Posted by **bubba** on Sunday, August 23, 2009 at 5:35pm.

two airplanes leave an airport at the same time. the velocity of the first airplane is 700 m/h at a heading of 52.2degrees . the velocity of the second is 600 m/h at a heading of 97 degrees . how far apart are they after 3.1 h? answer in units of m.

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**bobpursley**, Sunday, August 23, 2009 at 5:41pm
The angle between them is the difference of the two bearings, each has a leg of velocity*time

Using the law of cosines...

c^2=leg1^2 + leg2^2 -2*leg1*lleg2*cosC

where angle C is the difference of headings.

- physics -
**bubba**, Sunday, August 23, 2009 at 5:48pm
not sure if i calculated right but is 1562 m the right answer : /

- physics -
**bobpursley**, Sunday, August 23, 2009 at 6:05pm
what were your legs,and the angle?

I will check those.

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