How Fast is the Airplane Flying Relative to the Air?

In summary: The vector that resulted from the new direction of wind is 72i-hat + 156.74j-hat, which has a magnitude of 172.5km/hr.
  • #1
Teslapb
10
0

Homework Statement


After flying for 15 min in a wind blowing 42 km/h at an angle of 31° south of east, an airplane pilot is over a town that is 50 km due north of the starting point. What is the speed of the airplane relative to the air?

Homework Equations


Vx of wind=42km/hr*cos(31)
Vy of wind=42km/hr*sin(31)

The Attempt at a Solution


I found the velocity needed by the plant to travel that distance and stay in a straight line despite the wind and I found it to be -36i-hat+221.63j-hat. I then subtracted that from the components of the wind velocity and found the magnitude of the resulting vector. What am I missing here?
 
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  • #2
Teslapb said:
found it to be -36i-hat+221.63j-hat.
Fine so far, but just to be clear, is that the ground velocity or velocity relative to air?
Once you have found the velocity relative to the air, how do you turn that into a speed?
 
  • #3
I found that to be the velocity relative to the air. I then found the magnitude of that vector to me the relative speed.
 
  • #4
Teslapb said:
I found that to be the velocity relative to the air. I then found the magnitude of that vector to me the relative speed.
But originally you wrote
Teslapb said:
I then subtracted that from the components of the wind velocity
Did you do that or not?
 
  • #5
Yes I did that. That was the relative velocity to the air and then the magnitude of that is what I then used as my answer because it is asking for speed not velocity
 
  • #6
Teslapb said:
Yes I did that. That was the relative velocity to the air and then the magnitude of that is what I then used as my answer because it is asking for speed not velocity
We're going round in circles. You wrote:
Teslapb said:
I found the velocity needed by the plant to travel that distance and stay in a straight line despite the wind and I found it to be -36i-hat+221.63j-hat.
That is the velocity of the plane relative to the air, right? So...
Teslapb said:
I then subtracted that from the components of the wind velocity
... why did you do that?
 
  • #7
I see what you're saying now. You are right I did not subtract the components of the air velocity twice. I did so once and that is the answer I got in i j format. I then found the magnitude of that vector.
 
  • #8
Teslapb said:
I see what you're saying now. You are right I did not subtract the components of the air velocity twice. I did so once and that is the answer I got in i j format. I then found the magnitude of that vector.
Then you should have got the answer. What answer did you get? Do you believe it to be wrong? Do you know what it is supposed to be?
The only doubt I have with the question is whether the given wind direction is where it is headed or where it is coming from. In everyday usage, a West wind, for example, means a wind from the West.
 
  • #9
My previous answers into my online homework system have been 243, 200, 253.4, and 253.7 km/hr. I believe the last two to be correct. If you can find no other flaw in my work then I think you are correct, the wind is blowing FROM 31 degrees south of east.
.
 
  • #10
The vector that resulted from the new direction of wind is 72i-hat + 156.74j-hat, which has a magnitude of 172.5km/hr.
 
  • #11
Teslapb said:
My previous answers into my online homework system have been 243, 200, 253.4, and 253.7 km/hr. I believe the last two to be correct. If you can find no other flaw in my work then I think you are correct, the wind is blowing FROM 31 degrees south of east.
.
From your (36, 221.63) I get 224.53. How did you get those four numbers?
Teslapb said:
The vector that resulted from the new direction of wind is 72i-hat + 156.74j-hat, which has a magnitude of 172.5km/hr.
Reversing the wind direction is not going to change the magnitude of the EW component.
 
  • #12
After relooking at the problem I think that I mistakenly put 36 for i-hat when it would need to be -72 or +72 because the plane needs to match the wind velocity in the opposite direction. For example if the plane x velocity is +36 to cancel the -36 wind velocity it would be 36-(-36)=72. The other thing I noticed was that if the plane is going 221.63 km/hr to counteract the wind and get to the spot 50km away, then 221.63-(-21.63) would be 243.26. So I think the vector which I need to find the magnitude of is 72i-hat +243.26j-hat. Except that is 253.7 which the program marked as incorrect.
 
  • #13
Teslapb said:
I think that I mistakenly put 36 for i-hat when it would need to be -72 or +72 because the plane needs to match the wind velocity in the opposite direction
No, you were right the first time. If the wind's EW component is 36kmh to the E then to cancel that the plane's airspeed must have EW component 36kmh to the W.
 
  • #14
Right, and once you subtract those velocities to find the relative speed it becomes -36-36=72
 
  • #15
Teslapb said:
Right, and once you subtract those velocities to find the relative speed it becomes -36-36=72
The plane's airspeed is its relative speed. There is subtraction to be done.
 
  • #16
haruspex said:
The plane's airspeed is its relative speed. There is subtraction to be done.
So I do not know what that means.
 
  • #17
I submitted the answer of 224.5 and it was correct, I am just not sure why.
 
  • #18
Teslapb said:
So I do not know what that means.
Because I left out the word "no". There is no subtraction to be done.
 

FAQ: How Fast is the Airplane Flying Relative to the Air?

1. How does an airplane stay in the air?

An airplane stays in the air due to a combination of its design, including its wings and engines, and the principles of aerodynamics. When the wings of an airplane move through the air, they create lift, which balances out the weight of the airplane and keeps it in the air.

2. What is relative motion?

Relative motion is the perceived motion of an object in relation to another object. In the context of airplanes, relative motion refers to the movement of the airplane in relation to the ground or other objects in its surroundings.

3. Why do airplanes have different speeds at takeoff and cruising?

During takeoff, an airplane needs a higher speed to generate enough lift to become airborne. Once the airplane is cruising at a high altitude, it can maintain a lower speed due to the reduced air resistance at higher altitudes.

4. How does wind affect airplane and relative motion?

Wind can have a significant impact on an airplane's flight. A headwind, or wind blowing in the opposite direction of the airplane's motion, can slow down the airplane and make it harder to maintain a steady course. A tailwind, or wind blowing in the same direction as the airplane's motion, can increase its speed and make it easier to maintain a steady course.

5. What is the difference between airspeed and groundspeed?

Airspeed is the speed of the airplane in relation to the air around it. Groundspeed is the speed of the airplane in relation to the ground. Due to factors such as wind, airspeed and groundspeed may differ, with groundspeed being either faster or slower than airspeed.

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