Q4E - University Physics with Modern Physics

Question

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

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

(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?

Step-by-Step Solution

Answer

1. The x and y-components of the velocity of the squirrel, as functions of time, are $0.280 m/s+\left(0.072 m/s^2\right)t$ and $\left(0.057 m/s^3\right)t^2$ respectively.
2. The distance of the squirrel from its initial position origin is 3.31 m.
3. The magnitude and direction of the velocity of squirrel are 1.56 m/s and $65.8°$ respectively.

Step 1: Identification of given data

The given data can be listed below,

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

Step 2: Concept/Significance of velocity of an object

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.

Step 3: (a) Determination of v x t and v y t , the x and y -components of the velocity of the squirrel, as functions of 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

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

 

And 

 

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

 

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

Step 4: (b) Determination of the distance of the squirrel from its initial position.

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 $t=5 s$ 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.

Step 5: (c) Determination of the magnitude and direction of the squirrel’s velocity.

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.