Question

Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 44.3 ◦ . The velocity of the second is 590 m/h at a heading of 83◦ . How far apart are they after 2.9 h?

Answer 1

Answer:

So airplane will be 1324.9453 m apart after 2.9 hour

Explanation:

So if we draw the vectors of a 2d graph we see that the difference in angles is  = 83 - 44.3 = $83-44.3=38.7^{\circ}$

Distance traveled by first plane = 730×2.9 = 2117 m

And distance traveled by second plane = 590×2.9 = 1711 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 38.7.

Using the law of cosine, $d^2$ representing the distance between the planes, we see that:

$d^2=2117^2 + 1711^2 -2\times (2117)\times (1711)cos(38.7)=1755480.2482$

d = 1324.9453 m