91影视

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

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.

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-

$$\begin{gathered} {\mathrm{R}}_{\mathrm{x}}=鈥�122\mathrm{m}脳\sin 58掳 \\ {\mathrm{R}}_{\mathrm{y}}=122\mathrm{m}脳\cos 58掳 \end{gathered}$$

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,

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

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,

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

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,

$\tan 胃=\frac{B_y}{B_x}$

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

Substituting the values in the above equation,

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

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