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?

Short answer

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

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

Step-by-step solution

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 .

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.

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.