SOLUTION: Two trains leave railway stations at the same time. The first train travel due West in the second train due north. The first travels 5KM/hr faster than the second train. if afte

Algebra ->  Finance -> SOLUTION: Two trains leave railway stations at the same time. The first train travel due West in the second train due north. The first travels 5KM/hr faster than the second train. if afte      Log On


   

Question 1180675: Two trains leave railway stations at the same time. The first train travel due West
in the second train due north. The first travels 5KM/hr faster than the second train. if after two hours, they are 50 KM apart, find the average speed of each train.
Found 2 solutions by greenestamps, Edwin McCravy:
Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


Since the two trains are traveling in directions that differ by 90 degrees, the line joining the positions of the two trains at any time forms a right triangle with the paths of the trains. So solve using the Pythagorean Theorem.

x = slower train speed (km/h)
x+5 = faster train speed (km/h)
2x = distance (km) slower train travels in 2 hours
2x+10 = distance (km) faster train travels in 2 hours

After 2 hours, the right triangle has legs of 2x and 2x+10 and a hypotenuse of 50:

%282x%29%5E2%2B%282x%2B10%29%5E2+=+50%5E2
4x%5E2%2B%284x%5E2%2B40x%2B100%29+=+2500
8x%5E2%2B40x-2400+=+0
x%5E2%2B5x-300+=+0
%28x%2B20%29%28x-15%29+=+0

x+=+-20 or x+=+15

Obviously reject the negative solution, since the speeds of the trains are not negative.

ANSWER: The average speeds of the two trains are x=15km/h and x+5=20km/h

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Let the northbound train's rate be r km/h
Then the westbound train's rate is r+5 km/h

After 2 hours, using d = rt

the northbound train's distance is 2r
the westbound train's distance is 2(r+5)

We use the Pythagorean theorem to write an
expression for their distance apart, which
is the hypotenuse of a right triangle.

distance ap[art = sqrt%28%282%28r%2B5%29%29%5E2%2B%282r%29%5E2%29

Make the drawing



sqrt%28%282%28r%2B5%29%29%5E2%2B%282r%29%5E2%29%22%22=%22%2250

Square both sides:

%282%28r%2B5%29%29%5E2%2B%282r%29%5E2%22%22=%22%222500

That simplifies to 

8r%5E2%2B40r-2400%22%22=%22%220

Which conveniently can be divided through by 8

r%5E2%2B5r-300%22%22=%22%220

which is conveniently factorable as

%28r%2B20%29%28r-15%29%22%22=%22%220

r+20 = 0;   r-15 = 0
   r = -20;    r = 15

Ignore the negative answer.

The northbound train's rate is 15 km/h
The westbound train's rate is 15+5=20 km/h

Edwin