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.

- NotTim

- jamiebookeater

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- NotTim

not physics

- anonymous

Vector addition?

- NotTim

application of vector addition, yah.

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## More answers

- anonymous

Draw it out.
Find it's components.
Add their x componants and y componants separately.
Put them back together in a triangle.

- NotTim

i somehow figured out how to get resultant, but not the angle...

- NotTim

im not suppose to use component method

- anonymous

Use tan(y/x)

- anonymous

Oh I see. THen I'm afraid I can be of no help to you.

- NotTim

ok...oh well. thx u

- radar

draw it out, and remember when directions are concerned, 0 degrees is North.

- 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

yeah, i forgot it was bearings

- radar

O.K, good luck on this problem it is a good question

- 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

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

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

@Radar Hey....do you have time? I need you to please revisit this...

- NotTim

|dw:1333520833003:dw|

- 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

ur question is a hard 1. i tried 8 times and cant figure it out. sorry

- NotTim

ok. its alright.

- anonymous

Okay ive done this before. Let me dig up my notes.

- NotTim

hello?

- NotTim

computer went wonky.

- anonymous

Sorry. Were you given the answer?

- NotTim

yeah.its 154.0km/h at a bearing of 211.9 degrees.

- anonymous

Draw your version of the diagram. I wanna see if we are using the same one

- NotTim

i drew 2...

- NotTim

or do you want what the txtbook has>?

- anonymous

yeah that would be good

- NotTim

|dw:1333522792176:dw|

- Hero

I like how "not physics" is the first thing NotTim says.

- NotTim

you know, just in case.

- anonymous

So we are finding the resultant vector?

- NotTim

i already found it (somehow...I don't really know how). I'm looking for the angle now.

- NotTim

bump

- 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

argghh. that's component method. im not suppose to use component method!

- experimentX

since they don't have same directions, i don't think you have other choice ...

- NotTim

what about application of vector additions? using sin adn cos law?

- experimentX

let's try it.

- anonymous

\[\tan(\theta)=\frac{35}{150}, \theta=13.13\deg\]
subtract from 225

- NotTim

r^2=150^2+35^2-2(150)(35)cos90

- anonymous

theta is angle between resultant and original flight

- NotTim

wait...so trig ratios work, even with non-right angle triangles?

- experimentX

|dw:1333520171086:dw|

- 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

|dw:1333520250197:dw|

- anonymous

225-135=90

- NotTim

its 315...

- NotTim

im sure thats not an error in my grammar

- anonymous

blowing from a bearing of 315ยบ is the same as blowing at a bearing of 135ยบ

- experimentX

|dw:1333520364868:dw|

- anonymous

if there was wind ahead of you at a bearing of 315 degrees and it passed by you, it'd be going at 135ยบ

- NotTim

dock, i dont udnerstand. how would i derive that info?

- experimentX

i think that would be 180 - (315-225)
and apply cosine law ..
i think that might work

- anonymous

if i tell you wind is coming from in front of you, which way is it going to push you?

- anonymous

the same way its coming from?

- NotTim

no; I know the different of to and from (Radar explained that); how would I figure out:|dw:1333531452863:dw|

- anonymous

you have a magnitutude of 35 km/h wind, you shift the vector over to the end of the other vector

- NotTim

yeah, but how EXACTLY do i figure out the new angle/bearing formed is 135?

- anonymous

|dw:1333520788256:dw|
you could do as you did above.

- anonymous

draw a line from 315 bearing through the origin

- 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

its just intuitive, not really something you should have to prove

- anonymous

just the wording of the problem

- NotTim

no, but how do i figure that out myself? to get 135 from the equilibrant of 315?

- NotTim

i dont know if im making sense here...

- anonymous

they aren't giving you the true bearing of the wind's velocity

- anonymous

they are giving it to you from a different reference

- 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

its so far up now. what angle did i derive?

- NotTim

135

- NotTim

the wind blowing.

- anonymous

its just a reference frame. from an observer's point of view, the wind is at a bearing of 315ยบ. from the wind's perspective, the observer is at abearing of 135ยบ

- NotTim

How did you calculate 135degrees?

- anonymous

315-180

- NotTim

oohhhkay.

- anonymous

the observer (plane) and wind, if looking at each other, form a 180ยบ line of sight

- anonymous

or a straight line

- NotTim

ok.

- NotTim

now that i figured that out, what's our next step?

- 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

we werent taught arctan (to add to my problems)

- NotTim

or, that was long before

- anonymous

this will give you the angle between the resultant and the plane's direction

- NotTim

oh. trig ratios. tanangle=o/a right

- anonymous

arctan(35/150) is approximately 13.13ยบ

- NotTim

how do i know which one is my opposite and adjacent?

- anonymous

|dw:1333521664877:dw|

- NotTim

oh bugger. I thought we were looking for another angle. this clarfies much,.

- anonymous

as you can see, the magnitutde can also be calculated with pythagorean theorem

- NotTim

yes.

- NotTim

i was just seeing if i could do this if it werent a right angle triangle. apparently i couldnt.

- 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

thx with sticking with me here folks, greatly appreciated.

- anonymous

subtract that angle from the plane's bearing

- anonymous

yw

- 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

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.

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