Question 950545
Let {{{ s }}} = the speed in mi/hr of the 1st train
{{{ s + 32 }}} = the speed in mi/hr of the 2nd train
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From 1 PM to 3 PM is {{{ 2 }}} hrs
The 1st train's head start is:
{{{ d[1] = s*2 }}}
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Start a stopwatch when the 2nd train leaves
at 3 PM and stop the watch at 7 PM
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Let {{{ d }}} = the distance the 2nd train travels
from 3 PM to 7 PM
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Equation for the 1st train:
(1) {{{ d - d[1] = s*4 }}}
Equation for the 2nd train:
(2) {{{ d = ( s+32 )*4 }}}
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(1) {{{ d - 2s = 4s }}}
(1) {{{ d = 6s }}}
Plug this result into (2)
(2) {{{ 6s = ( s + 32 )*4 }}}
(2) {{{ 6s = 4s + 128 }}}
(2) {{{ 2s = 128 }}}
(2) {{{ s = 64 }}}
and
{{{ s + 32 = 96 }}}
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The speed of the 1st train is 64 mi/hr
The speed of the 2nd train is 96 mi/hr
check:
(1) {{{ d = 6s }}}
(1) {{{ d = 6*64 }}}
(1) {{{ d = 384 }}}
and
(2) {{{ d = ( s+32 )*4 }}}
(2) {{{ d = 96*4 }}}
(2) {{{ d = 384 }}}
OK