A pilot maintains a heading due west with an air speed of 240 km/h. After flying for 30 min, he finds himself over a town that he knows is 150 km west and 40 km south of his starting point.

a) What is the wind velocity, in magnitude and direction?
b) What heading should he now maintain, with the same air speed, to follow a course due west from the town?
Thank you!!

100 Km/h [S37W], [W19N]

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To solve this problem, we will use the concept of vector addition.

a) Let's assume that the wind is blowing in a direction of θ degrees and with a magnitude of w km/h. The pilot's airspeed is given as 240 km/h due west (270 degrees). The wind will affect the pilot's movement, causing him to deviate from his initial heading.

Let's break down the pilot's movement into two components:
1. The east-west component (x-axis) due to the airspeed of 240 km/h.
2. The north-south component (y-axis) due to the wind.

Considering the fact that the pilot maintained a heading due west, his east-west component remains the same at 240 km/h. This can be represented as:

240 km/h * cos(270 degrees) = -240 km/h (negative sign indicates westward motion)

To find the north-south component due to the wind, we can use the given information that the pilot finds himself 150 km west and 40 km south of his starting point after flying for 30 minutes. Using time, distance, and speed formulas:

East-West component due to wind:
-240 km/h * (30 min / 60 min) = -120 km (negative sign indicates westward motion)

South-North component due to wind:
40 km = -40 km (negative sign indicates southward motion)

Now, we can determine the magnitudes and direction of the components of the wind velocity vector:

Magnitude of the wind velocity:
Magnitude = sqrt((East-West component)^2 + (South-North component)^2)
Magnitude = sqrt((-120 km)^2 + (-40 km)^2) = sqrt(14400 km^2 + 1600 km^2) = sqrt(16000 km^2) = 400 km/h

Direction of the wind velocity vector:
Direction = arctan(South-North component / East-West component)
Direction = arctan(-40 km / -120 km) = arctan(1/3)

Therefore, the wind velocity is 400 km/h in the direction of approximately 18.43 degrees north of east.

b) To maintain a course due west from the town, the pilot needs to take into account the wind velocity and adjust his heading.

Since the wind is blowing in a direction of 18.43 degrees north of east, the pilot needs to point his aircraft slightly north of the west to compensate for the wind's effect. Let's call this new heading θ'.

To find the new heading (θ') that the pilot should maintain, we can use vector addition again:

New east-west component = airspeed * cos(θ')
New south-north component = airspeed * sin(θ')

We know that:
New east-west component = 240 km/h (due west)
New south-north component = wind velocity * sin(θ)

Solving for θ', we have:
240 km/h * cos(θ') = 240 km/h
wind velocity * sin(θ) = 240 km/h

Using the wind velocity obtained in part (a) (400 km/h) and the known angle θ (18.43 degrees), we can solve for θ':

400 km/h * sin(18.43 degrees) = 240 km/h * cos(θ')
θ' = arccos(400 km/h * sin(18.43 degrees) / 240 km/h)

Therefore, the pilot should maintain a heading of approximately θ' degrees (using the calculated value of θ') to follow a course due west from the town.