Question 878621
How much of a head start does the 1st ship have?
{{{ 15 }}} min = {{{ 1/4 }}} hr
{{{ d[1] = 25*(1/4) }}}
{{{ d[1] = 6.25 }}} km
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Start a stopwatch when the 2nd ship leaves
Let {{{ t }}} = the time in hrs on the stopwatch when
the 2nd ship catches the 1st ship
Let {{{ d }}} = the distance the 2nd ship travels
until it catches the 1st ship
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Equation for the 1st ship:
(1) {{{ d - 6.25 = 25t }}}
Equation for the 2nd ship:
(2) {{{ d = 28t }}}
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Substitute (2) into (1)
(1) {{{ 28t - 6.25 = 25t }}}
(1) {{{ 3t = 6.25 }}}
(1) {{{ t = 2.08333 }}}
Convert {{{ .08333 }}} hrs to min
{{{ .08333*60 = 5 }}} min
The 2nd ship will catches the 1st 
in 2 hrs 5 min
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check:
(2) {{{ d = 28t }}}
(2) {{{ d = 28*2.08333 }}}
(2) {{{ d = 58.333 }}} km
and
(1) {{{ d - 6.25 = 25t }}}
(1) {{{ d - 6.25 = 25*2.08333 }}}
(1) {{{ d = 58.333 }}} km
OK