Q61P - University Physics with Modern Physics

Question

As noted in Exercise 1.26, a spelunker is surveying a cave. She follows a passage 180 m straight west, then 210 m in a direction $45°$ east of south, and then 280 m at $30°$ east of north. After a fourth displacement, she finds herself back where she started. Use the method of components to determine the magnitude and direction of the fourth displacement. Draw the vector-addition diagram and show that it is in qualitative agreement with your numerical solution.

Step-by-Step Solution

Answer

 

Identification of given data

• Target variable is in fourth variable.
• The displacement vectors are,$\overrightarrow{\mathrm{A}},\overrightarrow{\mathrm{B}} \mathrm{and} \overrightarrow{\mathrm{C}}$.
• The unmeasured displacement vector is,$\overrightarrow{D}$.
• The resultant displacement vector is,$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}+\overrightarrow{D}$.
• As she ends up where she started, so resultant vector is,$\overrightarrow{R}=0$.

$$\begin{gathered} 0=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}+\overrightarrow{D} \\ \overrightarrow{D}=-\left(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}\right) \\ {D}_x=-\left(A_x+B_x+C_x\right) \\ {D}_y=-\left(A_y+B_y+C_y\right) \end{gathered}$$

Concept of displacement vectors

In mathematics, the term "displacement vector" is used. It's a thing that moves in a straight line. It indicates the direction as well as distance travelled in a single line.

These three variables are often used to show how fast and how far an item has been travelling in physics. 

Illustrate direction of girl with vector diagram and numerical solution

To direction of the girl can be evaluated as,

Resolving the displacement vectors,

$$\begin{gathered} {\mathrm{A}}_{\mathrm{x}}=-180\mathrm{m},{\mathrm{A}}_{\mathrm{y}}=0 \\ {\mathrm{B}}_{\mathrm{x}}=\mathrm{Bcos}315°=\left(210\mathrm{m}\right)\cos 315°=+148.5\mathrm{m} \\ {\mathrm{B}}_{\mathrm{y}}=\mathrm{Bsin}315°=\left(210\mathrm{m}\right)\sin 315°=-148.5\mathrm{m} \\ {\mathrm{C}}_{\mathrm{x}}=\mathrm{Ccos}60°=\left(280\mathrm{m}\right)\cos 60°=+140\mathrm{m} \\ {\mathrm{C}}_{\mathrm{y}}=\mathrm{Csin}60=\left(280\right)\sin 60°=+242.5\mathrm{m} \end{gathered}$$

The fourth unmeasured displacement vector’s direction can be evaluated as,

$$\begin{gathered} {\mathrm{D}}_{\mathrm{x}}=-\left({\mathrm{A}}_{\mathrm{x}}+{\mathrm{B}}_{\mathrm{x}}+{\mathrm{C}}_{\mathrm{x}}\right) \\ {\mathrm{D}}_{\mathrm{x}}=-\left(-180\mathrm{m}+148.5\mathrm{m}+140\mathrm{m}\right) \\ {\mathrm{D}}_{\mathrm{x}}=-108.5\mathrm{m} \\ {\mathrm{D}}_{\mathrm{y}}=-\left({\mathrm{A}}_{\mathrm{y}}+{\mathrm{B}}_{\mathrm{y}}+{\mathrm{C}}_{\mathrm{y}}\right) \\ {\mathrm{D}}_{\mathrm{y}}=-\left(0-148.5\mathrm{m}+242.5\mathrm{m}\right) \\ {\mathrm{D}}_{\mathrm{y}}=-94.0\mathrm{m} \end{gathered}$$

$$\begin{gathered} D=\sqrt{D_x^2+D_y^2} \\ D=\sqrt{{\left(-108.5m\right)}^2+{\left(-94.0m\right)}^2} \\ D=143.5m \\ \tan \theta =\frac{D_y}{D_x} \\ \tan \theta =\frac{-94.0m}{-108.5m} \\ \tan \theta =0.8664 \\ \theta ={\tan }^{-1}\left(0.8664\right) \\ \theta =40.9°+180°=220.9° \end{gathered}$$

Since, $\overrightarrow{D}$is in third quadrant both so $\overrightarrow{D_x} \mathrm{and} \overrightarrow{D_y}$are negative.

       

The direction of $\overrightarrow{D}$can be expressed in terms of,

$\begin{array}{ll}\varphi =\theta -180°& \\ \varphi =40.9°\approx 41°& \end{array}$

So, she will head South-West.