Question 648831
Let the speed of the 1st train = {{{s}}}
{{{ s + 24 }}} = the speed of the 2nd train
Let {{{ d[1] }}} = the head start  that the 1st train got
4 PM - 2PM is {{{2}}} hrs, so
{{{ d[1] = s*2 }}}
Let {{{ d }}} = the distance that the 2nd train has to go
to catch the 1st train
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7PM - 4PM = {{{ 3 }}} hrs
Use this time for both trains. Figure that
the 1st train already has the head start
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1st train's equation:
{{{ d - d[1] = s*3 }}}
(1) {{{ d - 2s = s*3 }}}
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2nd train's equation:
(2) {{{ d = ( s + 24 )*3 }}}
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(1) {{{ d = 5s }}}
and
(2) {{{ d = 3s + 72 }}}
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Substitute (1) into (2)
(2) {{{ 5s = 3s + 72 }}}
(2) {{{ 2s = 72 }}}
(2) {{{ s = 36 }}}
also
{{{ s + 24 = 36 + 24 }}}
{{{ s + 24 = 60 }}}
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The 1st train's speed is 36 mi/hr
The 2nd train's speed is 60 mi/hr
check:
{{{ d[1] = s*2 }}}
{{{ d[1] = 36*2 }}}
{{{ d[1] = 72 }}}
So, the 1st train had a head start of 72 mi
which took 2 hrs. Now it has 3 hrs to go
the rest of the way
2nd train's equation:
(2) {{{ d = ( s + 24 )*3 }}}
(2) {{{ d = 60*3 }}}
(2) {{{ d = 180 }}} mi
{{{ 180 - 72 = 108 }}} mi
The 1st train must go 108 mi in 3 hrs at 36 mi/hr
{{{ 108 = 36*3 }}}
{{{ 108 = 108 }}}
OK