Q65P - University Physics with Modern Physics

Question

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 km due north. How far and in what direction must you then fly to reach a private landing strip that is 32.0 km due west of the College Station airport?

Step-by-Step Solution

Answer

The magnitude of my fly to reach a private landing strip is, $C=60.9 \mathrm{km}$

The direction of my fly to reach a private landing strip is, $\theta =33.0°$ south-west.

Step 1: Identification of given data

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-

              

Step 2: Concept of displacement vector

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.

Step 3: Determine the magnitude and direction of fly to reach a private landing strip

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 .