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 2 years ago
A small aircraft, on a heading of 225 degrees, is cruising at 150km/h. IT is encountering a wind blowing from a bearing of 315 degrees at 35km/h. Determine the aircraft's ground velocity.
 2 years ago
A small aircraft, on a heading of 225 degrees, is cruising at 150km/h. IT is encountering a wind blowing from a bearing of 315 degrees at 35km/h. Determine the aircraft's ground velocity.

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NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0application of vector addition, yah.

LordHades
 2 years ago
Best ResponseYou've already chosen the best response.1Draw it out. Find it's components. Add their x componants and y componants separately. Put them back together in a triangle.

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0i somehow figured out how to get resultant, but not the angle...

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0im not suppose to use component method

LordHades
 2 years ago
Best ResponseYou've already chosen the best response.1Oh I see. THen I'm afraid I can be of no help to you.

radar
 2 years ago
Best ResponseYou've already chosen the best response.0draw it out, and remember when directions are concerned, 0 degrees is North.

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0i did....but i didnt draw it here. i got the magnitude, but not the angle. i think though that i might have read the answer wrong...im still double checking again. but thank you

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0yeah, i forgot it was bearings

radar
 2 years ago
Best ResponseYou've already chosen the best response.0O.K, good luck on this problem it is a good question

radar
 2 years ago
Best ResponseYou've already chosen the best response.0Oh yes, note that the wind is FROM a bearing of 315 degrees in other words the force is towards E 45 S.

radar
 2 years ago
Best ResponseYou've already chosen the best response.0Note also the angle between the aircraft and the wind vector is 90 degrees. as aircraft is flying on a bearing of W 45 S

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0hey, radar, i got 13.1 degrees (not considering bearings), as the angle between the resultant and the aircraft's cruising speed. Um. What should I do with that?

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0@Radar Hey....do you have time? I need you to please revisit this...

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0thats how i came to my angle. the correct answer is a BEARING of 211.9 degreees. my answer is not a bearing

lexi_rocks_lexi
 2 years ago
Best ResponseYou've already chosen the best response.0ur question is a hard 1. i tried 8 times and cant figure it out. sorry

bobobobobb
 2 years ago
Best ResponseYou've already chosen the best response.1Okay ive done this before. Let me dig up my notes.

bobobobobb
 2 years ago
Best ResponseYou've already chosen the best response.1Sorry. Were you given the answer?

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0yeah.its 154.0km/h at a bearing of 211.9 degrees.

bobobobobb
 2 years ago
Best ResponseYou've already chosen the best response.1Draw your version of the diagram. I wanna see if we are using the same one

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0or do you want what the txtbook has>?

bobobobobb
 2 years ago
Best ResponseYou've already chosen the best response.1yeah that would be good

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0I like how "not physics" is the first thing NotTim says.

bobobobobb
 2 years ago
Best ResponseYou've already chosen the best response.1So we are finding the resultant vector?

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0i already found it (somehow...I don't really know how). I'm looking for the angle now.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0find out the individual horizontal and vertical factors ... add them or substract them according to necessity ... and reconstruct them. this is how these kinds of problems are done

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0argghh. that's component method. im not suppose to use component method!

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0since they don't have same directions, i don't think you have other choice ...

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0what about application of vector additions? using sin adn cos law?

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1\[\tan(\theta)=\frac{35}{150}, \theta=13.13\deg\] subtract from 225

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0r^2=150^2+35^22(150)(35)cos90

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1theta is angle between resultant and original flight

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0wait...so trig ratios work, even with nonright angle triangles?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1333520171086:dw

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1it is a right triangle because the plane is flying at 225 degrees, wind coming from 315 degrees or going at a bearing of 135

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1333520250197:dw

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0im sure thats not an error in my grammar

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1blowing from a bearing of 315º is the same as blowing at a bearing of 135º

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1333520364868:dw

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1if there was wind ahead of you at a bearing of 315 degrees and it passed by you, it'd be going at 135º

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0dock, i dont udnerstand. how would i derive that info?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0i think that would be 180  (315225) and apply cosine law .. i think that might work

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1if i tell you wind is coming from in front of you, which way is it going to push you?

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1the same way its coming from?

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0no; I know the different of to and from (Radar explained that); how would I figure out:dw:1333531452863:dw

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1you have a magnitutude of 35 km/h wind, you shift the vector over to the end of the other vector

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0yeah, but how EXACTLY do i figure out the new angle/bearing formed is 135?

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1333520788256:dw you could do as you did above.

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1draw a line from 315 bearing through the origin

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, I understand that. I want to know how I can derive 135 from 315 when i get the equilibrant vector, like you did.

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1its just intuitive, not really something you should have to prove

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1just the wording of the problem

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0no, but how do i figure that out myself? to get 135 from the equilibrant of 315?

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0i dont know if im making sense here...

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1they aren't giving you the true bearing of the wind's velocity

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1they are giving it to you from a different reference

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0I understand that. Yes. Very much so. The difference between to and from. But the angle, how did you derive your angle like that?

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1its so far up now. what angle did i derive?

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1its just a reference frame. from an observer's point of view, the wind is at a bearing of 315º. from the wind's perspective, the observer is at abearing of 135º

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0How did you calculate 135degrees?

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1the observer (plane) and wind, if looking at each other, form a 180º line of sight

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0now that i figured that out, what's our next step?

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1you know the magnitude of the plane's velocity and window's velocity. they form a right angle. the picture i drew above shows this. use arctan

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0we werent taught arctan (to add to my problems)

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0or, that was long before

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1this will give you the angle between the resultant and the plane's direction

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0oh. trig ratios. tanangle=o/a right

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1arctan(35/150) is approximately 13.13º

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0how do i know which one is my opposite and adjacent?

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1333521664877:dw

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0oh bugger. I thought we were looking for another angle. this clarfies much,.

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1as you can see, the magnitutde can also be calculated with pythagorean theorem

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0i was just seeing if i could do this if it werent a right angle triangle. apparently i couldnt.

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0ok. we're near the end now. how do i figure out hte angle, as in bearings? we've got 13.13 degrees. wat next?

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0thx with sticking with me here folks, greatly appreciated.

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.1subtract that angle from the plane's bearing

NotTim
 2 years ago
Best ResponseYou've already chosen the best response.0thank you all. your assistance has been greatly appreciated. I wish you the best of luck in your studies, and that you are able to help others as greatly as you have done for me.

radar
 2 years ago
Best ResponseYou've already chosen the best response.0Had to leave, but I see that you got some excellent help and advice from Dockworker. I agree with the results. That angle of 13.13 degrees is not the bearing but it is the correct deviation from his original bearing. The bearing rose goes clockwise with 0 degrees north 180 degrees due south etc.
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