The position of a squirrel running in a park is given by

$\overrightarrow{\mathbf{r}}\mathbf{=}\left[\left(0.0280m/s\right)t+\left(0.0360m/s^2\right)t^2\right]\hat{\mathbf{i}}\mathbf{+}\left(0.0190m/s^3\right){\mathit{t}}^{\mathbf{3}}\hat{\mathbf{j}}$

(a) What are${\mathit{v}}_{\mathbf{x}}\mathit{t}$and${\mathit{v}}_{\mathbf{y}}\mathit{t}$, the x- and y-components of the velocity of the squirrel, as functions of time?

(b) At$\mathit{t}\mathbf{=}\mathbf{5}\mathbf{}\mathit{s}$, how far is the squirrel from its initial position?

(c) At$\mathit{t}\mathbf{=}\mathbf{5}\mathbf{}\mathit{s}$, what are the magnitude and direction of the squirrel’s velocity?

The given data can be listed below,

• The position of the squirrel is,$\overrightarrow{r}=\left[\left(0.0280m/s\right)t+\left(0.0360m/s^2\right)t^2\right]\hat{i}+\left(0.0190m/s^3\right)t^3\hat{j}$
• the time taken by the squirrel is $t=5s$

The change in displacement per unit of time is what "velocity," a vector quantity, is defined as. When motion occurs in a straight line, displacement is reduced to merely distance, and velocity is measured in meters per second as speed or distance traveled over time.

The components of instantaneous velocity are given by derivation of position with respect to time.

$v_{x,ins}=\frac{dr}{dt}$

Substitute the value in the above equation

$$\begin{gathered} v_x\left(t\right)=\frac{dx\left(t\right)}{dt} \\ =0.280m/s+\left(0.072m/s^2\right)t \end{gathered}$$

And

$$\begin{gathered} v_y\left(t\right)=\frac{dy\left(t\right)}{dt} \\ =\left(0.057m/s^3\right)t^2 \end{gathered}$$

Thus,the x and y-components of the velocity of the squirrel, as functions of time are $0.280m/s+\left(0.072m/s^2\right)t$and $\left(0.057m/s^3\right)t^2$respectively.

The x-component of the position of the squirrel at$t=5$ is given by,

$$\begin{gathered} x=\left(0.280\mathrm{m}/\mathrm{s}\right)\left(5\mathrm{s}\right)+\left(0.0360\mathrm{m}/{\mathrm{s}}^2\right){\left(5\mathrm{s}\right)}^2 \\ =2.3\mathrm{m} \end{gathered}$$

The y-component of the position of the squirrel at$t=5s$ is given by,

$$\begin{gathered} x=\left(0.0190\mathrm{m}/{\mathrm{s}}^3\right){\left(5\mathrm{s}\right)}^2 \\ =2.375\mathrm{m} \end{gathered}$$

The distance of squirrel from the origin is given by,

$r=\sqrt{x^2+y^2}$

Substitute all the values in the above,

$$\begin{gathered} r=\sqrt{{\left(2.3\mathrm{m}\right)}^2+{\left(2.375\mathrm{m}\right)}^2} \\ =3.31\mathrm{m} \end{gathered}$$

Thus, the distance of the squirrel from its initial position origin is 3.31 m.

The x and y-component of the velocity bis given by,

$$\begin{gathered} v_x=0.280\mathrm{m}/\mathrm{s}+\left(0.072\mathrm{m}/{\mathrm{s}}^2\right)\left(5\mathrm{s}\right) \\ =0.64\mathrm{m}/\mathrm{s} \\ {v}_y=0.057\mathrm{m}/{\mathrm{s}}^3{\left(5\mathrm{s}\right)}^2 \\ =1.425\mathrm{m}/\mathrm{s} \end{gathered}$$

The magnitude of the velocity of the squirrel is given by,

$v=\sqrt{v_x^2+v_y^2}$

Here,$v_x$ is the x-component of the velocity and$v_x$ is the y-component of the velocity.

Substitute all the values in the above,

$$\begin{gathered} v=\sqrt{{\left(0.64\mathrm{m}/\mathrm{s}\right)}^2+{\left(1.425\mathrm{m}/\mathrm{s}\right)}^2} \\ =1.56\mathrm{m}/\mathrm{s} \end{gathered}$$

The direction of the velocity is given by,

$$\begin{gathered} \tan \alpha =\frac{v_y}{v_x} \\ \alpha ={\tan }^{-1}\left(\frac{v_y}{v_x}\right) \end{gathered}$$

Substitute all the values in the above,

$$\begin{gathered} \alpha ={\tan }^{-1}\left(\frac{1.425}{0.64}\right) \\ =65.8° \end{gathered}$$

Thus, the magnitude and direction of the velocity of squirrel are 1.56 m/s and$65.8°$ respectively.