A squirrel has x- and y-coordinates$\left(1.1m,3.4m\right)$, at time${\mathit{t}}_{\mathbf{1}}\mathbf{=}\mathbf{0}$and coordinates$\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{3}\mathbf{m}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{m}\mathbf{)}$at time$\mathbf{}{\mathbf{t}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$. For this time interval, find

(a) the components of the average velocity, and

(b) the magnitude and direction of the average velocity.

The given data can be listed below,

• The x and y coordinate of a squirrel at the time$t_1=0$ is,$\left(1.1,3.4\right)\mathrm{m}$ .
• The x and y coordinate of a squirrel at the time$t_2=3.0\mathrm{s}$ is,role="math" localid="1664865967346" $\left(5.3,-0.5\right)\mathrm{m}$ .

The average velocity of an interval depends on the change in location and the change in time.

The x-component of average velocity is given by,

$x_{x,avg}=\frac{x_2-x_1}{t_2-t_1}$

Here, $x_2$is the x-coordinate at the time $t_2,x_1$ is the x-coordinate at the origin at the time$t_1$ and $t_2-t_1$is the time interval.

Substitute all the values in the above,

$$\begin{gathered} x_{x,avg}=\frac{\left(5.3-1.1\right)\mathrm{m}}{\left(3-0\right)\mathrm{s}} \\ =1.4\mathrm{m}/\mathrm{s} \end{gathered}$$

The y-component of average velocity is given by,

$v_{y,avg}=\frac{y_2-y_1}{t_2-t_1}$

$y_2$ is the y-coordinate at the time $t_2,y_1$is the y-coordinate at the origin at the time $t_1$, and $t_2-t_1$is the time interval.

Substitute all the values in the above,

$$\begin{gathered} v_{y,avg}=\frac{\left(-0.5-3.4\right)\mathrm{m}}{\left(3-0\right)\mathrm{s}} \\ =-1.3\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the x and y components of the average velocity are$\left(1.4,-1.3\right)\mathrm{m}/\mathrm{s}$

The magnitude of average velocity is given by,

$v_{avg}=\sqrt{{\left(V_{x,avg}\right)}^2+{\left(V_{y,avg}\right)}^2}$

Here, $V_{x,avg}$is the x-component of average velocity and $V_{y,avg}$is the y-component of average velocity.

Substitute all the values in the above,

$$\begin{gathered} V_{avg}=\sqrt{{\left(1.4\right)}^2+{\left(-1.3\right)}^2}\pm \mathrm{m}/\mathrm{s} \\ =1.91\mathrm{m}/\mathrm{s} \end{gathered}$$

The direction of average velocity is given by,

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

Substitute all the values in the above,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{-1.3\mathrm{m}/\mathrm{s}}{1.4\mathrm{m}/\mathrm{s}}\right) \\ =-42.9° \end{gathered}$$

Thus, the magnitude of the average velocity is$1.91\mathrm{m}/\mathrm{s}$ and the direction of average velocity is$-42.9°$ downward.