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@iPhone5 Hmmm... are you sure that's the answer Well I got 14.6 degrees W of N and 266.1 km/h.

Massive trigonometry ahead so be ready lol

First draw a diagram according to the problem. We know that the plane is heading West of North because, pretend that it is flying straight north, it will be flying at tan(50/250) or ~11 degrees West of North, less than the 25 degrees we wanted. So it has to start with West of north.

Let's set t = the angle the plane is currently flying. You should have, on the diagram, an arrow starting from the bottom right pointing to the origin, and label the angle between y-axis and the arrow t.

Now, the horizontal component of the plane is simply 250 km/h * sin(t) to the west, and the vertical component is 250 km/h * cos(t) to the north. Just simple trig <3

Don't forget the wind. The wind is blowing to the west at 50 km/h.

The total resulting vector, which is an arrow starting from the origin pointing to the upper left, with the angle between the arrow and the y-axis = 25 degrees, has a horizontal component of the sum of all the other horizontal components up there pointing to the west. This simply means x = 50 km/h + 250 km/h * sin(t) to the west.

Then for the vertical component. The wind is not blowing at an angle so it's simply y = 250 km/h * cos(t).

Now hopefully you saw a triangle up there. You will find that tan(25) = total x / total y. Solve for t and you will get the angle measure.

Here's the math </3

tan (25) = (50 km/h + 250 km/h * sin(t)) / (250 km/h * cos(t))

And we know that tan (25) = sin (25) / cos (25), and we can simplify a little on the right side, so:

sin (25) / cos (25) = (1 km/h + 5 km/h * sin(t)) / (5 km/h * cos(t))

Cross multiply then distribution, and I will ignore units for now because it gets messy

cos (25) + 5 sin(t) cos(25) = 5 * sin(25) * cos (t)

cos (25) = 5 ( sin(25) cos (t) - sin (t) cos (25)) <- Expanded form of sin (A - B) <3

cos(25) = 5 (sin (25 - t))

cos(25) / 5 = sin (25 - t)

sin^-1 (cos(25)/5) = 25 - t

25 - t = 10.4

t = 14.6 degrees <- ANSWER

Now for the easy part (No more trigs!)

The magnitude for ground speed is simply sqrt( x^2 + y^2). Pythagorean Theorem.

And we said earlier that x = 50 km/h + 250 km/h * sin(t), and y = 250 km/h * cos(t). And we just figured out what "t" is.

I will leave the rest to you You should get something close to 266.17

Edit: Ninja'd

[quote=iPhone5]4.6 degrees W of N 267 km/h < that is the answer. Idk how to get to it. QQ

edit: I already tried what you did. does not work out,.[/quote]

I got 14.6 degrees W of N with the resultant ground speed being 267 km/h

[quote=iPhone5]4.6 degrees W of N 267 km/h < that is the answer. Idk how to get to it. QQ

edit: I already tried what you did. does not work out,.[/quote]

Oh, sorry my bad! Completely derped on that one,
To get that answer, use vector sum, its really hard to explain this without drawing it but you will have to use sin and cosine law as this does not only involve a right triangle, but also an obtuse triangle.

If you roughly sketch out where the plane is headed, it should be some where along the west as we can assume, (even if it was not, we would be be exactly 180 degrees over the opposite angle once we do the calculation, this can then be recognized as an error and we can change the direction later) then add another vector line directly west of it. This will show that the sum of the two vectors should meet at exactly 25 degrees from WofN, meaning the angle between the horizontal and direction of flight should be 90-25.
Therefore using the Sin law : 250/sin(65)=50/sin(25-x) x being the angle WofN for the direction of flight. Then using the angle, you can solve for your total velocity relative to the ground.
Edit: Answer is 14.6 degrees WofN, not 4.6

i think 25 west of north is 90+25= 115 degrees measured from the horizontal axis.

The better question is, can the plane do donuts?

[quote=Bellonoid]Break the direction the plane is travelling into horizontal and vertical components using similar triangles. Then take the horizontal component of the plane's motion (travelling west) and subtract it by 50m/s. Then take the angle of new triangle that is formed by doing so.[/quote]

4.6 degrees W of N 267 km/h < that is the answer. Idk how to get to it. QQ

edit: I already tried what you did. does not work out,.

Break the direction the plane is travelling into horizontal and vertical components using similar triangles. Then take the horizontal component of the plane's motion (travelling west) and subtract it by 50m/s. Then take the angle of new triangle that is formed by doing so.

Wow, probably one of, if not the worst, place to find physics help. Try Stack Exchange?

something looks wrong here but for the most part I'm getting 13 degrees W of N, x = -55.6 y = 243.7.

65 degrees is what you start with because 90 - 25 = 65.
the plane needs to move 226.6 km/h N and 105.6 km/h W to move in the 65 degrees direction.
to make up for the 50.0 km/h W winds, the plane will reduce it's direction of W by 50.0 km/h; in this case the plane is now moving at 55.6 km/h W.

cos^-1(55.6/250) = 77 degrees N of W = 13 degrees W of N

[quote=OhMahGun]Does the plane have to travel at 250 km/h or can it travel at whatever speed it wants?[/quote]

250

post is short

Draw your vectors, do your horizontal and vertical components, add your components, find your answer.

Or maybe I'm wrong, I'm too lazy to do this question, zzz.

Does the plane have to travel at 250 km/h or can it travel at whatever speed it wants?

help plz

Oh, relative motion... How I don't miss you.

oh opps, since i already made this thead, can i please recieve help.

[quote=iPhone5]sorry i didnt know where to put this. I wanted to put it on a forum with low popularity so it wont be spammy.[/quote]

for future reference:
http://www.basilmarket.com/show/forum/6
there's a chat section where people would be glad to help you with your homework

sorry i didnt know where to put this. I wanted to put it on a forum with low popularity so it wont be spammy.

this is definitely fashion