Question

An aeroplane leaves an airport and flies due north at a speed of 1000km 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 1

Let A be the first aeroplane flied due north at a speed of 1000 km/hr and B be the second aeroplane flied due west at a speed of 1200 km/hr 

Distance covered by plane A in 1 1/2 Hours = 1000 × 3/2 = 1500 km 

Distance covered by plane B in 1 1/2 hours = 1200 × 3/2 = 1800 km 

Now, In right triangle ABC 

By using Pythagoras theorem, we have

AB² = BC² + CA² 

= (1800)² + (1500)² 

= 3240000 + 2250000 

= 5490000 

∴ AB² = 5490000

⇒ AB = 300 \(\sqrt{61}\) m 

Hence, the distance between two planes after 1 1/2 hours is 300 \(\sqrt{61}\) m