Two airplanes leave an airport at the same time. The first airplane is 650 m/h at 11.3◦.The second is 560 m/h at 97◦. How far apart are they after 2 h?

Question

Two airplanes leave an airport at the same

time. The velocity of the first airplane is

650 m/h at a heading of 11.3◦.The velocity

of the second is 560 m/h at a heading of 97◦.

How far apart are they after 2 h?

Answer in units of m.

Answer 1

Let's use this vector notation to make sure we're speaking the same language: vector A = <x component, y component> so if it goes up 3 in the y direction and over 5 in the x direction the vector would be <3,5>

Airplane 1: distance = 650mph * 2 hours = 1300 miles at 11.3 degrees = <1300*cos(11.3), 1300*sin(11.3)>

Airplane 2: distance = 560 mph * 2 hours = 1120 miles at 97 degrees = <1120*cos(97), 1120*sin(97)>

So those are the two position vectors, to find the distance between them we make a vector with the difference in the x direction and the distance in the y direction:

difference = < 1120cos(97) - 1300cos(11.3) , 1120sin(97) - 1300sin(11.3)> = <-1411.293, 856.922> miles

So, what does that mean? Airplane 2 is 1411.293 miles left of and 856.922 miles up above airplane 1. So how do we find the distance directly between them? well we have the x value and the y value and we want the hypotenuse between them so d = sqrt(x²+y²) = 1651.08 miles apart.

If that last step was confusing, think about a right triangle with a base of 3 and height of 5, it's 3 in the x and 5 in the y, the hypotenuse is the distance from the bottom left corner (the origin) and Pythagorean's theorem says the hypotenuse is sqrt (b^2 + h^2). This is also known as just taking the magnitude of the vector or the 2-norm or euclidean norm.