Relative Velocity with plane and wind

[Woolyabyss]

Homework Statement

An aircraft flew due east fro P to Q at u1 km/h.Wind speed from south west was v km/h.On the return journey from Q to P,due west, the aircraft's speed was u2 km/h, the wind velocity being unchanged.If the speed of the aircraft in still air was x km/h, x>v,show by resolving the perpendicular to PQ.or otherwise,that
u1 - u2 = v√2

Homework Equations

3. The Attempt at a Solution [/b/

Can somebody help me with this?I think I made a simple mistake.

vpw = (xcosa)i - (xsina)j

vw = (v√2/2)i + (v√2/2)j

vp = (xcosa + v√2/2)i +(xsina + v√2/2)

since the j component must be zero sina = 2v/2x .... cosa =√(4x^2 -2v^2)/2x

vp =x√(4x^2 -2v^2)/2x + v√2/2 = (v√2 + √(4x^2 -2v^2))/2 = u1


return journey

since the plane is traveling in the opposite direction

vpw = (-xcosa)i - (xsina)j

vp = (-xcosa + v√2/2)i +(-xsina + v√2/2)j

since j is zero cosa =√(4x^2-2v^2)/2x

u2=vp= (v√2 - √4x^2 -2v^2)/2 i

u1 - u2 = (√(4x^2 - 2v^2) +√(2v^2))/2 - (√(4x^2 -2v^2) - √(2v^2))/2

If I simplify it doesn't work out.Should I have multiplied u2 by minus 1 since its going in the minus i direction?

 

[Simon Bridge]

Probably the easiest way to check your working is to draw out the vector triangles and do the geometry.

 

[Woolyabyss]

Sorry I am not exactly sure what you mean.I get the correct answer if I use the vector as opposed to the magnitude of the vector.

 

[Simon Bridge]

The solution involves adding and subtracting vectors - there are two main ways of doing this. You have been resolving each vector into components N-S and E-W.
Have you tried sketching the vectors out using head-to-tail?

 

[Woolyabyss]

I just figured where I went wrong thanks anyway.