Question 183239
Each train has it's own equation
{{{d[A] = r[A]*t[A]}}}
and
{{{d[B] = r[B]*t[B]}}}
given:
{{{r[A] = 40}}} mi/hr
{{{r[B] = 50}}} mi/hr
Assume I have a stopwatch, and I start it when
when train B passes the station 12 min, or {{{12/60}}}
of an hour after train A has passed. I will stop
the stopwatch when they meet.
How much of a headstart does train A have when B
passes the station? Let {{{d[h]}}} = headstart distance
{{{d[h] = r[A]*t[h]}}}
{{{d[h] = 40*(12/60)}}}
{{{d[h] = 8}}}mi
the distance A has to travel compared to B is:
{{{d[A] = d[B] - 8}}}mi
Their time of travel will be the same
{{{t[A] = t[B]}}} (I'll call it {{{t}}})
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I can rewrite the equations
{{{d[A] = r[A]*t[A]}}}
{{{d[B] = r[B]*t[B]}}}
----------------------
(1) {{{d[B] - 8 = 40t}}}
(2) {{{d[B] = 50t}}}
This is 2 equations and 2 unknowns so it's solvable
(1) {{{d[B] - 40t = 8}}}
(2) {{{d[B] - 50t = 0}}}
Subtract (2) from (1)
(1) {{{d[B] - 40t = 8}}}
(2) {{{-d[B] + 50t = 0}}}
(3) {{{10t = 8}}}
(3) {{{t = 4/5}}} hr (or {{{.8*60 = 48}}}min) 
This is the time that passes since train B goes through
station at 3:22 PM {{{22 + 48 = 70}}}
The trains will meet at 4:10 PM
check answer:
(1) {{{d[B] - 8 = 40t}}}
(2) {{{d[B] = 50t}}}
------------------------
(1) {{{d[B] - 8 = 40*(4/5)}}}
(1) {{{d[B] = 32 + 8}}}
(1) {{{d[B] = 40}}} mi
(2) {{{d[B] = 50*(4/5)}}}
(2) {{{d[B] = 40}}} mi
OK