Question 1039763
You can think of this as one of the cars standing
still and the other covering the distance between them
traveling at the sum of their speeds
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Let {{{ s }}} = the speed of the slower car
{{{ s + 10 }}} = the speed of the faster car
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convert {{{ 36 }}} min to hrs
{{{ 36/60 = .6 }}}
{{{ 1 + .6 = 1.6 }}} hrs
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{{{ 112 = ( s + s + 10 )*1.6 }}}
{{{ 112 = ( 2s + 10 )*1.6 }}}
{{{ 112 = 3.2s + 16 }}}
{{{ 96 = 3.2s }}}
{{{ s = 30 }}}
and
{{{ s + 10 = 40 }}}
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Car leaving Vernonville ( slower car ) averages 30 mi/hr
Car leaving Newport ( faster car averages 40 mi/hr
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check answer:
{{{ d[1] = 30*1.6 }}}
{{{ d[1] = 48 }}} mi
and
{{{ d[2] = 40*1.6 }}}
{{{ d[2] = 64 }}} mi
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{{{ d[1] + d[2] = 48 + 64 }}}
{{{ d[1] + d[2] = 112 }}}
OK