Question 696528
How much of a head start does the 1st person have?
3:00 PM minus 11:00 AM is {{{ 4 }}} hrs
{{{ 48*4 = 192 }}} miles
Start a stop watch when the 2nd person leaves at 3 PM
Let {{{ t }}} = time on the stop watch 
when they are {{{ 604 }}} mi apart
Let {{{ d }}} = The distance that the 2nd person has to
travel until they are {{{ 604 }}} mi apart
( they are already {{{ 192 }}} mi apart )
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1st person's equation:
(1) {{{ 604 - 192 - d = 48t }}}
2nd person's equation:
(2) {{{ d =55t }}}
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(1) {{{ 412 - d = 48t }}}
Substitute (2) into (1)
(1) {{{ 412 - 55t = 48t }}}
(1) {{{ 103t = 412 }}}
(1) {{{ t = 4 }}}
This is the time on the stop watch which started at 3 PM
They will be 604 miles apart at 7 PM
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check answer:
(1) {{{ 604 - 192 - d = 48*4 }}}
(1) {{{ 604 - 192 - d = 192 }}}
(1) {{{ d = 604 - 384 }}}
(1) {{{ d = 220 }}}
and
(2) {{{ d =55t }}}
(2) {{{ d = 55*4 }}}
(2) {{{ d = 220 }}}
OK
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The 2nd person travels 220 mi after 1st person
travels 192 mi and an additional {{{ 48*4 = 192 }}} mi
{{{ 220 + 192 + 192 = 604 }}}
OK