Q69P - University Physics with Modern Physics

Question

You are lost at night in a large, open field. Your GPS tells you that you are 122.0 cm from your truck, in a direction 58.0^° east of south. You walk 72.0 m due west along a ditch. How much farther, and in what direction, must you walk to reach your truck?

Step-by-Step Solution

Answer

The magnitude the man walks is, $B=71.9\mathrm{m}$.

The direction the man walk is, ${\theta }_B=64.1°$.

Step 1: Identification of given data

• The resultant magnitude of truck, $R_x=122.0\mathrm{m},58°$east-west .
• The direction magnitude of person is, $A_x=72.0\mathrm{m}$ west .

Step 2: Concept of direction of vector

A vector's orientation, or the angle it creates with the x-axis, determines its direction. With an arrow at the top and another fixed point at other end, a vector may be shown.

The vector's direction may be determined by looking at the arrowhead.

Step 3: Determination of the magnitude and direction of man to reach the truck

The free body diagram has been provided below-

                           

 

The equation of the resultant magnitude of the truck in the x and y direction can be expressed as-

 

$$\begin{gathered} {\mathrm{R}}_{\mathrm{x}}=‐122 \mathrm{m}\times \sin 58° \\ {\mathrm{R}}_{\mathrm{y}}=122 \mathrm{m}\times \cos 58° \end{gathered}$$

 

The magnitude of man to reach the truck in x-direction can be evaluated using,

$$\begin{gathered} R_x=A_x+B_x \\ {B}_x=R_x-A_x \end{gathered}$$

Here, $B_x$ is the distance required by the man to reach the truck in the x direction

 

Substituting the values in the above equation, 

 

$$\begin{gathered} B_x=\left(-122.0\mathrm{m}\right)\left(\sin 58.0°\right)-\left(-72.0\mathrm{m}\right) \\ =-31.462\mathrm{m} \end{gathered}$$

 

The magnitude of man to reach the truck in y-direction can be evaluated using,

$$\begin{gathered} R_y=A_y+B_y \\ {B}_y=R_y-A_y \end{gathered}$$ 

 

Here, $B_y$ is the distance required by the man to reach the truck in the y direction

 

Substituting the values in the above equation, 

$$\begin{gathered} B_y=\left(122.0 \mathrm{m}\right)\left(\cos 58.0°\right)-0 \\ =64.450 \mathrm{m} \end{gathered}$$

The resultant magnitude of man to reach the truck is,

$B=\sqrt{B_x^2+B_y^2}$

 

Here, B is the resultant magnitude to reach to the truck

 

Substituting the values in the above equation, 

$$\begin{gathered} B=\sqrt{{\left(31.462\mathrm{m}\right)}^2+{\left(64.450\mathrm{m}\right)}^2} \\ =71.9\mathrm{m} \end{gathered}$$

 

Thus, the man will travel 71.9m to reach the truck.

The equation of the direction of man is given by,

 

$\tan \theta =\frac{B_y}{B_x}$

 

Here, $\tan \theta$ is the direction of the man

 

Substituting the values in the above equation, 

$$\begin{gathered} \tan \theta =\frac{64.650m}{-31.462m} \\ \tan \theta =2.05486 \\ \theta ={\tan }^{-1}\left(2.05486\right) \\ =-64.1° \mathrm{or} 115.9° \mathrm{north}-\mathrm{east} \end{gathered}$$

Thus, the man will travel $64.1°$ north-east to reach the truck.