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.
:
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 = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
In this equation: x = s; a = 1; b =14; c = -240
{{{s = (-14 +- sqrt(14^2 - 4 * 1 * -240))/(2*1) }}} 
:
{{{s = (-14 +- sqrt(196 - (-960) ))/2 }}}
{{{s = (-14 +- sqrt(196 + 960 ))/2 }}}
{{{s = (-14 +- sqrt(1156 ))/2 }}}
The positive solution is what we want here:
{{{s = (-14 + 34)/2 }}}
{{{s = 20/2}}}
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((5*10)^2 = (5*24)^2)}}} = 130