Speed and Distance relative to still air

Question

You are flying a plane from Windsor to Gatio (N65E).

The wind is coming from N20W at 60 km/h.

What should your speed relative to the still air and heading be if you are to make the 500km trip in 1h 15min? 

I am very confused on how to figure out the formulas to this question. I know how to do my diagram, but I don't understand how to set up this question to find the correct answer. The textbook said that "Answers may vary"??

Answer 1

The actual velocity will be the vector sum of the wind velocity plus

the (sought) velocity of the plane relative to still air.

Let

w be the given wind velocity,

v be the velocity of the plane relative to still air (to be computed),

u be the desired velocity from Winsor to Gatio,

α = 20° be the angle the vector w is making with the axis pointing South,

β (to be computed) be the angle the vector v is making with the axis pointing North,

γ = 65° be the angle the vector u is making with the axis pointing North.

We are given

|w| = 60km/h

|u| = 500km/1.25h = 400km/h

We have the vector sum

u = v + w

That translates into an equation for the horizontal direction East-West and

into an equation for the vertical direction North-South.

|u|sin(γ) = |v|sin(β) + |w|sin(α)

|u|cos(γ) = |v|cos(β) - |w|cos(α)

We solve the two equations for |v| and β:

|v|sin(β) = |u|sin(γ) - |w|sin(α)

|v|cos(β) = |u|cos(γ) + |w|cos(α)

Dividing the two equations

tan(β) = (|u|sin(γ) - |w|sin(α))/(|u|cos(γ) + |w|cos(α))

β = arctan((|u|sin(γ) - |w|sin(α))/(|u|cos(γ) + |w|cos(α)))

Substituting actual numbers

β = arctan((400×sin(65°) - 60×sin(20°))/(400×cos(65°) + 60×cos(20°))) = 56.6°

To compute the speed |v| we can use the equation for either direction

|v| = (|u|sin(γ) - |w|sin(α))/sin(β)

Substituting actual numbers

|v| = (400×sin(65°) - 60×sin(20°))/sin(56.6°) = 410

So the plane must fly with speed of about 410 km/h in the direction N56.6E.

I do not understand in what way "Answers may vary", except possibly for rounding.