Question 972623
For this problem, you can think of one car
standing still and the other one 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 }}} hrs
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I can say:
{{{ 176 = ( s + s + 10 )*1.6 }}}
{{{ 176 = ( 2s + 10 )*1.6 }}}
{{{ 176 = 3.2s + 16 }}}
{{{ 3.2s = 160 }}}
{{{ s = 50 }}}
and
{{{ s + 10 = 60 }}}
The average speeds are 50 mi/hr and 60 mi/hr
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check:
With both of the cars moving instead:
Slower car:
{{{ d[1] = 50*1.6 }}}
{{{ d[1] = 80 }}}
Faster car:
{{{ d[2] = 60*1.6 }}}
{{{ d[2] = 96 }}}
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{{{ d[1] + d[2] = 80 + 96 }}}
{{{ d[1] + d[2] = 176 }}} mi
OK