Question

Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 65.3 ◦ . The velocity of the second is 590 m/h at a heading of 102◦ . How far apart are they after 1.7 h? Answer in units of m.

Answer 1

Plane will 741.6959 m apart after 1.7 hour

Step-by-step explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is =
`102^(\circ)-65.3^(\circ)=36.7^(\circ)`

Speed of first plane = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

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

`r^2=1241^2+1003^2-2* 1003* 1241cos(36.7)=550112.8295`

r = 741.6959 m