An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/hr. How far apart will be the two planes after 1.5 hours?
Answers
SOLUTION :
Given : Speed of aeroplane due north = 1000 km/h
Speed of aeroplane due west = 1200 km/h
Time = 1.5 hrs
Distance traveled by the plane flying towards north in 1.5 hrs(OA) = 1000 × 1.5 = 1500 km
[ Distance = speed × time]
Distance traveled by the plane flying towards west in 1.5 hrs (OB) = 1200 × 1.5 = 1800 km
Let these distances are represented by OA and OB in a figure .
In ∆AOB,
AB² = OA² + OB²
AB =√(OA² + OB²)
[By using Pythagoras theorem]
Distance between these planes after 1.5 hrs (AB) = √(1500)² + (1800)²
= √(2250000 + 3240000)
= √5490000
= √(9 × 610000)
Distance between these planes after 1.5 hrs (AB) = 300√61
AB = 300 × 7.8102
AB = 2,343.06 km
AB = 2,343 km (approx)
Hence, the distance between these planes will be 2,343 km (approx) after 1.5 hrs.
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Answer:
Step-by-step explanation:
Given :-
Speed of aeroplane due north = 1000 km/h
Speed of aeroplane due west = 1200 km/h
Time = 1.5 hrs
To Find :-
Distance between these planes.
Formula to be used :-
Pythagoras Theorem
Solution :-
Distance travelled by first aeroplane towards north = 1000 × 1.5 = 1500 km
Similarly, Distance travelled by second aeroplane towards west = 1200 × 1.5 = 1800 km
In △ABC,
On applying Pythagoras Theorem,
⇒ AC² = AB² + BC²
⇒ AC² = (1500)² + (1800)²
⇒ AC² = 300 √61 km
Hence, The Distance between these planes is 300 √61 km.