Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

(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.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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:

$V_{p e} = V_{p a} + V_{a e}$

The component table will be as below:

	X	Y
V_pa	$(- V_{p a}) \cos 5^{°}$	$(- V_{p a}) \sin 5^{°}$
V_ae	42.3	-15.4
Resultant	$(- V_{p e}) \cos 45^{°}$	$(- V_{p e}) \sin 45^{°}$

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

$X c o m p o n e n t V_{x} = V_{i} \cos θ V_{x} = (- V_{p a}) \cos θ V_{x} = (- V_{p a}) \cos 5^{°}$

$Y c o m p o n e n t V_{y} = V_{i} \sin θ V_{y} = (- V_{p a}) \sin θ V_{y} = (- V_{p a}) \sin 5^{°}$

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

$\begin{array}{rcl} X C o m p o n e n t \\ V_{x} & = & V_{i} \cos θ \\ V_{x} & = & (45) \cos 20^{°} \\ V_{x} & = & 42.3 \end{array}$

$\begin{array}{rcl} Y C o m p o n e n t \\ V_{y} & = & V_{i} \sin θ \\ V_{y} & = & (- 45) \sin 40^{°} \\ V_{y} & = & - 15.4 \end{array}$

Step 3: Angle of the resultant vectors

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

$X c o m p o n e n t V_{x} = V_{p e} \cos θ V_{p e x} = (- V_{p e}) \cos 45^{°} Y c o m p o n e n t V_{p e y} = V_{i} \sin θ V_{y} = (- V_{p e}) \sin 45^{°}$

Hence, putting the value in the equation

$(- V_{p a}) \cos 45^{°} + 42.3 = (- V_{p e}) \cos 45^{°} . . . . . . (1) A n d e q u a t i o n t o i s : (- V_{p a}) \sin 45^{°} + (- 15.4) = (- V_{p e}) \sin 45^{°} . . . . . . (2)$

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.

Step 4: Speed of plane relative to the earth

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

$V_{p e} = \frac{V_{p a} \sin 5^{°} + 15.4}{\sin 45^{°}} V_{p e} = 29.6 \frac{m}{s}$

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

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Velocity of the plane relative to the earth

Step 3: Angle of the resultant vectors

Step 4: Speed of plane relative to the earth

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Velocity of the plane relative to the earth

Step 3: Angle of the resultant vectors

Step 4: Speed of plane relative to the earth

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