You leave the airport in College Station and fly 23.0 km in a direction$\mathbf{34}\mathbf{°}$south of east. You then fly 46.0 kmdue north. How far and in what direction must you then fly to reach a private landing strip that is 32.0 kmdue west of the College Station airport?

The given data can be listed below as-

• The direction of the vector A is in the east direction that is, $\overrightarrow{A}=23.0\mathrm{km}$at an angle of $34°$.
• The direction of the vector B is in the north direction that is, $\overrightarrow{A}=46.0\mathrm{km}$.
• The direction of the resultant vector R is in the west direction that is, $\overrightarrow{R}=32.0\mathrm{km}$

The diagram is given as-

Mathematicians use the term "displacement vector." A vector is what it is. It indicates the direction and total distance in a straight line.

Speed, acceleration as well as distance travelled are often shown in physics using this method.

The free body diagram has been provided below-

The magnitude of fly to reach a private landing strip in x-direction is,

$C_x=R_x-A_x-B_x$

Here, $C_x$is the magnitude of the fly to reach a private landing strip in the x-direction, $R_x$is the resultant vector in the x direction $A_x$and$B_x$is the vector A and B in the x direction.

Substituting the values in the above equation,

$$\begin{gathered} C_x=\left(-32\mathrm{km}\right)-\left(23.0\mathrm{km}\right)\cos 34.0°-0 \\ =-51.07\mathrm{km} \end{gathered}$$

The magnitude of fly to reach a private landing strip in y-direction is,

$C_y=R_y-A_y-B_y$

Here, $C_y$is the magnitude of the fly to reach a private landing strip in the y-direction, $R_y$is the resultant vector in the y direction and $A_y$and $B_y$is the vector A and B in the y direction.

Substituting the values in the above equation,

$$\begin{gathered} C_y=0-\left(-23\sin 34.0°\right)\mathrm{km}-\left(46.0\mathrm{km}\right) \\ =-33.14\mathrm{km} \end{gathered}$$

The resultant magnitude of fly to reach a private landing strip is,

$C=\sqrt{C_x^2+C_y^2}$

Here, C is the resultant magnitude

Substituting the values in the above equation,

$$\begin{gathered} C=\sqrt{{\left(-51.07\mathrm{km}\right)}^2+{\left(-33.14\mathrm{km}\right)}^2} \\ =60.9\mathrm{km} \end{gathered}$$

The direction of fly to reach a private landing strip can be evaluated as,

$\tan \theta =\frac{C_y}{C_x}$

Here,$\tan \theta$ is the angle to reach to the private landing strip

Substituting the values in the above equation,

$$\begin{gathered} \tan \theta =\frac{-33.14\mathrm{km}}{-51.07\mathrm{km}} \\ \theta ={\tan }^{-1}\left(\frac{-33.14\mathrm{km}}{-51.07\mathrm{km}}\right) \\ =33.0° \end{gathered}$$

Thus, the direction of fly to reach a private landing strip is $33.0°$south-west .