Chapter 3, Problem 86P

(a)

To determine

The change of direction requires compensating for the wind.

Answer to Problem 86P

The change of direction requires compensating for the wind is ${9.45}^{\circ }\text{north of west}$.

Explanation of Solution

In this case, let $x$ in the east direction and $y$ in the north direction.

Air speed of the small plane is $30\text{m/s}$ and wind blowing is $10\text{m/s},{30}^{\circ }\text{south of west}$.

Write the expression forthe airspeed of the small plane,

  ${\overrightarrow{\text{v}}}_{PA}=v_{PA}\cos \theta \widehat{\text{x}}+v_{PA}\sin \theta \widehat{\text{y}}$        (I)

Here, ${\overrightarrow{\text{v}}}_{PA}$ is the airspeed of the small plane, $v_{PA}$ is the velocity of the plane relative to the air and $\theta$ is the change of direction.

Write the expression forthe wind flow,

  ${\overrightarrow{\text{v}}}_{AG}=v_{AG}\cos \varphi \widehat{\text{x}}+v_{AG}\sin \varphi \widehat{\text{y}}$        (II)

Here, ${\overrightarrow{\text{v}}}_{AG}$ is the wind flow, $\varphi$ is the change in angle and $v_{AG}$ is the velocity of the air relative to the ground.

In this case,

    ${\left(v_{PA}\right)}_y+{\left(v_{AG}\right)}_y=0$        (III)

Conclusion:

Substitute $30\text{m/s}$ for $v_{PA}$ in (I),

  ${\overrightarrow{\text{v}}}_{PA}=-30\cos \theta \widehat{\text{x}}+30\sin \theta \widehat{\text{y}}$

Substitute $10\text{m/s}$ for $v_{AG}$ and ${30}^{\circ }$ for $\theta$ in (II),

  $\begin{array}{c}{a}_{\text{rad}}=-10\text{m/s}\cos {30}^{\circ }\widehat{\text{x}}-10\text{m/s}\sin {30}^{\circ }\widehat{\text{y}}\\ =\left(-8.66\widehat{\text{x}}-5\widehat{\text{y}}\right)\text{m/s}\end{array}$

Substitute $30\sin \theta$ for ${\left(v_{PA}\right)}_y$ and $-5$ for $\left(v_{AG}\right)$ in (III),

    $\begin{array}{c}30\sin \theta +\left(-5\right)=0\\ \sin \theta =\frac{5}{30}\\ \theta ={\sin }^{-1}\left(\frac{1}{6}\right)\\ ={9.45}^{\circ }\text{north}\text{of}\text{east}\end{array}$

Therefore, the change of direction requires compensating for the wind is ${9.45}^{\circ }\text{north of west}$.

(b)

To determine

The ground speed of the airplane after she changes the direction of the heading.

Answer to Problem 86P

The ground speed of the airplane is $38.26\text{m/s}$.

Explanation of Solution

From the part (a) equation (I),

    ${\overrightarrow{\text{v}}}_{PA}=v_{PA}\cos \theta \widehat{\text{x}}+v_{PA}\sin \theta \widehat{\text{y}}$

Write the expression for the velocity of the plane relative to the ground.

    $\left|{\overrightarrow{\text{v}}}_{PG}\right|=\left|{\left(v_{PA}\right)}_x\right|+\left|{\left(v_{AG}\right)}_x\right|$        (IV)

Conclusion:

Substitute ${9.45}^{\circ }$ for $\theta$ and $30\text{m/s}$ for $v_{PA}$ in equation (I) from part (a)

    $\begin{array}{c}{\overrightarrow{\text{v}}}_{PA}=-30\cos {9.45}^{\circ }\widehat{\text{x}}+30\sin {9.45}^{\circ }\widehat{\text{y}}\\ =\left(-29.6\widehat{\text{x}}+4.9\widehat{\text{y}}\right)\text{m/s}\end{array}$

Substitute $-8.66\text{m/s}$ for ${\left(v_{PA}\right)}_x$ and $-29.6\text{m/s}$ for ${\left(v_{AG}\right)}_x$ in (IV),

    $\begin{array}{c}\left|{\overrightarrow{\text{v}}}_{PG}\right|=\left|-8.66\text{m/s}\right|+\left|-29.6\text{m/s}\right|\\ =38.26\text{m/s}\end{array}$

Therefore, the ground speed of the airplane is $38.26\text{m/s}$.