Q2E - University Physics with Modern Physics

Question

A rhinoceros is at the origin of coordinates at time ${\mathit{t}}_{\mathbf{1}}\mathbf{=}\mathbf{0}$. For the time interval from ${\mathit{t}}_{\mathbf{1}}\mathbf{=}\mathbf{0}$ to ${\mathit{t}}_{\mathbf{2}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{s}$, the rhino’s average velocity has x-component - 3.8 m/s and y-component 4.9 m/s . At time ${\mathit{t}}_{\mathbf{2}}\mathbf{=}\mathbf{12}\mathbf{ }\mathit{s}$, 

(a) what are the x and y-coordinates of the rhino? 

(b) How far is the rhino from the origin?

Step-by-Step Solution

Answer

1. The coordinates of the rhino at the time $t_2$ is $\left(-45.6, 58.8\right) \mathrm{m}$ .
2. The distance of a rhino from its origin is $74.41 \mathrm{m}$.

Step 1: Identification of given data

The given data can be listed below,

• The x-component of the rhino’s velocity is, $V_x=-3.8 \mathrm{m}/\mathrm{s}$
• The y component of the rhino’s velocity is,  
• The time interval is given is, $t_1=0$ to $t_2=12 \mathrm{s}$  

Step 2: Concept/Significance of Average velocity

The entire displacement of an object divided by the whole time elapsed is its average velocity. In simpler terms, it's the pace at which an object moves from one location to another.

Step 3: (a) Determination of the x and y -coordinates of the rhino at

The x-component of the velocity is given by,

 

 $v_x=\frac{x_2-x_1}{t_2-t_1}$                                                                                                                   …(i)

 

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

 

The y-component of the velocity is given by,

 

$v_y=\frac{y_2-y_1}{t_2-t_1}$                                                                                                                  …(ii)

 

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

 

The coordinates of rhino at origin are (0, 0).

 

Substitute all the values in the above two-equation.

 

$$\begin{gathered} v_x=\frac{x_2-0}{12-0} \\ 12v_x=x_2 \end{gathered}$$

 

Substitute all the values in the above equation as:

 

$$\begin{gathered} x_2=12v \\ =12 \mathrm{s}\left(-3.8 \mathrm{m}/\mathrm{s}\right) \\ =-45.6 \mathrm{m} \end{gathered}$$ 

 

And, y-component of rhino at time 12 s is given by,

 

$$\begin{gathered} v_y=\frac{y_2-0}{12-0} \\ 12v_y=y_2 \end{gathered}$$

 

Substitute all the values in the above equation.

$$\begin{gathered} y_2=12v_y \\ =12 \mathrm{s}\left(4.9 \mathrm{m}/\mathrm{s}\right) \\ =58.8 \mathrm{m} \end{gathered}$$

Thus, the coordinates of the rhino at the time $t_2$ is $\left(-45.6, 58.8\right) \mathrm{m}$ .

Step 4: (b) Determination of the distance of the rhino from the origin

The distance of the rhino from the origin is given by,

 

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ 

 

Here, $x_2$ is the x-coordinate of rhino at the time $t_2$ , $x_1$ is the x-coordinate of rhino at the origin at the time $t_1$ , $y_2$ is the y-coordinate of rhino at the time $t_2$ , and $y_1$ is the y-coordinate of rhino at the origin at time $t_1$.

 

Substitute all the values in the above equation.

 

$$\begin{gathered} d=\sqrt{{\left(-45.6 \mathrm{m}-0\right)}^2+{\left(58.8 \mathrm{m}-0\right)}^2} \\ =74.41 \mathrm{m} \end{gathered}$$

 

Thus, the distance of a rhino from its origin is $74.41 \mathrm{m}$ .