SOLUTION: Plane A leaves airport 1 and heads for airport 2 which is a distance of 2,700 km. At the same time, plane B leaves airport 2 and heads for airport 1. Due to headwinds and tailwinds

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Question 472741: Plane A leaves airport 1 and heads for airport 2 which is a distance of 2,700 km. At the same time, plane B leaves airport 2 and heads for airport 1. Due to headwinds and tailwinds, Plane A's average speed is 100 kph greater than that of plane B. If the two planes pass each other after 3 hours, find the speed of each plane.
Answer by monika_p(71) About Me  (Show Source):
You can put this solution on YOUR website!
The first thing that you should understand here is that if the planes start at the same time then at the point at which they meet, the time will be the same t=3hr. In addition, the total distance travelled by the two planes will be equal to the sum of their individual distances travelled x+y=2700

speed of plane A Va=Vb+100kph the plane travelled 2700-y distance in t=3 hr
speed of plane B is Vb the plane travelled 2700-x distance in t=3 hr
using formula of the speed v=d%2Ft
for the plane A we write
Va=+x%2Ft
Va=%282700-y%29%2F3
Vb%2B100=%282700-y%29%2F3
Vb=%282700-y%29%2F3-100
for the plane B
Vb=y%2Ft
Vb=y%2F3
Now compare two equations Vb=Vb
%282700-y%29%2F3-100=y%2F3 multiply both sides by 3
2700-y-300=y
2y=2400
y=1200
then x= 1500

having the distance x and y the planes travelled in t=3hr you can find speed of each plane
Va= x/t= 1500/3=500 kph
Vb=y/t=1200/3=400kph