What is the horizontal distance between the two airplanes at 1:00pm? An airplane flying west at 400 km/h goes over another airplane flying south at 500 km/h at 11:00 am.
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calendarEducator since 2008
write3,662 answers
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First we will calculate the distance each airplane travels by 1:00 pm.
The first plane:
Travels west with speed (s1) = 400 at 11:00 am
Then the time by 1:00 pm is T1 = 2 hours.
Then, we know that the distance (D) is given by the formula:
D = S*T
==> D 1= 400*2 = 800 km ...........(1)
The second airplane:
Given that the speed is S2 = 500
Then time is also 2 hours ==> T2= 2
==> D2 = T2* S2 = 500*2 = 1000 km.............(2)
Now we need to find the distance between them at 1:00.
The planes travels in two perpendicular paths.
Then, the distance between them is the hypotenuse of a right angle triangle.
==> D^2 = D1^2 + D2^2 = (800)^2 + (1000)^2 = 1,640,000
==> D = 1280.62 km
Then, the distance between the planes by 1:00pm is 1280.62 km.
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calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
The first airplane is flying at 400km/h and going towards the west. The second airplane is traveling at 500 km/h and flying towards the south. They both pass each other at 11:00 a.m. At 1:00 pm, 2 hours have passed.
The first plane has flown 400*2 = 800 km toward the west and the second airplane has flown 500*2 = 1000 km toward the south.
The distance between them can be found using the Pythagorean Theorem as sqrt (1000^2 + 800^2)
=> 100*sqrt (100 + 64)
=> 100* sqrt 164
=> 1280.6 km
The required horizontal separation between the airplanes at 1: 00 pm is 1280.6 km.
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