mathrm{j}}$

The position vector of point D, $\stackrel{\rightharpoonup }{{\mathrm{r}}_D}=\left(45\mathrm{m}\right)\hat{\mathrm{i}}+\left(45\mathrm{m}\right)\overbrace{\mathrm{j}}$

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:

${\overrightarrow{v}}_{avg}=\frac{\nabla \overrightarrow{r}}{\nabla t}$ … (i)

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

The expression to determine the angle of velocity is given as:

$\theta ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)$ … (ii)

Here, $v_x$and $v_y$are the x and y components of velocity.

The displacement from point A to point B is,

$$\begin{gathered} \nabla {\overrightarrow{\mathrm{r}}}_{\mathrm{AB}}={\overrightarrow{\mathrm{r}}}_{\mathrm{B}}-\overrightarrow{{\mathrm{r}}_{\mathrm{A}}} \\ =\left\{\left(30\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(-45\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}}\right\}-\left\{\left(15\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(-15\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}}\right\} \\ =\left(15\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(-30\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}} \end{gathered}$$

Using equation (i), the average velocity is,

$$\begin{gathered} {\overrightarrow{\mathrm{v}}}_{\mathrm{avg}}=\frac{\left(15\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(-30\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}}}{5.0\times 60\mathrm{s}} \\ =\left(0.05\mathrm{m}/\mathrm{s}\right)\stackrel{⏜}{\mathrm{i}}+\left(-0.10\mathrm{m}/\mathrm{s}\right)\stackrel{⏜}{\mathrm{j}} \end{gathered}$$

The magnitude of average velocity is,

$$\begin{gathered} \left|{\mathrm{v}}_{\mathrm{avg}}\right|=\sqrt{\left(0.05\mathrm{cm}/\mathrm{s}\right)+{\left(-10\mathrm{cm}/\mathrm{s}\right)}^2} \\ =11.2\mathrm{cm}/\mathrm{s} \end{gathered}$$

The displacement from point A to point C is,

$$\begin{gathered} \nabla {\stackrel{⏜}{\mathrm{r}}}_{\mathrm{AC}}={\overrightarrow{\mathrm{r}}}_{\mathrm{C}}-{\overrightarrow{\mathrm{r}}}_{\mathrm{a}} \\ =\left\{\left(20\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(-15\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}}\right\}-\left\{\left(15\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(-15\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}}\right\} \\ =\left(5\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(0\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}} \end{gathered}$$

Using equation (i), the average velocity is,

$$\begin{gathered} {\overrightarrow{\mathrm{v}}}_{\mathrm{avg}}=\frac{\left(5\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(0\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}}}{10\times 60\mathrm{s}} \\ =\left(0.0083\mathrm{m}/\mathrm{s}\right)\stackrel{⏜}{\mathrm{i}} \\ =\left(0.83\mathrm{cm}/\mathrm{s}\right)\stackrel{⏜}{\mathrm{i}} \end{gathered}$$

The magnitude of average velocity is,

$\left|{\overrightarrow{\mathrm{v}}}_{\mathrm{avg}}\right|=0.83\mathrm{cm}/\mathrm{s}$

The displacement from point A to point D is,

$$\begin{gathered} \nabla {\overrightarrow{\mathrm{r}}}_{\mathrm{AD}}={\overrightarrow{\mathrm{r}}}_{\mathrm{D}}-{\overrightarrow{\mathrm{r}}}_{\mathrm{A}} \\ =\left\{\left(45\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(45\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}}\right\}-\left\{\left(15\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(-15\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}}\right\} \\ =\left(30\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(60\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}} \end{gathered}$$

Using equation (i), the average velocity is,

$$\begin{gathered} {\overrightarrow{\mathrm{v}}}_{\mathrm{avg}}=\frac{\left(30\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(60\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}}}{15\times 60\mathrm{s}} \\ =\left(0.033\mathrm{m}/\mathrm{s}\right)\stackrel{⏜}{\mathrm{i}}+\left(0.067\mathrm{m}/\mathrm{s}\right)\stackrel{⏜}{\mathrm{j}} \\ =\left(3.3\mathrm{cm}/\mathrm{s}\right)\stackrel{⏜}{\mathrm{i}}+\left(6.7\mathrm{cm}/\mathrm{s}\right)\stackrel{⏜}{\mathrm{j}} \end{gathered}$$

The magnitude of average velocity is,

$$\begin{gathered} \left|{\overrightarrow{\mathrm{v}}}_{\mathrm{avg}}\right|=\sqrt{\left(3.3\mathrm{cm}/\mathrm{s}\right)+{\left(6.7\mathrm{cm}/\mathrm{s}\right)}^2} \\ =7.5\mathrm{cm}/\mathrm{s} \end{gathered}$$

Thus, the magnitude of least velocity is 0.83 cm/s .

Since the least average velocity has only x-component therefore its direction is towards positive x-axis.

Thus, the angle of the average velocity with the least magnitude is $0°$.

From the calculations made in step 3, the average velocity having greatest magnitude is from point A to point B. its magnitude is 11.2 cm/s

Thus, the magnitude of the average velocity with the greatest magnitude is 11.2 cm/s .

Using equation (ii), the angle for the average velocity with greatest magnitude is calculated as:

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{-10cm/s}{5cm/s}\right) \\ =-63.4° \end{gathered}$$

Thus, the angle of the average velocity with the greatest magnitude is$-63.4°$ .