Question:
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1(1/2) hours?
Answer:
Distance travelled by aeroplane in 3/2 hours towards north direction = 1000*3/2
= 500*3
= 1500 km
So AB = 1500
Distance travelled by aeroplane in 3/2 hours towards north direction = 1200*3/2
= 600*3
= 1800 km
So BC = 1800
Now from Pythagorus theorem
AC2 = AB2 + BC2
=> AC2 = (1500)2 + (1800)2
=> AC2 = 2250000 + 3240000
=> AC2 = 5490000
=> AC = √5490000
=> AC =√(32 * 1002 * 61)
=> AC = 3*100* √61
=> AC = 300√61
So distance between two planes after 3/2 hours = 300√61 km
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