Airplane Traveling East against Wind

In summary, the plane is travelling at 95.26 m/s with respect to the ground and is heading east of due north.
  • #1
zjamal2
2
0

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!
 
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  • #2
welcome to pf!

hi zjamal2! welcome to pf! :wink:
zjamal2 said:
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 :redface:

call the velcocity of the plane wrt the air VPA, and the velcocity of the air wrt the ground VAG

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

you want VPG, which is VPA + VAG

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 :smile:
 
  • #3
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
 
  • #4
hi zjamal2! :smile:

(just got up :zzz: …)
zjamal2 said:
But then again you mentioned that G is a vector

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

GA (or VGA) 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" :wink:
 
  • #5

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.
 

1. How does wind affect an airplane traveling east?

Wind can have a significant impact on the speed and stability of an airplane traveling east. If the wind is blowing in the same direction as the airplane's travel, it can increase its speed and make the flight smoother. However, if the wind is blowing in the opposite direction, it can slow down the airplane and cause turbulence.

2. What is the reason for the difference in flight time when traveling east against the wind?

The difference in flight time when traveling east against the wind is due to a concept called airspeed. Airspeed is the speed of the airplane relative to the air around it. When there is a headwind (wind blowing against the direction of travel), the airspeed decreases, causing the flight to take longer. On the other hand, when there is a tailwind (wind blowing in the same direction as travel), the airspeed increases, resulting in a shorter flight time.

3. Are there any safety concerns for airplanes traveling east against the wind?

Yes, there can be safety concerns for airplanes traveling east against the wind. Strong headwinds can make it challenging for the airplane to maintain its desired altitude, and the pilot may need to make adjustments to compensate for the wind. Additionally, severe turbulence caused by strong winds can be hazardous for both passengers and the airplane.

4. How do pilots prepare for flying east against the wind?

Pilots are trained to take into account the wind conditions when planning a flight, especially when traveling east against the wind. They use weather reports and forecasts to determine the strength and direction of the wind, and then make adjustments to the flight plan accordingly. This may include altering the altitude, speed, or route of the flight.

5. Can airplanes fly faster when traveling east with a strong tailwind?

Yes, airplanes can fly faster when traveling east with a strong tailwind. A tailwind can increase the airspeed, allowing the airplane to cover more distance in a shorter amount of time. However, pilots must ensure that they maintain a safe airspeed and do not exceed the aircraft's limits, even with a strong tailwind.

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