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 how far apart will be the two plane after 1 into 1 by 2 hours​

Answer 1

Answer:

the distance would be between them after 1/2 hours is 100m

Step-by-step explanation:

distance covered by first plane -

1000×1/2 = 500

distance covered by second one -

1200×1/2 = 600

therefore, 600-500 = 100

Answer 2

$\mathfrak{\huge{\red{\underline{\underline{Answer:-}}}}}$

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 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