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Carnegie Mellon University

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Problem 23 Hard Difficulty

A jet airliner moving initially at $3.00 \times 10^{2} \mathrm{mi} / \mathrm{h}$ due east enters a region where the wind is blowing $1.00 \times 10^{2} \mathrm{mi} / \mathrm{h}$ in a direction $30.0^{\circ}$ north of east. (a) Find the components of the velocity of the jet airliner relative to the air, $\overrightarrow{\mathbf{v}}_{\mathrm{jA}}$ . (b) Find the components of the velocity of the air relative to Earth, $\overrightarrow{\mathbf{v}}_{\mathrm{AF}}(\mathrm{c})$ Write an equation analogous to Equation 3.11 for the velocities $\overrightarrow{\mathbf{v}}_{\mathrm{JA}}, \overrightarrow{\mathbf{v}}_{\mathrm{AE}},$ and $\overrightarrow{\mathbf{v}}_{\mathrm{JE}}$ (d) What are the speed and direction of the aircraft relative to the ground?

Answer

a. $\left( \vec { v } _ { 1 A } \right) _ { z } = 3.00 \times 10 ^ { 2 } \mathrm { mi } / \mathrm { h }$
$\left( \vec { v } _ { 1 A } \right) _ { v } = 0$
b. $\left( \vec { v } _ { A B } \right) _ { x } = 86.6 \mathrm { mi } / \mathrm { h }$
$\left( \vec { v } _ { A E } \right) _ { y } = 50.0 \mathrm { mi } / \mathrm { h }$
c. $\vec { v } _ { 1 A } = \vec { v } _ { 1 E } - \vec { v } _ { A E }$
d. $3.90 \times 10 ^ { 2 }$ mi $/ \mathrm { h }$ , at $7.36 ^ { \circ }$ north of east

Discussion

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Video Transcript

so for party. We know that here the jet is moving 300 miles an hour due east due to air. So we're going to say that East is positive and north as positive, and Weston and South would, of course, be negative. And so we can say north, he's positive, and then Southwest Negative. And so from this we can say that then the X component of the velocity of the jet relative to the air is gonna be equaling 3.0 times 10 to the second miles per hour. And we can say that then the why component of the velocity of the jet relative to the air is gonna be equaling zero miles per hour. So those would be a two answers for per day. Now, for part B, we know that the velocity of the air relative to the Earth we can say velocity of the air relative to earth would be 1.0 times 10 to the second miles per hour at 30 degrees north of east. So here we can say the axe component is gonna be equaling the magnitude 100 miles an hour or 1.0 times 10 times 10 to the second. This would be multiplied by co sign of 30 degrees, and then the why component of the velocity of the air relative to the earth will be exact. Same thing 100 miles per hour, multiplied by sine of 30 degrees. And so for the X, we have 86.6 miles per hour and for the why we have 50.0 miles per hour. So those would be your two answers for part B. For part C. It's quite easy to figure out the equation. We know that the jet is essentially affected by the air and so relative to the Earth. We can actually say that the velocity of the jet relative to the earth will be equaling the velocity of the jet relative the air plus the velocity of the air relative to the Earth. Another way of saying this is simply that the velocity of the jet relative to the air is equaling the velocity of the jet relative to earth, minus the velocity of the air relative to the earth. Either answer is correct. And finally, four part d. We can say that, then the velocity of the jet relative to the earth in the X direction would be equaling the velocity of the jet relative to the air in the X direction, plus the velocity of the air relative to the earth in the X direction. So this is equaling. This would be 300 miles per hour, plus 86.6 miles per hour. So we can say that then the velocity of the jet relative to the earth and the ex direction is equaling 386.6 miles per hour. Now we can say that. Then we're actually have to convert to three. So we're actually enough around this to 3 87 Now we can do the exact same thing with the wide component and with the wide components a bit easier because we know that there isn't any why component for the velocity of the jet relative to the air. So essentially, the why component of the velocity of the jet relative to the earth would simply be equaling 50.0 miles per hour, exactly the same as the white component of the velocity of the air relative to the earth. And so we can then find the magnitude. We can then say that the magnitude of the velocity of the jet relative to the earth would be equaling the square root. Uh, we can say three 87 miles per hour quantity squared, plus 50 0.0 miles per hour quantity squared. The velocity, The magnitude of the velocity of the jet relative to the earth will be equaling approximately 390 miles per hour. So this would be our magnitude. And then for our direction, it would simply be our 10 of the why component divided by the X component. So this would be 50 divided by 387 and this is giving us 7.36 degrees. And this would be north of East, considering we have a positive X and a positive wide component. So those would be heir magnitude a new direction for Part D. That is the end of the solution. Thank you for watching

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