# Plane relative velocity

1. Feb 10, 2013

### Toranc3

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

An airplane pilot sets a compass course due west and maintains an airspeed of 220 km/h. After flying for a time of 0.500 h, she finds herself over a town a distance 120 km west and a distance 20 km south of her starting point.

Find the wind velocity
2. Relevant equations

Vp/e= vp/a +va/e

3. The attempt at a solution\

vp/e= velocity of plane relative to earth
vp/a = velocity of plane relative to air
va/e= velocity of air relative to earth
Vp/e= vp/a +va/e

This is all I have and I am stuck.

Vp/e=220km/h+va/e
I also did the pythagorean theorem with the distances given to get my resultant.

R=sqrt[ (120km)^(2) + (20km)^(2) ]
R=122 km.
Not sure what to do with this though. Could somebody point me in the right direction?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Feb 10, 2013
2. Feb 10, 2013

### Bostonpancake0

try working with the x and y components of the velocity acting on the airplane. once you have found the x and y components of the winds velocity u should be able to find its resultant speed and directon.

3. Feb 10, 2013

### Toranc3

hmm could i get some more hints?

4. Feb 11, 2013

### Bostonpancake0

try setting up and x and y axis, and work with the velocities of the air relative to the plane and vise versa using information provided. See how you used pythag for the displacement to find the resultant displacement. Try instead find the velocity of the air in the x plane and y plane and use pythag to solve. If you dont know what i mean I will post my working. (by the way what is the answer, just to be sure i am correct haha)

5. Feb 11, 2013

### Toranc3

44.7km/h 63.43 degrees south of west. Could I see your drawing too?

6. Feb 11, 2013

### Bostonpancake0

sorry if its messy

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7. Feb 11, 2013

### Bostonpancake0

sorry theta is meant to be on other side

8. Feb 12, 2013

### Toranc3

thanks buddy

9. Feb 12, 2013

### HallsofIvy

Staff Emeritus
You don't have to have a coordinate system for problems like this. You can use the Pythagorean theorem to find the length of the "actual course" of the air plane and arctan(20/120) gives the angle below the horizontal (west). So you know the lengths of two sides of the triangle (110 km west on the compass course and the length above) and the angle between those sides, arctan(20/120). You can use the cosine law to find the third side of the triangle and the sine law to find the angles.