- #1
bluefire90
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Two airplanes taxi as they approach the terminal. Plane 1 taxies with a speed of 11.3 m/s due north. Plane 2 taxies with a speed of 7.6 m/s in a direction 22.2° north of west. What is the direction of plane 1 relative to plane 2? Give the angle counterclockwise relative to the north. What is the direction of plane 2 relative to plane 1?
*I solved for magnitude first then velocity*
What is the magnitude of the velocity of plane 1 relative to plane 2?
To solve for the magnitude I found the vector components of plane 2:
Y component: 7.6 * sin(22.2°) = 2.87 m/s
X component: 7.6 * cos(22.2°)= 7.04 m/s
To find the final magnitude of the resultant vector I just used the pythagorean theorem after subtracting the plane 1 components FROM the plane 2 components.
2.87 m/s - 11.3 m/s = -8.43 m/s
7.04 m/s - 0 m/s = 7.04 m/s
Now to find the magnitude I used the pythagorean theorem.
Using the pythagorean theorom I got 10.98m/s as the relative velocity. This velocity is the same for plane 2 relative to plane 1.
To find the direction of plane 1 to plane 2 I just used the arctan formula
arctan (7.04/-8.43) = -39.86°
Now since its counterclockwise from north I just added to 360 to get 320° which is the correct answer.
I can't seem to figure out how to get the direction of plane 2 relative to plane 1 however and I've tried using the law of cosines and various other methods but I'm not sure where to go. Any help would be appreciated.
*I solved for magnitude first then velocity*
What is the magnitude of the velocity of plane 1 relative to plane 2?
To solve for the magnitude I found the vector components of plane 2:
Y component: 7.6 * sin(22.2°) = 2.87 m/s
X component: 7.6 * cos(22.2°)= 7.04 m/s
To find the final magnitude of the resultant vector I just used the pythagorean theorem after subtracting the plane 1 components FROM the plane 2 components.
2.87 m/s - 11.3 m/s = -8.43 m/s
7.04 m/s - 0 m/s = 7.04 m/s
Now to find the magnitude I used the pythagorean theorem.
Using the pythagorean theorom I got 10.98m/s as the relative velocity. This velocity is the same for plane 2 relative to plane 1.
To find the direction of plane 1 to plane 2 I just used the arctan formula
arctan (7.04/-8.43) = -39.86°
Now since its counterclockwise from north I just added to 360 to get 320° which is the correct answer.
I can't seem to figure out how to get the direction of plane 2 relative to plane 1 however and I've tried using the law of cosines and various other methods but I'm not sure where to go. Any help would be appreciated.
… that would be 40º …