How far apart are they after 1.4 h? Answer in units of m.

Question

Two airplanes leave an airport at the same time. The velocity of the first airplane is 740 m/h at a heading of 63.4 ◦ . The velocity of the second is 650 m/h at a heading of 184◦ . How far apart are they after 1.4 h? Answer in units of m.

Answer 1

The first airplane travels a distance of 740(1.4) = 1036 miles at the 63.4° heading.

The second airplane travels a distance of 650(1.4) = 910 miles at the 184° heading.

Making a sketch we see that this is just a problem of solving a triangle using the Law of Cosines:

d² = 910² + 1036² - 2(910)(1036)cos(120.6°)

d² = 2861203.77

d = √2861203.77

d = 1692 miles

You can accomplish the same thing by dividing each vector into x-components and y-components using cosine and sine ratios, adding the x-components together, adding the y-components together, and calculating the distance using the Pythagorean Theorem but it's a little more complicated that way I think.