A small aircraft leaves Calgary on a heading of E30deg S (30 degrees south of east) travelling 200 km/h. it is affected by a 40 km/h wind blowing toward S60deg W (60 degrees west of south). After one hour the plane’s position relative to Calgary will be?

A) 183 km E45deg S
B) 183 km S30deg W
C) 204 km E45deg S
D) 204 km S30deg W

What did you get? I would break up each vector into S and E components,add,then combine.

To find the position of the aircraft after one hour, we can break down the movement into two components: the aircraft's own velocity and the wind's effect.

First, let's calculate the displacement caused by the aircraft's velocity. The aircraft is traveling at a speed of 200 km/h in a direction of E30°S. This means that after one hour, it would have traveled 200 km in the E30°S direction.

Next, we need to consider the effect of the wind. The wind is blowing towards S60°W at a speed of 40 km/h. Since the wind is pushing against the aircraft's motion, we need to subtract its effect. To do this, we can calculate the wind's displacement after one hour.

Using basic trigonometry, we can determine that the wind's displacement in the E30°S direction would be 40 km/h × sin(60°) = 34.64 km. It will also have a displacement in the direction perpendicular to E30°S, which can be calculated as 40 km/h × cos(60°) = 20 km (because the angle between E30°S and the perpendicular direction is 90° - 60° = 30°).

Now, we can find the total displacement of the aircraft by adding the displacements caused by its own velocity and the wind. In the E30°S direction, the total displacement is 200 km - 34.64 km = 165.36 km. In the perpendicular direction, the total displacement is -20 km. (We use a negative sign because the wind is pushing against the motion of the aircraft.)

Finally, we can use these displacements to determine the aircraft's position relative to Calgary. The position will be 165.36 km in the E30°S direction and -20 km in the perpendicular direction. We can use Pythagoras' theorem to calculate the magnitude of this position. The square of the magnitude (M) is equal to (165.36 km)^2 + (-20 km)^2, which evaluates to 28086.8896 km^2. Taking the square root of this value gives us M = 167.74 km.

To determine the direction, we can use trigonometry again. The direction of the position is given by the angle θ, where tan(θ) = (-20 km) / (165.36 km). Thus, θ ≈ -6.87°.

Now that we have the magnitude and direction of the aircraft's position relative to Calgary, we can express the answer as 167.74 km E6.87°S. However, the answer choices are given in different formats, so we'll need to convert it. Adding 180° to the direction gives us 173.13°, which can be rounded to the nearest degree as 173°. Therefore, the position of the aircraft after one hour is approximately 167.74 km E6.87°S, or in the closest answer choice, option C) 204 km E45°S.