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Two trains leave the railway station at the same time.
The first travels due west and second train due north.
The speed of the first train is 5 km faster than the second train.
After two hours they are 50 km apart. Find the average speed of the trains
Let s = the speed of the northbound train
(s+5) = the speed of the westbound train
This is a right triangle problem: a^2 + b^2 = c^2
The distance between the trains is the hypotenuse
dist = speed * time
The time is 2 hrs, so we have
a = 2s; northbound train distance
b = 2(s+5) = (2s+10); westbound distance
c = 50; distance between the two trains
(2s)^2 + (2s+10)^2 = 50^2
4s^2 + 4s^2 + 40s + 100 = 2500
Arrange as a quadratic equation
4s^2 + 4s^2 + 40s + 100 - 2500 = 0
8s^2 + 40s - 2400 = 0
Simplify, divide by 8:
s^2 + 5s - 300 = 0
(s - 15)(s + 20) = 0
The positive solution is what we want here
s = 15 mph is the speed of the northbound train
5 + 15 = 20 mph is the speed of the westbound train
Check this; find the distance (d) between the trains using these distances
Northbound traveled 2(15) = 30 mi
Westbound traveled 2(20) = 40 mi
d = 50