Q71P - University Physics with Modern Physics

Question

An airplane pilot sets a compass course due west and maintains an airspeed of $\mathbf{220}\mathbf{ }\mathit{K}\mathit{m}\mathbf{/}\mathit{h}\mathbf{ }$. After flying for $\mathbf{0}\mathbf{.}\mathbf{500}\mathit{h}$, she finds herself over a town $\mathbf{120}\mathbf{ }\mathit{K}\mathit{m}$ west and $\mathbf{20}\mathbf{ }\mathit{K}\mathit{m}$ south of her starting point. (a) Find the wind velocity (magnitude and direction). (b) If the wind velocity is $\mathbf{40}\mathbf{ }\mathit{K}\mathit{m}\mathbf{/}\mathit{h}$due south, in what direction should the pilot set her course to travel due west? Use the same airspeed of $\mathbf{220}\mathbf{ }\mathit{K}\mathit{m}\mathbf{/}\mathit{h}\mathbf{.}$

Step-by-Step Solution

Answer

(a) The $Magnitude V_{A/E}=44.7 Km/h$ and direction of wind velocity is $63.4°$south of west .

 

(b) The course should be set by the pilot towards $10.5°$north of west.

 

Step 1: Relative velocity formula

From the formula of relative velocity of any object,

 

If any object P is moving related to A and A is moving related to E, then the velocity of P related to E is,

$V_{P/E}=V_{P/A}+V_{A/E}$

For horizontal motion, style="max-width: none; vertical-align: -14px;" $V_{P/A_h}=-220Km/h$

For vertical motion,$V_{P/A_v}=0Km/h$

Step 2: Given

For horizontal motion,

$V_{P/A_h}=-220Km/h$

For vertical motion,

$V_{P/A_v}=0 Km/h$

Step 3: (a) Wind velocity magnitude and direction

The wind velocities for horizontal and vertical motion are respectively,

$$\begin{gathered} V_{P/E_h}=-\frac{120}{0.500}Km/h \\ {V}_{P/E_h}=-240Km/h\left(west\right) \\ {V}_{P/E_v}=-\frac{20}{0.500}Km/h \\ {V}_{P/E_v}=-40Km/h\left(south\right) \end{gathered}$$

With the given information the following figure can be drawn

From both the horizontal component of velocity that is $V_{P/A_h}=-220Km/h$ and $V_{P/E_h}=-240Km/h\left(west\right)$

$$\begin{gathered} V_{P/E_h}=V_{P/A_h}+V_{A/E_h} \\ {V}_{P/E_h}=-240Km/h-(-220Km/h) \\ {V}_{P/E_h}=-20 Km/h \end{gathered}$$

The vertical components of velocity $V_{P/A_V}=0Km/h$ and $V_{P/E_V}=-40Km/h\left(south\right)$

$$\begin{gathered} V_{P/E_V}=V_{P/A_V}+V_{A/E_V} \\ {V}_{A/E_V}=-40Km/h-(-0Km/h) \\ {V}_{A/E_V}=-40Km/h \end{gathered}$$

Now the velocity magnitude will be 

$$\begin{gathered} V_{A/E}=\sqrt{{\left(V_{A/E_V}\right)}^2+{\left(V_{A/E_h}\right)}^2} \\ {V}_{A/E}=\sqrt{{\left(-40 Km/h\right)}^2+{\left(-20Km/h\right)}^2} \\ {V}_{A/E}=44.7Km/h \end{gathered}$$

The angle or direction will be,

$$\begin{gathered} \tan \varphi =\frac{-40Km/h}{-20Km/h} \\ \varphi =63.4° \end{gathered}$$

The $Magnitude V_{A/E}=44.7Km/h$and direction of wind velocity is $63.4°$south of west .

Step 4: (b) the direction in which the pilot should set her course

According to vector rule,

$V_{P/E}=V_{P/A}+V_{A/E}$

$V_{P/A} and V_{A/E}$ are $220Km/h and 40Km/h\left(south\right)$

The vector diagram of will be and the direction will be at right angle as shown in figure

From the above figure,

$$\begin{gathered} \sin \varphi =\frac{V_{A/E}}{V_{P/E}} \\ \sin \varphi =\frac{40Km/h}{220Km/h} \\ \varphi =10.5° \end{gathered}$$

The course should be set by the pilot toward $10.5°$snorth of west.