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 11 2 hours?

Answer 1

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 Pythagoras 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

Answer 2

$\huge{\boxed{\sf{ANSWER}}}$

Speed of 1st plane = 1000 km

Time = 1 1/2 = 3/2 hours

Distance covered = Speed × Time

= 1000 × 3/2

⇒ 1500 km

Speed of 2nd plane = 1200 km

Time = 3/2

Distance covered = Speed × Time

= 1200 × 3/2

⇒ 1800 km

Now, by using pythagoras theorem we get :-

AC² = AB² + BC²

AC² = ( 1500 )² + ( 1800 )²

AC² = 2250000 + 3240000

AC² = 5490000

AC = √549 × 10 × 10 × 10 × 10

AC = √3 × 3 × 61 × 10 × 10 × 10 × 10

AC = 3 × 10 × 10 √61

⇒ 300√61 km.  ANS

HENCE, TWO PLANES WOULD BE 300√61 km AWAY FROM EACH OTHER.