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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
 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
 2 years ago

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

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

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

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

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

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

NotTimBest ResponseYou've already chosen the best response.0
i 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
 2 years ago

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

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

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

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

NotTimBest ResponseYou've already chosen the best response.0
hey, 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?
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

experimentXBest ResponseYou've already chosen the best response.0
find 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
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

DockworkerBest ResponseYou've already chosen the best response.1
its 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
 2 years ago

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

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

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

DockworkerBest ResponseYou've already chosen the best response.1
you 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
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

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

NotTimBest ResponseYou've already chosen the best response.0
thank 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.
 2 years ago

radarBest ResponseYou've already chosen the best response.0
Had 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.
 2 years ago
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