NotTim
  • NotTim
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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
NotTim
  • NotTim
not physics
anonymous
  • anonymous
Vector addition?
NotTim
  • NotTim
application of vector addition, yah.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Draw it out. Find it's components. Add their x componants and y componants separately. Put them back together in a triangle.
NotTim
  • NotTim
i somehow figured out how to get resultant, but not the angle...
NotTim
  • NotTim
im not suppose to use component method
anonymous
  • anonymous
Use tan(y/x)
anonymous
  • anonymous
Oh I see. THen I'm afraid I can be of no help to you.
NotTim
  • NotTim
ok...oh well. thx u
radar
  • radar
draw it out, and remember when directions are concerned, 0 degrees is North.
NotTim
  • NotTim
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
NotTim
  • NotTim
yeah, i forgot it was bearings
radar
  • radar
O.K, good luck on this problem it is a good question
radar
  • radar
Oh yes, note that the wind is FROM a bearing of 315 degrees in other words the force is towards E 45 S.
radar
  • radar
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
NotTim
  • NotTim
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?
NotTim
  • NotTim
@Radar Hey....do you have time? I need you to please revisit this...
NotTim
  • NotTim
|dw:1333520833003:dw|
NotTim
  • NotTim
thats how i came to my angle. the correct answer is a BEARING of 211.9 degreees. my answer is not a bearing
anonymous
  • anonymous
ur question is a hard 1. i tried 8 times and cant figure it out. sorry
NotTim
  • NotTim
ok. its alright.
anonymous
  • anonymous
Okay ive done this before. Let me dig up my notes.
NotTim
  • NotTim
hello?
NotTim
  • NotTim
computer went wonky.
anonymous
  • anonymous
Sorry. Were you given the answer?
NotTim
  • NotTim
yeah.its 154.0km/h at a bearing of 211.9 degrees.
anonymous
  • anonymous
Draw your version of the diagram. I wanna see if we are using the same one
NotTim
  • NotTim
i drew 2...
NotTim
  • NotTim
or do you want what the txtbook has>?
anonymous
  • anonymous
yeah that would be good
NotTim
  • NotTim
|dw:1333522792176:dw|
Hero
  • Hero
I like how "not physics" is the first thing NotTim says.
NotTim
  • NotTim
you know, just in case.
anonymous
  • anonymous
So we are finding the resultant vector?
NotTim
  • NotTim
i already found it (somehow...I don't really know how). I'm looking for the angle now.
NotTim
  • NotTim
bump
experimentX
  • experimentX
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
NotTim
  • NotTim
argghh. that's component method. im not suppose to use component method!
experimentX
  • experimentX
since they don't have same directions, i don't think you have other choice ...
NotTim
  • NotTim
what about application of vector additions? using sin adn cos law?
experimentX
  • experimentX
let's try it.
anonymous
  • anonymous
\[\tan(\theta)=\frac{35}{150}, \theta=13.13\deg\] subtract from 225
NotTim
  • NotTim
r^2=150^2+35^2-2(150)(35)cos90
anonymous
  • anonymous
theta is angle between resultant and original flight
NotTim
  • NotTim
wait...so trig ratios work, even with non-right angle triangles?
experimentX
  • experimentX
|dw:1333520171086:dw|
anonymous
  • anonymous
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
experimentX
  • experimentX
|dw:1333520250197:dw|
anonymous
  • anonymous
225-135=90
NotTim
  • NotTim
its 315...
NotTim
  • NotTim
im sure thats not an error in my grammar
anonymous
  • anonymous
blowing from a bearing of 315º is the same as blowing at a bearing of 135º
experimentX
  • experimentX
|dw:1333520364868:dw|
anonymous
  • anonymous
if there was wind ahead of you at a bearing of 315 degrees and it passed by you, it'd be going at 135º
NotTim
  • NotTim
dock, i dont udnerstand. how would i derive that info?
experimentX
  • experimentX
i think that would be 180 - (315-225) and apply cosine law .. i think that might work
anonymous
  • anonymous
if i tell you wind is coming from in front of you, which way is it going to push you?
anonymous
  • anonymous
the same way its coming from?
NotTim
  • NotTim
no; I know the different of to and from (Radar explained that); how would I figure out:|dw:1333531452863:dw|
anonymous
  • anonymous
you have a magnitutude of 35 km/h wind, you shift the vector over to the end of the other vector
NotTim
  • NotTim
yeah, but how EXACTLY do i figure out the new angle/bearing formed is 135?
anonymous
  • anonymous
|dw:1333520788256:dw| you could do as you did above.
anonymous
  • anonymous
draw a line from 315 bearing through the origin
NotTim
  • NotTim
Yes, I understand that. I want to know how I can derive 135 from 315 when i get the equilibrant vector, like you did.
anonymous
  • anonymous
its just intuitive, not really something you should have to prove
anonymous
  • anonymous
just the wording of the problem
NotTim
  • NotTim
no, but how do i figure that out myself? to get 135 from the equilibrant of 315?
NotTim
  • NotTim
i dont know if im making sense here...
anonymous
  • anonymous
they aren't giving you the true bearing of the wind's velocity
anonymous
  • anonymous
they are giving it to you from a different reference
NotTim
  • NotTim
I understand that. Yes. Very much so. The difference between to and from. But the angle, how did you derive your angle like that?
anonymous
  • anonymous
its so far up now. what angle did i derive?
NotTim
  • NotTim
135
NotTim
  • NotTim
the wind blowing.
anonymous
  • anonymous
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
NotTim
  • NotTim
How did you calculate 135degrees?
anonymous
  • anonymous
315-180
NotTim
  • NotTim
oohhhkay.
anonymous
  • anonymous
the observer (plane) and wind, if looking at each other, form a 180º line of sight
anonymous
  • anonymous
or a straight line
NotTim
  • NotTim
ok.
NotTim
  • NotTim
now that i figured that out, what's our next step?
anonymous
  • anonymous
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
NotTim
  • NotTim
we werent taught arctan (to add to my problems)
NotTim
  • NotTim
or, that was long before
anonymous
  • anonymous
this will give you the angle between the resultant and the plane's direction
NotTim
  • NotTim
oh. trig ratios. tanangle=o/a right
anonymous
  • anonymous
arctan(35/150) is approximately 13.13º
NotTim
  • NotTim
how do i know which one is my opposite and adjacent?
anonymous
  • anonymous
|dw:1333521664877:dw|
NotTim
  • NotTim
oh bugger. I thought we were looking for another angle. this clarfies much,.
anonymous
  • anonymous
as you can see, the magnitutde can also be calculated with pythagorean theorem
NotTim
  • NotTim
yes.
NotTim
  • NotTim
i was just seeing if i could do this if it werent a right angle triangle. apparently i couldnt.
NotTim
  • NotTim
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?
NotTim
  • NotTim
thx with sticking with me here folks, greatly appreciated.
anonymous
  • anonymous
subtract that angle from the plane's bearing
anonymous
  • anonymous
yw
NotTim
  • NotTim
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.
radar
  • radar
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.