Chapter 3, Problem 19P

A commuter airplane starts from ar. airport and takes the route shown in Figure P3.19. The plane first flies to city A, located 175 km away in a direction 30.0° north of east. Next, it flies for 150 km 20.0° west of north, to city B. Finally, the plane flies 190 km due west, to city C. Find the location of city C relative to the location of the starling point.

Figure P3.19

To determine

 The location and direction of the city from the starting point.

Answer to Problem 19P

Solution:

The location and direction of the city C is $\text{245}\text{km}$ and $21.3°$ West of North respectively.

Explanation of Solution

Given Info:

The magnitude of vector OA is $175\text{km}$, the magnitude of vector AB is $150\text{km}$, and the magnitude of vector BC is $190\text{km}$.

Write the formula to calculate the x-component of OA.

$x_1=\text{OAcos}{\theta }_1$

• $x_1$ is the horizontal or x-component of the vector OA
• OA is the magnitude of the vector OA
• ${\theta }_1$ is the angle made by the OA vector with respect to the x- axis

Substitute $30.0°$ for ${\theta }_1$ and $175\text{km}$ for OA to calculate $x_1$.

$\begin{array}{c}{x}_1=\left(175\text{km}\right)\text{cos}\left(30.0°\right)\\ =\left(175\text{km}\right)\left(0.866\right)\\ =151.55\text{km}\end{array}$

Write the formula to calculate the x-component of AB.

$x_2=\text{ABcos}{\theta }_2$

• $x_2$ is the horizontal or x-component of the vector AB
• AB is the magnitude of the vector AB
• ${\theta }_2$ is the angle made by the AB vector with respect to the x- axis

Substitute $110.0°$ for ${\theta }_2$ and $150\text{km}$ for AB to calculate $x_2$.

$\begin{array}{c}{x}_2=\left(150\text{km}\right)\text{cos}\left(110.0°\right)\\ =-\left(150\text{km}\right)\left(0.34\right)\\ =-51.3\text{km}\end{array}$

Write the formula to calculate the x-component of BC.

$x_3=\text{BCcos}{\theta }_3$

• $x_3$ is the horizontal or x-component of the vector BC
• BC is the magnitude of the vector BC
• ${\theta }_3$ is the angle made by the BC vector with respect to the x- axis

Substitute $180.0°$ for ${\theta }_3$ and $190\text{km}$ for BC to calculate $x_3$.

$\begin{array}{c}{x}_3=\left(190\text{km}\right)\text{cos}\left(180.0°\right)\\ =-\left(190\text{km}\right)\left(1\right)\\ =-190\text{km}\end{array}$

Write the formula to calculate the total x component.

$x=x_1+x_2+x_3$

• x is the total x component.

Use $151.55\text{km}$ for $x_1$, $-51.3\text{km}$ for $x_2$ and $-190\text{km}$ for $x_3$ in the above expression for x.

$\begin{array}{c}x=151.55\text{km}-51.3\text{km}-190\text{km}\\ =-89.75\text{km}\end{array}$

Write the formula to calculate the y-component of OA.

$y_1=\text{OAsin}{\theta }_1{}^{\prime }$

• $y_1$ is the vertical or y-component of the vector OA
• ${\theta }_1{}^{\prime }$ is the angle made by the OA vector with respect to the y- axis

Substitute $30.0°$ for ${\theta }_1{}^{\prime }$ and $175\text{km}$ for OA to calculate $y_1$.

$\begin{array}{c}{y}_1=\left(175\text{km}\right)\sin \left(30.0°\right)\\ =\left(175\text{km}\right)\left(0.5\right)\\ =87.5\text{km}\end{array}$

Write the formula to calculate the y-component of AB.

$y_2=\text{ABsin}{\theta }_2{}^{\prime }$

• $x_2$ is the vertical or y-component of the vector AB
• ${\theta }_2{}^{\prime }$ is the angle made by the AB vector with respect to the y- axis

Substitute $110.0°$ for ${\theta }_2{}^{\prime }$ and $150\text{km}$ for AB to calculate $y_2$.

$\begin{array}{c}{y}_2=\left(150\text{km}\right)\text{sin}\left(110.0°\right)\\ =\left(150\text{km}\right)\left(0.94\right)\\ =140.95\text{km}\end{array}$

Write the formula to calculate the y-component of BC.

$y_3=\text{BCsin}{\theta }_3{}^{\prime }$

• $y_3$ is the vertical or y-component of the vector BC
• ${\theta }_3{}^{\prime }$ is the angle made by the BC vector with respect to the y- axis

Substitute $180.0°$ for ${\theta }_3{}^{\prime }$ and $190\text{km}$ for BC to calculate $y_3$.

$\begin{array}{c}{y}_3=\left(190\text{km}\right)\text{sin}\left(180.0°\right)\\ =\left(190\text{km}\right)\left(0\right)\\ =0\text{km}\end{array}$

Write the formula to calculate the total y component.

$y=y_1+y_2+y_3$

• y is the total y component.

Use $87.5\text{km}$ for $y_1$, $140.95\text{km}$ for $y_2$ and $0\text{km}$ for $y_3$ in the above expression for y.

$\begin{array}{c}y=87.5\text{km+}140.95\text{km+}0\text{km}\\ =228.45\text{km}\end{array}$

Write the formula to calculate the location of the city.

$r=\sqrt{x^2+y^2}$

• r is the location of the city from the starting point

Substitute $-89.75\text{km}$ for x and $228.45\text{km}$ for y to calculate r.

$\begin{array}{c}r=\sqrt{{\left(-89.75\text{km}\right)}^2+{\left(228.45\text{km}\right)}^2}\\ =\sqrt{8010.5+52212.5}\\ =245.4\text{km}\\ \approx 245\text{km}\end{array}$

Write the formula to calculate the direction of the city.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\left|x\right|}{y}\right)\\ ={\tan }^{-1}\left(\frac{\left|-89.5\text{km}\right|}{228.5\text{km}}\right)\\ ={\tan }^{-1}\left(0.39\right)\\ =21.3°\end{array}$

Conclusion:

Therefore, the location and direction of the city C is $\text{245}\text{km}$ and $21.3°$ West of North respectively