Question 819351
How much of a head start did Sarah get?
{{{ d[1] = 30*(12/60) = 6 }}} miles
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Start a stopwatch when Becky leaves
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Let {{{ d }}} = the distance that Becky travels
until she catches Sarah.
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Let {{{ t }}} = the time on the stopwatch when
Becky catches Sarah.
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Equation for Sarah:
(1) {{{ d - 6 = 30t }}}
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Equation for Becky:
(2) {{{ d = 50t }}}
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Substitute (2) into (1)
(1) {{{ 50t - 6 = 30t }}}
(1) {{{ 20t = 6 }}}
(1) {{{ t = 3/10 }}} hrs
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{{{ (3/10)*60 = 18 }}} min
Becky overtakes Sarah at 8:27 + 18 min = 8:45
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(2) {{{ d = 50t }}}
(2) {{{ d = 50*(3/10) }}}
(2) {{{ d = 15 }}}
They are 15 miles from Littlesburgh
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check:
(1) {{{ d - 6 = 30t }}}
(1) {{{ 15 - 6 = 30*(3/10) }}}
(1) {{{ 9 = 9 }}}
OK