# Homework Help: Relative Velocity Vector question

1. Aug 5, 2010

### rvnt

1. The problem statement, all variables and given/known data
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

2. Relevant equations

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

3. 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

2. Aug 6, 2010

### AC130Nav

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.

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3. Aug 7, 2010

### rvnt

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?

4. Aug 7, 2010

### Redbelly98

Staff Emeritus
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).

5. Aug 7, 2010

### AC130Nav

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.