Question

A passenger plane flew out of Dallas, Texas, and flew 1454.55 km east and 390.27 km north. Which airport did the plane fly to?

Answer 1

The addition of vectors allows finding that the arrival airport is 1506.02 km and 15.02º North of the East, from Dalas airport.  

Displacement is a vector quantity that must be calculated with vector algebra. Its magnitude can be calculated with the Pythagorean Theorem and using trigonometry its direction can be found

Let's calculate the magnitude of the vector from the departure airport (Dalas) to the arrival airport, in the attached we can see a diagram of this displacement  

 

             d² = x² + y²

             d = $\sqrt{1454.55^2 + 390.27^2}$

             d = 1506.02 km

The direction of this airport can be calculated using trigonometry

            tan θ = y / x

            θ = tan -1 y / x

            θ = tan-1 390.27 / 1454.55

            θ = 15.02º

angle can be written 15.02º North of East

In conclusion with the addition of vectors we can find that the arrival airport is 1506.02 km and 15.02º North of the East of Dalas airport.

Learn more about adding vectors here:

brainly.com/question/24643991