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 1/2 hour?
Answers
Let O be the position of an airport.
In 1 ½ hour the distance travelled by aeroplane when it flies due North at a speed of 1000 km/h
OA = 1000 × 3/2 = 1500 km
OA = 1500 km
[Distance = speed × time]
In 1 ½ hour the distance travelled by aeroplane when it flies due west at a speed of 1200 km/h
OB = 1200 × 3/2 = 1800 km
OB =1800 km
[Distance = speed × time]
In ∆ AOB
AB² = OB² + OA²
The directions North and West are perpendicular to each other.
[By Pythagoras theorem]
AB² = (1800)² + (1500)²
AB² = 3240000 + 2250000
AB² = 5490000 km
AB = √ 5490000
AB √ 90000 × 61
AB = 300√ 61 km
Hence, the distance between two planes after 1 ½ hour is 300√ 61 km.
HOPE THIS WILL HELP YOU..
Dear Student,
Answer:Distance between the two planes = 740.94 km
Solution: let A be the location of both planes at airport,
first fly towards north,from A to B at a speed 1000 km/h
second fly west at the speed of 1200 km/h
First we have to calculate the distance between AB and AC,thus we can apply pythagorus theorem to calculate the distance BC and hence the distance between two planes.
Time given = 1.5 hour
Speed 1000 km/h
Distance covered towards north AB = speed × time
= 1000×1.5
= 1500 km
Distance covered towards west AC = speed × time
= 1200×1.5
= 1800 km
In right angle triangle BC² = AC²+AB²
BC² = (1500)²+(1800)²
= 2250000+ 3240000
BC² = 549000
BC = √549000
BC = 740.94 km
Distance between the two planes = 740.94 km
hope it helps you.
