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Q60E

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Found in: Page 124

### College Physics (Urone)

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.

See the step by step solution

## Step 1: Given data

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.

## Step 2: Velocity of the plane 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 Vpa $\left(-{V}_{pa}\right)\mathrm{cos}{5}^{°}$ $\left(-{V}_{pa}\right)\mathrm{sin}{5}^{°}$ Vae 42.3 -15.4 Resultant $\left(-{V}_{pe}\right)\mathrm{cos}{45}^{°}$ $\left(-{V}_{pe}\right)\mathrm{sin}{45}^{°}$

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

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

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

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

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

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

## Step 3: Angle of the resultant vectors

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

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

Hence, putting the value in the equation

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

After solving the above two equations, we can get the value of Vpa and Vpe.

Hence, Vpa= 63.5 m/s.

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

## Step 4: Speed of plane relative to the earth

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

${V}_{pe}=\frac{{V}_{pa}\mathrm{sin}{5}^{°}+15.4}{\mathrm{sin}{45}^{°}}\phantom{\rule{0ex}{0ex}}{V}_{pe}=29.6m}{s}$

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