SOLUTION: Trains A and B leave the same city at right angles at the same time. Train A travels 14 km/h faster than train B. After 5 h, they are 130 km apart. Find the speed of each train.

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Question 218291: Trains A and B leave the same city at right angles at the same time. Train A travels 14 km/h faster than train B. After 5 h, they are 130 km apart. Find the speed of each train.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Trains A and B leave the same city at right angles at the same time.
Train A travels 14 km/h faster than train B. After 5 h, they are 130 km apart.
Find the speed of each train.
:
These are slow, slow, trains!
:
Use Pythagorean: a^2 + b^2 = c^2,
The dist between the trains is the hypotenuse, (c) = 130 mi
;
Let s = speed of B
then
(s+14) = speed of A
therefore:
Train A dist: 5(s+14)
Train B dist: 5s
;
(5(s+14))^2 + (5s)^2 = (130)^2
(5s + 70)^2 + 25s^2 = 16900
FOIL
(25s^2 + 700s + 4900) + 25s^2 = 16900
:
50s^2 + 700s + 4900 - 16900 = 0
:
50s^2 + 700s - 12000 = 0
Simplify divide 50:
s^2 + 14s - 240 = 0
:
Use the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
In this equation: x = s; a = 1; b =14; c = -240
s+=+%28-14+%2B-+sqrt%2814%5E2+-+4+%2A+1+%2A+-240%29%29%2F%282%2A1%29+
:
s+=+%28-14+%2B-+sqrt%28196+-+%28-960%29+%29%29%2F2+
s+=+%28-14+%2B-+sqrt%28196+%2B+960+%29%29%2F2+
s+=+%28-14+%2B-+sqrt%281156+%29%29%2F2+
The positive solution is what we want here:
s+=+%28-14+%2B+34%29%2F2+
s+=+20%2F2
s = 10 km/hr speed of train B
and
10 + 14 = 24 km/hr speed of train A
:
:
Check this on a calc: enter sqrt%28%285%2A10%29%5E2+=+%285%2A24%29%5E2%29 = 130