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Old 17th Jul 2008, 16:10
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Ouseburn
 
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If aircraft A took of at 10am flying at 170 mph and aircraft B took of at 10.45am flying at 230 mph. When would they meet and how far would they have travelled?

Va (Speed of Aircraft A) = 170 mph
Vb (Speed of Aircraft B) = 230 mph
Departure time of A: 10:00
Departure time of A: 10:45
45 minutes = 0.75 hours


After 45 minutes Distance travelled by Aircraft A: D = Va x 45 minutes = 170 x 0.75 = 127.5 miles

Now calculate relative speed of aircraft B: V = Vb - Va = 230 - 170 = 60 mph

Now calculate time taken for Aircraft B to catch up. ie time taken for Aircraft B to travel 127.5 miles at 60 mph: T = 127.5 / 60 = 2.125 hours


Now calculate total distance Aircraft A has travelled: Da = 170 x 2.875 (2.125 + 0.75) = 488.75 miles

Total distance Aircraft B has travelled: Db = 230 x 2.125 = 488.75 miles

Therefore, total distance both aircraft have now travelled: 488.75 + 488.75 = 977.5 miles


Time taken to meet: T = Db / Vb = 488.75 / 230 = 2.125 hours after Aircraft B departs.
This is equivalent of 127.5 minutes (2.125 x 60). So 10:45 + 127.5 minutes = 12:52:30 (12:52pm and 30 seconds).
Or
T = Da / Va = 488.75 / 170 = 2.875 hours after Aircraft A departs. 10:00 + 172.5 = 12:52:30

Last edited by Ouseburn; 19th Jul 2008 at 13:45.
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