SOLUTION: Trains A and B leave the same city at right angles at the same time. Train B travels 5 mph faster than Train A. After 2 hours, they are 50 miles apart. Find the speed of each train
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Question 564616: Trains A and B leave the same city at right angles at the same time. Train B travels 5 mph faster than Train A. After 2 hours, they are 50 miles apart. Find the speed of each train. Found 2 solutions by ad_alta, ankor@dixie-net.com:Answer by ad_alta(240) (Show Source):
You can put this solution on YOUR website! Let A have speed 'a.' Remember distance is rate times time, and the Pythagorean theorem. We then have (2a)^2+(2(a+5))^2=50^2. a=15, so the other rate is 20.
You can put this solution on YOUR website! Trains A and B leave the same city at right angles at the same time.
Train B travels 5 mph faster than Train A.
After 2 hours, they are 50 miles apart. Find the speed of each train.
:
let s = speed of train A
then
(s+5) = speed of train B
After 2 hrs:
train A dist = 2s
train B dist = 2(s+5)
:
We can treat this as a pythag, problem where a^2 + b^2 = c^2
Where
a = 2s
b = 2(s+5)
c = 50
:
(2s)^2 + (2(s+5))^2 = 50^2
4s^2 + (2s + 10)^2 = 2500
4s^2 + 4s^2 + 40s + 100 = 250
A quadratic equation
8s^2 + 40s + 100 - 2500 = 0
8s^2 + 40s - 2400 = 0
Simplify divide by 8
s^2 + 5s - 300 = 0
This will factor to
(s+20)(s-15) = 0
the positive solution
s = 15 mph is Train A
and
15 + 5 = 20 mph is train B
:
:
Check this on calc:
enter results: 50
