Q66P - University Physics with Modern Physics

Question

On a training flight, a student pilot flies from Lincoln, Nebraska, to Clarinda, Iowa, next to St. Joseph, Missouri, and then to Manhattan, Kansas

(Fig. P1.66). The directions are shown relative to north: $\mathbf{0}\mathbf{°}$is north, $\mathbf{90}\mathbf{°}$ is east, $\mathbf{180}\mathbf{°}$is south, and $\mathbf{270}\mathbf{°}$is west. Use the method of components to find (a) the distance she has to fly from Manhattan to get back to Lincoln, and (b) the direction (relative to north) she must fly to get there. Illustrate your solutions with a vector diagram.

                                   

Step-by-Step Solution

Answer

a) The distance that the student flies from Manhattan to get back to Lincoln is, $D=189\mathrm{km}$

b) the direction relative to north-west is $79.53°$

Step 1: Identification of given data

• The magnitude of the training flight isfor the vector A is, $\overrightarrow{A}=147 \mathrm{km}, 85° \mathrm{east}$

• The magnitude of the training flight is for the vector B is, $\overrightarrow{B}=106 \mathrm{km}, 167° \mathrm{south}‐\mathrm{east}$
• The magnitude of the training flight is for the vector C is, $\overrightarrow{C}=166 \mathrm{km}, 235° \mathrm{south}‐\mathrm{west}$

The figure has been provided below-

                               

Step 2: Concept of displacement vector

The displacement vector is just the original position vector minus the end position vector.

This is the quickest path from the beginning location of the moving body to the end position.

Step 3: Determine the distance that the student should fly to get from Manhattan

a)

 

The diagram has been provided below-

                                           

The magnitude of flight in x-direction is,

 

$D_x=-\left[\overrightarrow{A}\sin 85°+\overrightarrow{B}\sin 13°-\overrightarrow{C}\sin 55°\right]$

 

Here, $D_x$ is the magnitude of the flight in the x-direction,

 

Substituting the values in the above equation,

 

$$\begin{gathered} D_x=-\left[\left(147 \mathrm{km}\right)\sin 85°+\left(106 \mathrm{km}\right)\sin 13°-\left(166 \mathrm{km}\right)\sin 55°\right] \\ {D}_x=-34.3 \mathrm{km} \end{gathered}$$

 

The magnitude of flight in y-direction is,

$D_y=-\left[\overrightarrow{A}\cos 85°-\overrightarrow{B}\cos 13°-\overrightarrow{C}\cos 55°\right]$

Here, $D_y$ is the magnitude of the flight in the y-direction,

 

Substituting the values in the above equation,

 

$$\begin{gathered} D_y=-\left[\left(147 \mathrm{km}\right)\cos 85°-\left(106 \mathrm{km}\right)\cos 13°-\left(166 \mathrm{km}\right)\cos 55°\right] \\ {D}_y=185.7 \mathrm{km} \end{gathered}$$

 

The resultant magnitude of flight is,

 

$$\begin{gathered} D=\sqrt{D_x^2+D_y^2} \\ D=\sqrt{{\left(34.30 \mathrm{km}\right)}^2+{\left(-185.7 \mathrm{km}\right)}^2} \\ D=189 \mathrm{km} \end{gathered}$$

 

Thus, The distance that the student flies from Manhattan to get back to Lincoln is, $D=189 \mathrm{km}$

Step 4: Determine the direction of flight that it should cover to get from Manhattan

b)

 

The direction of flight can be evaluated by,

 

$\phi ={\tan }^{-1}\left(\frac{D_y}{D_x}\right)$

 

Substituting the values in the above equation,

 

$$\begin{gathered} \phi ={\tan }^{-1}\left(\frac{185.7 \mathrm{km}}{34.3 \mathrm{km}}\right) \\ =79.53° \end{gathered}$$

 

Since, magnitude of direction in x-direction is less than zero and in y-direction it is more than zero.

Thus, the direction relative to north-west is $79.53°$.