Final answer:
The magnitude of the relative speed between the two airplanes is approximately 300.65 km/hr, calculated by subtracting their x-components of velocity and using the Pythagorean theorem with the unchanged y-component.
Step-by-step explanation:
To calculate the magnitude of the relative speed between two airplanes, one flying due east at 600 km/hr and the other flying 30 degrees west of north at 500 km/hr, we can use vector addition. The first airplane has a velocity vector pointing directly east (positive x-direction), while the second airplane's velocity vector is split into a northern component (positive y-direction) and a western component (negative x-direction).
To find the western component of the second airplane's velocity, we calculate 500 km/hr × cos(30), which gives us 433 km/hr west. For the northern component, we calculate 500 km/hr × sin(30), which gives us 250 km/hr north. Then, we can calculate the relative speed by subtracting the x-components and combining them with the y-component (which remains unchallenged) using the Pythagorean theorem:
Relative speed = √((600 km/hr - 433 km/hr)² + (250 km/hr)²)
Relative speed = √(167 km/hr)² + (250 km/hr)²)
Relative speed = √(27889 + 62500)
Relative speed = √90389
Relative speed = approximately 300.65 km/hr