Question

An aeroplane leaves an airport and flies due north at a speed of 1000 kmph. At the same time another aeroplane leaves the same airport and flies due west at a speed of 1200 kmph. How far apart will the two planes be after 1\(\frac{1}{2}\) hour?

Answer 1

Given: Speed of the first plane due north = 1000 kmph. 

Speed of the second plane due west = 1200 kmph. 

Distance = Speed × Time 

Distance travelled by the first plane in 

1\(\frac{1}{2}\) hrs = 1000 × 1\(\frac{1}{2}\) = 1000 × \(\frac{3}{2}\) = 1500 km. 

Distance travelled by the second plane in 

1\(\frac{1}{2}\) hrs = 1200 × \(\frac{3}{2}\) = 1800 km. 

From the figure, △ABC is a right triangle; ∠A = 90°. 

AB² + AC² = BC² 

[Pythagoras theorem] 

1500² + 1800² = BC²

2250000 + 3240000 = BC² 

∴ BC = \(\sqrt{5490000}\)

= 100 ×\(\sqrt{549}\) m 

≅ 100 × 23.43 

≅ 2243km.