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Q70E

Expert-verifiedFound in: Page 125

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Construct Your Own Problem Consider an airplane headed for a runway in a cross wind. Construct a problem in which you calculate the angle the airplane must fly relative to the air mass in order to have a velocity parallel to the runway. Among the things to consider are the direction of the runway, the wind speed and direction (its velocity) and the speed of the plane relative to the air mass. Also calculate the speed of the airplane relative to the ground. Discuss any last minute maneuvers the pilot might have to perform in order for the plane to land with its wheels pointing straight down the runway.**

- $69.3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$
- $81.{8}^{\circ}.$

**Velocity is the rate of change in position of an item in motion as seen from a specific frame of reference and measured by a specific time standard.**

The velocity of the plane relative to the earth will be considered resultant.

This problem is the relative velocity problem; we can use the relative velocity formula.

Hence the velocity can be calculated by the following equation.

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

The component table will be as below:

| X | Y |

V | $\left(-70\right)\mathrm{cos}\theta $ | $\left(-70\right)\mathrm{sin}\theta $ |

V | 10 | 0 |

Resultant | 0 (plane is landing parallel to runway) | $\left(-70\right)\mathrm{sin}\theta $ |

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

$Xcomponent\phantom{\rule{0ex}{0ex}}{V}_{pax}={V}_{i}\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}{V}_{pax}=\left(-70\right)\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}Ycomponent\phantom{\rule{0ex}{0ex}}{V}_{pay}={V}_{i}\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}{V}_{pay}=\left(-70\right)\mathrm{sin}\theta $

Hence, solving the resultant equation to get the value of theta:

$\begin{array}{rcl}\left(-70\right)\mathrm{cos}\theta +10& =& 0\\ \left(70\right)\mathrm{cos}\theta & =& 10\\ \mathrm{cos}\theta & =& \frac{10}{70}\\ \theta & =& 81.{8}^{\xb0}.\end{array}$

The angle of the direction will be $81.{8}^{\xb0}$ .

Hence the resultant vector will be:

$\begin{array}{rcl}\stackrel{\rightharpoonup}{R}& =& \sqrt{{\left(\sum _{}X\right)}^{2}+{\left(\sum _{}Y\right)}^{2}}\\ \stackrel{\rightharpoonup}{R}& =& \sqrt{{\left(0\right)}^{2}+{\left(-70\mathrm{sin}81.8\right)}^{2}}\\ \stackrel{\rightharpoonup}{R}\left({V}_{pe}\right)& =& 63.3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

The velocity of the plane relative to the earth is $63.3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ .

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