Question 947523
Let {{{ s }}} = the speed of the boat in still water
Let {{{ c }}} = the speed of the current
{{{ s - c }}} = the boat's speed going upstream
{{{ s + c }}} = the boat's speed going downstream
Let {{{ t }}} = time in hrs for the whole trip
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The one-way distance the boat travels is 
{{{ 1.12 + 1.48 = 2.6 }}} km
the distance for the whole trip is:
{{{ 2*2.6 = 5.2 }}} km
The effect of the current cancels for the
round trip, so
{{{ t = 5.2/s }}}
This is also the time for the log to travel
{{{ 1.12 }}} km at the speed of the current
{{{ t = 1.12/c }}}
By substitution:
{{{ 5.2/s = 1.12/c }}}
{{{ 5.2c = 1.12s }}}
{{{ c = .2154s }}}
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I am also told that:
{{{ 1.48 = ( s - c )*( 65.6/60 ) }}}
{{{ 1.48*( 60/65.6 ) = s - c }}}
{{{ s - c = 1.3537 }}}
By substitution:
{{{ s - .2154s = 1.3537 }}}
{{{ .7846s = 1.3537 }}}
{{{ s = 1.7253 }}}
and, since
{{{ c = .2154s }}}
{{{ c = .2154*1.7253 }}}
{{{ c = .3716 }}}
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The time to go back downstream is:
{{{ t[1] = 2.6 / ( s + c ) }}}
{{{ t[1] = 2.6 / ( 1.7253 + .3716 ) }}}
{{{ t[1] = 2.6 / 2.097 }}} 
{{{ t[1] = 1.24 }}} hrs
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check answer:
{{{ 2.6 = ( s - c )*t[2] }}}
{{{ 2.6 = ( 1.7253 - .3716 )*t[2] }}}
{{{ 2.6 = 1.3537t[2] }}}
{{{ t[2] = 1.9207 }}}
and
{{{ 2.6 = ( s + c )*t[3] }}}
{{{ 2.6 = ( 1.7253 + .3716 )*t[3] }}}
{{{ 2.6 = 2.0969t[3] }}}
{{{ t[3] = 1.24 }}}
{{{ t[2] + t[3] = 1.9207 + 1.24 }}}
{{{ t = 3.1607 }}} ( this is off )
and, also,
{{{ t = 1.12/c }}}
{{{ t = 1.12/.3716 }}}
{{{ t = 3.014 }}}
and also,
{{{ t = 5.2/s }}}
{{{ t = 5.2/1.7253 }}}
{{{ t = 3.014 }}}
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My calculations went off somewhere
I think my method is OK
Check my math and probably get
another opinion, too