91Ó°ÊÓ

From the figure,

The position vector of point A, $\stackrel{â‡\P Ä}{{\mathrm{r}}_{\mathrm{A}}}=\left(15\mathrm{m}\right)\hat{\mathrm{i}}+(-15\mathrm{m})\stackrel{Áåž}{\mathrm{j}}$

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

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

The position vector of point D, $\stackrel{â‡\P Ä}{{\mathrm{r}}_D}=\left(45\mathrm{m}\right)\hat{\mathrm{i}}+\left(45\mathrm{m}\right)\stackrel{Áåž}{\mathrm{j}}$

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:

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

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

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

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

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

The displacement from point A to point B is,

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

Using equation (i), the average velocity is,

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

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,

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

Using equation (i), the average velocity is,

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

The magnitude of average velocity is,

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

The displacement from point A to point D is,

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

Using equation (i), the average velocity is,

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

The magnitude of average velocity is,

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

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:

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

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