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put this solution on YOUR website!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
Then
(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
:
Factors to
(s - 15)(s + 20) = 0
:
The positive solution is what we want here
s = 15 mph is the speed of the northbound train
then
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 =
d = 50
