An airplane flies 340 km due east, makes a

right-angle turn, and then flies 390 km due
north.
Find the magnitude of the plane’s displace-
ment from its starting point. Neglect the
curvature of the earth.
Answer in units of km

340^2+390^2=267700

take the square root of 267700 n it equals abt 517.397

To find the magnitude of the plane's displacement from its starting point, we can use the Pythagorean theorem. The plane's displacement is the straight line distance between its starting point and its final position.

Given that the airplane flew 340 km due east and then 390 km due north, we can draw a right-angled triangle with the eastward distance as the adjacent side and the northward distance as the opposite side.

Using the Pythagorean theorem, the magnitude of the displacement (d) is given by:

d² = (340 km)² + (390 km)²

d² = 115600 km² + 152100 km²

d² = 267700 km²

Taking the square root of both sides, we find:

d ≈ √(267700 km²)

d ≈ 517.36 km

Therefore, the magnitude of the plane's displacement from its starting point is approximately 517.36 km.