Question 1013597
Let {{{ d }}} = distance in miles N to V car travels until they meet
{{{ 112 - d }}} = distance in miles V to N car travels until they meet
{{{ s }}} = speed of V to N car in mi/hr
{{{ s + 10 }}} = speed of N to V car in mi/hr
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1 hr 36 min = {{{ 1 + 36/60 }}} hrs
{{{ 1 + 36/60 = 1.6 }}} hrs
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Equation for N to V car:
(1) {{{ d = ( s + 10 )*1.6 }}}
Equation for V to N car:
(2) {{{ 112 - d = s*1.6 }}}
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(2) {{{ d = 112 - 1.6s }}}
Substitute (2) into (1)
(1) {{{ 112 - 1.6s = ( s + 10 )*1.6 }}}
(1) {{{ 112 - 1.6s = 1.6s + 16 }}}
(1) {{{ 3.2s = 112 - 16 }}}
(1) {{{ 3.2s = 96 }}}
(1) {{{ s = 30 }}}
and
{{{ s + 10 = 40 }}}
The car leaving Newport averages 40 miles per hour
The car leaving Vernonville averages 30 miles per hour
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check:
(1) {{{ d = ( s + 10 )*1.6 }}}
(1) {{{ d = ( 30 + 10 )*1.6 }}}
(1) {{{ d = 64 }}}
and
(2) {{{ 112 - d = s*1.6 }}}
(2) {{{ 112 - d = 30*1.6 }}}
(2) {{{ d = 112 - 30*1.6 }}}
(2) {{{ d = 112 - 48 }}}
(2) {{{ d = 64 }}}
OK