Airplane Traveling East against Wind

[zjamal2]

Homework Statement

You are on an airplane traveling due east at 110 m/s with respect to the air. The air is moving with a speed 43 m/s with respect to the ground at an angle of 30° west of due north.

1)What is the speed of the plane with respect to the ground?
>>> I answered this question and got 95.26m/s by finding the x-component of the plane's velocity. Meaning 110 * cos(30) = 95.26m/s

2)What is the heading of the plane with respect to the ground? (Let 0° represent due north, 90° represents due east). _________° East of due North
>>> Here is where I am confused. I understand that involves finding the components of the wind itself meaning 43 * cos(30) and 43 * sin(30) but aside from that I don't really know what to do with that information at all.

Homework Equations

Unsure. This is trigonometry-related and aside from basic sine cosine stuff I really don't remember anything or know what is needed to find the angle of something like this.

The Attempt at a Solution

Mentioned above. Thanks much in advance!

 

[tiny-tim]

welcome to pf!

hi zjamal2! welcome to pf!

1)What is the speed of the plane with respect to the ground?
>>> I answered this question and got 95.26m/s by finding the x-component of the plane's velocity. Meaning 110 * cos(30) = 95.26m/s

no

call the velcocity of the plane wrt the air V_PA, and the velcocity of the air wrt the ground V_AG …

velocities are vectors, so they obey the laws of vector addition …

you want V_PG, which is V_PA + V_AG …

there are two ways to do this …

i] draw a vector triangle of the three velocities, with vertices labelled P A and G, and with arrows along each side pointing the correct way, and use the sine and cosine rules

ii] use x and y components (separately)

choose either way (or both), and show us what you get ​

 

[zjamal2]

Thanks for your reply,
I'm new to this concept of vectors but am trying to make the vector triangle.

I realize that vector P's direction is towards the East and has a magnitude of 110mph.
But then again you mentioned that G is a vector and that makes me wonder what it could be since the ground has no magnitude or direction. The Air vector would be aimed Northwest since the Wind is moving at an angle 30 degrees from due north with a magnitude of 43 mph.

Sorry but still struggling with these basics. Please help me make more sense out of this. Thanks

 

[tiny-tim]

hi zjamal2!

(just got up :zzz: …)

But then again you mentioned that G is a vector

no, G is just a letter that stands for "ground"

GA (or V_GA) is a vector

you're missing the point:

all velocities are relative​

when we add velocities, we add relative velocities​

there is no "Air vector" or "Air velocity", only "the velocity of the Air relative to the ground" or "the velocity of the Air relative to the plane"

 

[rwisz]

To find the heading of the plane with respect to the ground, we need to find the angle between the plane's direction of motion and the north direction. This can be done by using the inverse tangent function (arctan).

First, we need to find the x and y components of the plane's velocity vector. As you mentioned, the x-component is 110 m/s * cos(30) = 95.26 m/s. The y-component can be found by using the sine function: 110 m/s * sin(30) = 55 m/s.

Next, we need to find the x and y components of the wind velocity vector. The x-component is 43 m/s * cos(30) = 37.19 m/s. The y-component is 43 m/s * sin(30) = 21.5 m/s.

Now, we can find the total velocity vector of the plane with respect to the ground by adding the x and y components of the plane's velocity and the wind's velocity. This can be done using the Pythagorean theorem: V = √(Vx^2 + Vy^2) = √((95.26 + 37.19)^2 + (55 + 21.5)^2) = 111.87 m/s.

Next, we can find the angle between the plane's velocity and the north direction by using the inverse tangent function: θ = arctan(Vy/Vx) = arctan(76.5/132.45) = 29.26°.

Therefore, the heading of the plane with respect to the ground is 29.26° east of due north.