Question 938248
{{{ 2s + s }}} = speed going downstream
{{{ 2s - s }}} = speed going upstream
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Going downstream:
(1) {{{ 3 = ( 2s + s )*t[1] }}}
Going upstream:
(2) {{{ 3 = ( 2s - s )*t[2] }}}
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(1) {{{ 3 = 3s*t[1] }}}
(2) {{{ 3 = s*t[2] }}}
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(1) {{{ t[1] = 1/s }}}
(2) {{{ t[2] = 3/s }}}
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{{{ t[1] + t[2] = 1/s + 3/s }}}
{{{ T(s) = 1/s + 3/s }}}
{{{ T(s) = 4/s }}}