Question 524332
The 1st train to leave has a head start of
{{{ d[1] = 80*3 }}}
{{{ d[1] = 240 }}} mi
Start a stopwatch when the 2nd train leaves
the station. Both trains will now travel for
the same amount of time.
Let {{{t}}} = the time on stopwatch for both trains
Let {{{d}}} = the distance the 2nd train travels
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For 2nd train:
(1) {{{ d = 100t }}}
For 1st train:
(2) {{{ d - 240 = 80t }}}
Substitute (1) into (2)
(2) {{{ 100t - 240 = 80t }}}
(2) {{{ 20t = 240 }}}
(2) {{{ t = 12 }}}
The 2nd train will catch up to the 1st 12 hours after
the 2nd train leaves station at 6:00 + 3 hrs = 9:00 PM
That would be 9:00 AM the next morning.
check answer:
(1) {{{ d = 100t }}}
(1) {{{ d = 100*12 }}}
(1) {{{ d = 1200 }}} ( distance from station to where they meet )
and
(2) {{{ d - 240 = 80t }}}
(2) {{{ d - 240 = 80*12 }}}
(2) {{{ d = 240 + 960 }}}
(2) {{{ d = 1200 }}}
OK