Question 182580
For both boats, the time travelling is the same
For cabin cruiser:
(1) {{{d[1] = r[1]*t}}}
For power boat:
(2) {{{d[2] = r[2]*t}}}
given:
{{{d[1] = 20}}}
{{{d[2] = 40}}}
{{{r[1] = r[2] - 5}}}
Rewriting (1) and (2):
(1) {{{20 = (r[2] - 5)*t}}}
(2) {{{40 = r[2]*t}}}
This is 2 equations and 2 unknowns, so it's solvable
(1) {{{20 = r[2]*t - 5t}}}
(2) {{{40 = r[2]*t}}}
Subtract (1) from (2)
(2) {{{40 = r[2]*t}}}
(1) {{{-20 = -r[2]*t + 5t}}}
{{{20 = 5t}}}
{{{t = 4}}} hrs
------------------
(1) {{{20 = r[1]*t}}}
(1) {{{20 = r[1]*4}}}
{{{r[1] = 5}}} mi/hr
--------------------
(2) {{{40 = r[2]*t}}}
(2) {{{40 = r[2]*4}}}
{{{r[2] = 10}}} mi/hr
--------------------
The cabin cruiser goes 5 mi/hr and
the power boat goes 10 mi/hr