Question 180585
Let {{{r[b]}}}= the speed of the boat in still water
Let {{{r[c]}}}= the speed of the current
For the trip upstream:
(1) {{{d[u] = r[u]*t[u]}}}
For the trip downstream:
(2) {{{d[d] = r[d]*t[d]}}}
given:
{{{d[u] = 10}}} mi
{{{d[d] = 10}}} mi
{{{t[u] = 1}}} hr
{{{t[d] = .5}}} hr
upstream:
{{{r[u] = r[b] - r[c]}}}
{{{r[d] = r[b] + r[c]}}}
------------------------
(1) {{{d[u] = r[u]*t[u]}}}
(1) {{{10 = r[u]*1}}}
(1) {{{10 = (r[b] - r[c])*1}}}
(1) {{{10 = r[b] - r[c]}}}
and
(2) {{{d[d] = r[d]*t[d]}}}
(2) {{{10 = r[d]*.5}}}
(2) {{{10 = (r[b] + r[c])*.5}}}
(2) {{{10 = .5r[b] + .5r[c]}}}
-------------------------------
Subtract (2) from (1)
(1) {{{10 = r[b] - r[c]}}}
(2) {{{-10 = -.5r[b] - .5r[c]}}}
(3) {{{0 = .5r[b] - 1.5r[c]}}}
{{{1.5r[c] = .5r[b]}}}
{{{3r[c] = r[b]}}}
Now I can rewrite (1)
(1) {{{10 = r[b] - r[c]}}}
(1) {{{10 = 3r[c] - r[c]}}}
(1) {{{10 = 2r[c]}}}
{{{r[c] = 5}}} mi/hr
and, since
(1) {{{10 = r[b] - r[c]}}}
(1) {{{10 = r[b] - 5}}}
(1) {{{r[b] = 15}}} mi/hr
The speed of ther boat in still water is 15 mi/hr
The speed of the current is 5 mi/hr
check:
(2) {{{10 = (r[b] + r[c])*.5}}}
{{{10 = (15 + 5)*.5}}}
{{{10 = 20*.5}}}
{{{10 = 10}}}
OK