Question 1019822
What is the bullet train's head start
in mi/hr ?
The difference between 5 AM and 6 AM is 1 hr
{{{ d[1] = 50*1 }}}
{{{ d[1] = 50 }}} mi
Start a stopwatch when the passenger train leaves 
at 6 AM
Let {{{ t }}} = time on stopwatch in hrs when
passenger train overtakes the bullet train
Let {{{ d }}} = distance in miles passenger train
travels until it catches up with bullet train
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Equation for bullet train:
(1) {{{ d - 50 = 50t }}}
Equation for passenger train:
(2) {{{ d = 55t }}}
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Substitute (2) into (1)
(1) {{{ 55t - 50 = 50t }}}
(1) {{{ 5t = 50 }}}
(1) {{{ t = 10 }}}
The passenger train overtakes the bullet train
in 10 hrs
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Check:
(2) {{{ d = 55t }}}
(2) {{{ d = 55*10 }}}
(2) {{{ d = 550 }}}
and
(1) {{{ d - 50 = 50t }}}
(1) {{{ d - 50 = 50*10 }}}
(1) {{{ d = 50 + 500 }}}
(1) {{{ d = 550 }}}
OK