Answer:
The speed of the wind is 118.10km/h in the direction 36.07° north of east.
Question:
A light plane is headed due south with a speed relative to still air of 165 km/h . After 1.00 h, the pilot notices that they have covered only 135 km and their direction is not south but southeast 45.0. What is the wind velocity?
Explanation:
Resultant distance moved by the plane = distance moved by plane in still air + distance moved as a result of wind.
R = Dp + Dw
Dp = speed of plane in still air × time of travel = 165km/h × 1 h = 165km south
R = 135km South east
Resolving the distance to x and y components.
Let north represent positive y component and east represent positive x component.
x- component
135cosθ = 0 + Dwx
Dwx = 135cos45 = 95.46km
y- component
-135sinθ = -165 + Dwy
Dwy = 165 - 135sin45
Dwy = 69.54 km
Dw = √(95.46^2 + 69.54^2)
Dw = 118.10 km
Speed = distance/time = 118.10km/1h = 118.10km/h
Angle
Tanθ = Dwy/Dwx = 69.54/95.46
θ = taninverse(69.54/95.46)
θ = 36.07°
Since both x and y component are positive (north and east)
The speed of the wind is 118.10km/h in the direction 36.07° north of east.