Chapter 4, Problem 55PQ

(a)

To determine

Find the time required to complete the trip if the jetliner is experiencing a headwind.

Answer to Problem 55PQ

The time required to complete the trip if the jetliner is experiencing a headwind is $\text{1}\text{.34 h}$.

Explanation of Solution

The jetliner is travelling through the air and the air can move relative to ground. The jetliner displacement to the south is $880\text{ km}$. Consider the north as positive y and east as positive x.

Write the equation for the velocity of airplane relative to ground.

    ${\left({\overrightarrow{v}}_J\right)}_G={\left({\overrightarrow{v}}_J\right)}_W+{\left({\overrightarrow{v}}_W\right)}_G={\left({\overrightarrow{v}}_J\right)}_W-{\left({\overrightarrow{v}}_G\right)}_W$                                                           (I)

Here, ${\left({\overrightarrow{v}}_J\right)}_G$ is the velocity of jetliner with respect to ground, ${\left({\overrightarrow{v}}_J\right)}_W$ is the velocity of jetliner with respect to wind, ${\left({\overrightarrow{v}}_W\right)}_G$ is the velocity of wind with respect to ground and ${\left({\overrightarrow{v}}_G\right)}_W$ is the velocity of ground with respect to wind.

Write the equation for time interval.

    $\Delta t=\frac{\Delta y}{{\left(v_J\right)}_{G,y}}$                                                                                     (II)

Here, $\Delta t$ is the time interval, $\Delta y$ is the displacement of the jetliner and ${\left(v_J\right)}_{G,y}$ is the y component of velocity of jetliner with respect to ground.

Conclusion:

Substitute $\text{710 km/h}$ for ${\left({\overrightarrow{v}}_J\right)}_W$ and $55\text{km/h}$ for ${\left({\overrightarrow{v}}_G\right)}_W$ in equation I.

    $\begin{array}{c}{\left({\overrightarrow{v}}_J\right)}_G=\text{710 km/h}-55\text{km/h}\\ =655\text{km/h}\end{array}$

Substitute $655\text{km/h}$ for ${\left(v_J\right)}_{G,y}$ and $880\text{ km}$ for $\Delta y$ in equation II.

    $\begin{array}{c}\Delta t=\frac{880\text{ km}}{655\text{km/h}}\\ =1.34\text{ h}\end{array}$

Therefore, the time required to complete the trip if the jetliner is experiencing a headwind is $\text{1}\text{.34 h}$.

(b)

To determine

Find the time required to complete the trip if the jetliner is experiencing a tailwind.

Answer to Problem 55PQ

The time required to complete the trip if the jetliner is experiencing a tailwind is $\text{1}\text{.15 h}$.

Explanation of Solution

The wind is blowing at south with speed of $\text{55 km/h}$.

Write the equation for the velocity of airplane relative to ground.

    ${\left({\overrightarrow{v}}_J\right)}_G={\left({\overrightarrow{v}}_J\right)}_W+{\left({\overrightarrow{v}}_W\right)}_G$                                                           (III)

Conclusion:

Substitute $\text{710 km/h}$ for ${\left({\overrightarrow{v}}_J\right)}_W$ and $55\text{km/h}$ for ${\left({\overrightarrow{v}}_W\right)}_G$ in equation III.

    $\begin{array}{c}{\left({\overrightarrow{v}}_J\right)}_G=\text{710 km/h}+55\text{km/h}\\ =765\text{km/h}\end{array}$

Substitute $765\text{km/h}$ for ${\left(v_J\right)}_{G,y}$ and $880\text{ km}$ for $\Delta y$ in equation II.

    $\begin{array}{c}\Delta t=\frac{880\text{ km}}{765\text{km/h}}\\ =1.15\text{ h}\end{array}$

Therefore, the time required to complete the trip if the jetliner is experiencing a tailwind is $\text{1}\text{.15 h}$.

(c)

To determine

Find the time required to complete the trip if the jetliner is experiencing a crosswind.

Answer to Problem 55PQ

The time required to complete the trip if the jetliner is experiencing a crosswind is $\text{1}\text{.24 h}$.

Explanation of Solution

The wind is blowing at west with speed of $\text{55 km/h}$. Therefore, the airplane should directly travel south to reach the destination, so it must travel somewhat east and south so that the wind’s west component in the wind’s velocity is cancelled by the east component of the airplane velocity.

If the airplane heads with an angle $\theta$ measured east of south, then the x component of the velocity must be equal in magnitude to the wind speed.

    $\begin{array}{c}\left(710\text{ km/h}\right)\sin \theta =55.0\text{ km/h}\\ \sin \theta =\frac{55}{710}\\ \theta ={\sin }^{-1}\left(\frac{55}{710}\right)\\ =4.44°\end{array}$

There is no north south component in the velocity of wind, the south component of airplane velocity is equal to its velocity relative to ground.

Then,

    $\begin{array}{c}{\left(v_J\right)}_{G,y}=\left(710\text{ km/h}\right)\cos \theta \\ =\left(710\text{ km/h}\right)\cos \left(4.44°\right)\\ =708\text{ km/h}\end{array}$

Conclusion:

Substitute $708\text{km/h}$ for ${\left(v_J\right)}_{G,y}$ and $880\text{ km}$ for $\Delta y$ in equation II.

    $\begin{array}{c}\Delta t=\frac{880\text{ km}}{708\text{km/h}}\\ =1.24\text{ h}\end{array}$

Therefore, the time required to complete the trip if the jetliner is experiencing a crosswind is $\text{1}\text{.24 h}$.