Q1E - University Physics with Modern Physics

Question

A squirrel has x- and y-coordinates $\left(1.1m, 3.4 m\right)$, 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.

Step-by-Step Solution

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

Step 2: Concept/Significance of average velocity

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

Step 3: (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 $\left(1.4,-1.3\right) \mathrm{m}/\mathrm{s}$

Step 4: (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.