Question 1019878
15 min is {{{ 1/4 }}} hr
Her husband's head start is:
{{{ d[1] = 40*(1/4) }}}
{{{ d[1] = 10 }}} mi
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Let {{{ d }}} = distance Karen drives until
she catches husband
Let {{{ t }}} = time in hrs it takes
her to catch husband
Start a stopwatch when Karen leaves
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Husband's equation:
(1) {{{ d - 10 = 40t }}}
Karen's equation:
(2) {{{ d = 60t }}}
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(2) {{{ t = d/60 }}}
and
(1) {{{ d - 10 = 40*( d/60 ) }}}
(1) {{{ 60d - 600 = 40d }}}
(1) {{{ 20d = 600 }}}
(1) {{{ d = 30 }}}
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She catches him 30 mi from home
which is less than 45 mi, so the answer
is yes.
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check:
(2) {{{ t = d/60 }}}
(2) {{{ t = 30/60 }}}
(2) {{{ t =1/2 }}}
and
(1) {{{ d - 10 = 40t }}}
(1) {{{ 30 - 10 = 40t }}}
(1) {{{ 40t = 20 }}}
(1) {{{ t = 1/2 }}}
OK