Relative Velocity Vector question

Click For Summary

Homework Help Overview

The problem involves an airplane's velocity in relation to wind conditions, specifically addressing relative velocity vectors. The airplane is flying due south at a specified speed while a wind from the southwest affects its trajectory. Participants are tasked with calculating the airplane's ground speed, the displacement from its intended path, and the necessary heading adjustment to maintain a due south course.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants attempt to calculate the resultant velocity using the Pythagorean theorem, questioning whether this method is appropriate given the wind's direction.
  • Others express confusion about the validity of their calculations and the implications of the wind's effect on the airplane's speed.
  • Several participants suggest breaking down the vectors into components to better understand the relationships between the airplane's velocity and the wind.
  • Questions arise regarding the assumptions made in the problem setup, particularly about the nature of the triangle formed by the vectors.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the problem. Some have provided guidance on using vector components to approach the calculations, while others are still grappling with the initial assumptions and calculations. There is no explicit consensus on the correct method yet, but productive dialogue is occurring.

Contextual Notes

Participants note that the problem may involve complexities due to the wind's direction and the resulting vector triangle not being a right triangle. There is also mention of potential misunderstandings regarding the teacher's expectations for the calculations.

rvnt
Messages
14
Reaction score
0

Homework Statement


An airplane is heading due south at a speed of 600km/h. If the wind begins to blow from the southwest at a speed of 100km/h (average), calculate:
a) The velocity (magnitude and direction) of the plane relative to the ground
b) How far from its intended position will it be after 10min if the pilot takes no corrective action
c) In what direction should the pilot aim the plane so that it will fly due south


Homework Equations



V= Square root of: Vx^2 + Vy^2

The Attempt at a Solution


a) Square root of 600km/h^2 + 100km/h^2 = 608.27 km/h
b) 100km/h * 0.166 h= 16.6 km
c) Sin (600/100km/h)= 0.1045 degrees West of south
 
Physics news on Phys.org
rvnt said:

Homework Statement


An airplane is heading due south at a speed of 600km/h. If the wind begins to blow from the southwest at a speed of 100km/h (average), calculate:
a) The velocity (magnitude and direction) of the plane relative to the ground
b) How far from its intended position will it be after 10min if the pilot takes no corrective action
c) In what direction should the pilot aim the plane so that it will fly due south


Homework Equations



V= Square root of: Vx^2 + Vy^2

The Attempt at a Solution


a) Square root of 600km/h^2 + 100km/h^2 = 608.27 km/h
b) 100km/h * 0.166 h= 16.6 km
c) Sin (600/100km/h)= 0.1045 degrees West of south

a) Groundspeed cannot be greater than airspeed if you have a quartering headwind.
b) I'd use .167, however you're correct as the problem is stated. You're sure the teacher didn't ask for distance off course?
c) Not sure how you got this. Off by two orders of magnitude.

Please look at the attached graphic.
 

Attachments

  • WindTriangle.jpg
    WindTriangle.jpg
    13 KB · Views: 2,878
I understand the vector diagram properly now but I still am troubled with a). I realize the answer 608.28km/h is too high but what am I doing wrong? Square root of 600km/h^2 + 100km/h^2 is the only way I can think of to approach this questions?
 
rvnt said:
I understand the vector diagram properly now but I still am troubled with a). I realize the answer 608.28km/h is too high but what am I doing wrong? Square root of 600km/h^2 + 100km/h^2 is the only way I can think of to approach this questions?
That only works for a right triangle. But the triangle (Labelled "wind triangle") posted by AC130Nav is not a right triangle.

Instead, you'll need to combine the wind and the due-southward vectors. One way to do this is: figure out horizontal and vertical components for each of the two vectors, then add the components to get the resultant vector (Labelled "resultant track" in AC's figure).
 
rvnt said:
I understand the vector diagram properly now but I still am troubled with a). I realize the answer 608.28km/h is too high but what am I doing wrong? Square root of 600km/h^2 + 100km/h^2 is the only way I can think of to approach this questions?

If you draw a horizontal line on my left diagram through the intersection of the resultant track and the 100 kph lines, you will break the wind triangle into two right triangles, one of which has 45 degree corners. You can calculate the distances and then reverse calculate the speed along the resultant track in 10 minutes or simply pretend he did this for an hour and the 600 kph is 600k reduced by the side of the 45 degree right triangle, which is also the other side of the upper half right triangle whose hypotenuse is then the groundspeed along the resultant track in k and kph.

You'll need trig for the course correction.
 

Similar threads

Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
4K
Replies
25
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K