Question 937773
Let {{{ s }}} = the speed of the cruise ship in mi/hr
{{{ 9s }}} = the cruise ship's head start in miles
{{{ s + 18 }}} = the speed of the submarine in mi/hr
Let {{{ d }}} = distance in miles that the submarine travels
until it catches the cruise ship
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Start measuring time in hrs when submarine leaves
Cruise ship's equation:
(1) {{{ d - 9s = s*5 }}}
Submarine's equation:
(2) {{{ d = ( s + 18 )*5 }}}
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Substitute (2) into (1)
(1) {{{ ( s + 18 )*5 - 9s = s*5 }}}
(1) {{{ 5s + 90 - 9s = 5s }}}
(1) {{{ 9s = 90 }}}
(1) {{{ s = 10 }}}
and
{{{ s + 18 = 10 + 18 }}}
{{{ s + 18 = 28 }}}
The submarine's average speed was 28 mi/hr
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check:
(2) {{{ d = ( 10 + 18 )*5 }}}
(2) {{{ d = 28*5 }}}
(2) {{{ d = 140 }}} mi
and
(1) {{{ d - 9s = s*5 }}}
(1) {{{ d - 9*10 = 10*5 }}}
(1) {{{ d = 90 + 50 }}}
(1) {{{ d = 140 }}} mi
OK