Question

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.

Short answer

1. The x and y-component of the average velocity are $\left(1.4,-1.3\right)\mathrm{m}/\mathrm{s}$.
2. The magnitude of the average velocity is$1.91\mathrm{m}/\mathrm{s}$ and the direction of average velocity is$-42.9°$ downward.

Step-by-step solution

Identification of given data

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" $\left(5.3,-0.5\right)\mathrm{m}$ .

Concept/Significance of average velocity

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

(a) Determination of the components of the average velocity

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$\left(1.4,-1.3\right)\mathrm{m}/\mathrm{s}$

(a) Determination of the magnitude and direction of the average velocity

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.