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Q60E

Expert-verifiedFound in: Page 124

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(****a) Another airplane is flying in a jet stream that is blowing at ** ** m/sin a direction ** **south of east. Its direction of motion relative to the Earth is ** **south of west, while its direction of travel relative to the air is ** ** south of west. What is the airplane’s speed relative to the air mass? **

**(b) What is the airplane’s speed relative to the Earth?**

The velocity of the plane with respect to air is 63.5 m/s. The velocity of the plane relative to the earth is 29.6 m/s.

The velocity of air relative to the earth is 45 m/s. It is in 20 degree.

**The sum of the plane's velocity relative to the air and the air's velocity relative to the earth equals the plane's velocity relative to the earth.**

The velocity of the plane with respect to the earth:

${V}_{pe}={V}_{pa}+{V}_{ae}$

The component table will be as below:

| X | Y |

V | $\left(-{V}_{pa}\right)\mathrm{cos}{5}^{\xb0}$ | $\left(-{V}_{pa}\right)\mathrm{sin}{5}^{\xb0}$ |

V | 42.3 | -15.4 |

Resultant | $\left(-{V}_{pe}\right)\mathrm{cos}{45}^{\xb0}$ | $\left(-{V}_{pe}\right)\mathrm{sin}{45}^{\xb0}$ |

Here we need to find the x component and the y component.

$Xcomponent\phantom{\rule{0ex}{0ex}}{V}_{x}={V}_{i}\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}{V}_{x}=\left(-{V}_{pa}\right)\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}{V}_{x}=\left(-{V}_{pa}\right)\mathrm{cos}{5}^{\xb0}$

$Ycomponent\phantom{\rule{0ex}{0ex}}{V}_{y}={V}_{i}\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}{V}_{y}=\left(-{V}_{pa}\right)\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}{V}_{y}=\left(-{V}_{pa}\right)\mathrm{sin}{5}^{\xb0}$

Hence, the velocity of air w.r.t to the earth component :

$\begin{array}{rcl}& & XComponent\\ {V}_{x}& =& {V}_{i}\mathrm{cos}\theta \\ {V}_{x}& =& \left(45\right)\mathrm{cos}{20}^{\xb0}\\ {V}_{x}& =& 42.3\end{array}$

$\begin{array}{rcl}& & YComponent\\ {V}_{y}& =& {V}_{i}\mathrm{sin}\theta \\ {V}_{y}& =& \left(-45\right)\mathrm{sin}{40}^{\xb0}\\ {V}_{y}& =& -15.4\end{array}$

Now we have given the angle of the resultant vector; hence

$Xcomponent\phantom{\rule{0ex}{0ex}}{V}_{x}={V}_{pe}\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}{V}_{pex}=\left(-{V}_{pe}\right)\mathrm{cos}{45}^{\xb0}\phantom{\rule{0ex}{0ex}}Ycomponent\phantom{\rule{0ex}{0ex}}{V}_{pey}={V}_{i}\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}{V}_{y}=\left(-{V}_{pe}\right)\mathrm{sin}{45}^{\xb0}$

Hence, putting the value in the equation

$\left(-{V}_{pa}\right)\mathrm{cos}{45}^{\xb0}+42.3=\left(-{V}_{pe}\right)\mathrm{cos}{45}^{\xb0}......\left(1\right)\phantom{\rule{0ex}{0ex}}Andequationtois:\phantom{\rule{0ex}{0ex}}\left(-{V}_{pa}\right)\mathrm{sin}{45}^{\xb0}+\left(-15.4\right)=\left(-{V}_{pe}\right)\mathrm{sin}{45}^{\xb0}......\left(2\right)$

After solving the above two equations, we can get the value of V_{pa} and V_{pe}.

Hence, V_{pa}= 63.5 m/s.

The velocity of the plane with respect to air is 63.5 m/s.

b) The speed of the plane relative to the earth should be

${V}_{pe}=\frac{{V}_{pa}\mathrm{sin}{5}^{\xb0}+15.4}{\mathrm{sin}{45}^{\xb0}}\phantom{\rule{0ex}{0ex}}{V}_{pe}=29.6\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

The velocity of the plane relative to the earth is 29.6 m/s.

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