Two airplanes leave an airport at the same time. the velocity of the first plane is 670 m/h at a heading of 48°. the velocity of the second is 560 m/h at a heading of 132°. how far apart are they after 2.2h?

   First plane is 1819.58 miles is far from second plane after 2. hours.

Explanation:

     Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.

 Velocity of the first plane is 670 m/h at a heading of 48° = (670 cos 48 i + 670 sin 48 j)      

                                = (448.32 i + 497.91 j) m/h

 Velocity of the second is 560 m/h at a heading of 132° = ( 560 cos 132 i + 560 sin 132 j)

                                 = (-374.71 i + 416.16 j) m/h

We know the equation displacement = Velocity * Time taken

Displacement of first plane after 2.2 hr = (448.32 i + 497.91 j)*2.2

                               = (986.304 i + 1095.402 j) miles.

 Displacement of second plane after 2.2 hr = (-374.71 i + 416.16 j)*2.2

                               = (-824.362 i + 915.552 j) miles.

 Displacement between two planes = (986.304 i + 1095.402 j) - (-824.362 i + 915.552 j)

                                            = (1810.666 i + 179.868 j) mi

 Distance between two planes after 2.2 hours =  

  So first plane is 1819.58 miles is far from second plane after 2. hours.

true is the correct answer

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