# 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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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.**

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.**