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005 10.0 points
Two airplanes leave an airport at the same
time. The velocity of the first airplane is
700 m/h at a heading of 69.8°. The velocity
of the second is 620 m/h at a heading of 107.
How far apart are they after 1.9 h?
Answer in units of m.



Answer :

Answer:

V1 = 700 cos 69.8 i + 700 sin 69.8 j           vector components of V1

V1 = 241.7 i + 656.9 j          in miles/hour

V2 = 620 cos 107  i + 620 sin 107 j

V2 = -181.3 i + 592.9 j

V = V1 - V2 = 423 i + 64 j        rate of planes separation

1.9 V = 803.7 i + 121.6 j     miles   separation of planes after 1.9 hr

D = (803.7^2 + 121.6^2)^1/2 = 812.8 miles

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