Vectors plane question using components

AI Thread Summary
An airplane needs to fly 1000 km northeast while facing a westward wind of 40 km/hr and has a maximum speed of 650 km/hr. The pilot's direction is determined by resolving the velocity components of the plane and wind. The equations derived from the components lead to the relationship sin θ - cos θ = -0.0615, which can be simplified using the identity sin(a-b). Solving this gives an angle of approximately 42.5 degrees for the pilot to steer the plane. Verifying the x and y components confirms that this angle will achieve the desired northeast trajectory.
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Homework Statement



An airplane wishes to fly from A to B which is located 1000 km northeast of A.
The maximum speed of the plane in still air is 650 km/hr.
There is a wind blowing West of 40 km/hr
In what direction should the pilot steer the plane to complete the trip as fast as possible?

Homework Equations




The Attempt at a Solution



I drew a diagram where an arrow from origin is 45 degrees above the east axis. Then at the head of that arrow is the head of the west arrow pointing west. and the two tails are connected for the plane's direction with the unknown angle between the plane velocity and the east axis.

velocity of plane in still air (vp)= 650(cos θ, sin θ)
velocity of wind (vw)= (-40, 0)
total velocity = |vtot| (cos 45, sin 45) = vp + vw
where |vtot| is the magnitude of the total velocity

Computing x component gives me:
|vtot| cos 45 = 650cosθ - 40

Computing y component give me:
|vtot|sin45 = 650sinθ

I've isolated for |vtot| but i don't know what to do after that.
 
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Computing x component gives me:
|vtot| cos 45 = 650cosθ - 40

Computing y component give me:
|vtot|sin45 = 650sinθ
Put the two together to get (since sin 45 = cos 45):
650cosθ - 40 = 650sinθ

That boils down to sin θ - cos θ = -.0615
A bit awkward, isn't it?
You could graph sin θ - cos θ and look for when the y coordinate is -.0615.

You know, it looks kind of like the right side of
sin(a-b) = sin(a)*cos(b) - cos(a)*sin(b) with b = 45 degrees
sin(a-45) = sin(a)*cos(45) - cos(a)*sin(45)
sin(a-45) = cos(45)*[sin(a) - cos(a)]
[sin(a) - cos(a)] = sin(a-45)/cos(45)
 
Is there any other way to compute without the use of graphing?

Could you explain further on why you use sin(a-b)?
 
We have sin θ - cos θ = -.0615.
We can use the identity sin(a) - cos(a) = sin(a-45)/cos(45) to simplify this to: sin(θ-45)/cos(45) = -.0615
sin(θ-45) = -.0615*cos(45) = -.0435
θ-45 = arcsin(-.0435)
θ-45 = -2.49
θ = 42.5 degrees
It would be worth checking this with your x and y components 650cosθ - 40 and 650sinθ to see if they are equal (indicating NE for the total of the two velocity vectors)
 
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