The plane traveled 541.6 km during this time.
The Breakdown
Finding the displacement of the plane due to its velocity of 100 km/hr for 3 hours. Since the plane is flying due north, its displacement vector is:
d1 = 100 km/hr × 3 hr × north = 300 km north
v_wind,north = 30 km/hr × cos(315°) ≈ -21.2 km/hr (southward component)
v_wind,east = 30 km/hr × sin(315°) ≈ -21.2 km/hr (westward component)
Note that we use a negative sign for the northward component since the wind is blowing southward.
The displacement vector due to the wind is then:
d2 = (v_wind,north + v_plane,north) × t + v_wind,east * t
= (-21.2 km/hr + 100 km/hr) × 3 hr × north + (-21.2 km/hr) × 3 hr × west
≈ 237.6 km north + 63.6 km west
To find the total displacement of the plane, we add the two displacement vectors:
d_total = d1 + d2
= 300 km north + 237.6 km north + 63.6 km west
≈ 537.6 km north + 63.6 km west
The distance traveled by the plane is the magnitude of the total displacement vector:
|d_total| = sqrt((537.6 km)² + (63.6 km)2)
≈ 541.6 km
Therefore, the plane traveled 541.6 km during this time.