<PRE><B>Question 564616</B>
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 {{{sqrt(30^2 + 40^2)}}} results: 50</PRE>