An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour.
An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after \(1\frac{1}{2}\) hours?
Speed of first aeroplane = 1000 km/hr
Distance covered by first aeroplane flying due north in \(1\frac{1}{2}\) hours (OA) = 100 × 3/2 km = 1500 km
Speed of second aeroplane = 1200 km/hr
Distance covered by second aeroplane flying due west in \(1\frac{1}{2}\) hours (OB) = 1200 × 3/2 km = 1800 km
In right angle ΔAOB, by Pythagoras Theorem,
AB^{2} =^{ }AO^{2} + OB^{2}
⇒ AB^{2} =^{ }(1500)^{2} + (1800)^{2}
⇒ AB = √(2250000 + 3240000)
= √5490000
⇒ AB = 300√61 km
Hence, the distance between two aeroplanes will be 300√61 km.

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