Question 985990
Orson rides his power boat up and down a canal.
 The water in the canal flows at 6 mph.
 Orson takes 5 hours longer to travel 360 miles against the current than he does to travel 360 miles with the current.
 What is the speed of Orson's boat in still water?
:
let s = his speed in still water
then
(s-6) = his effective speed upstream
and
(s+6) = his effective speed downstream
:
Write a time equation, time = dist/speed
time up - time down = 5 hrs
{{{360/((s-6))}}} - {{{360/((s+6))}}} = 5
multiply equation by (s-6)(s+6)
(s-6)(s+6)*{{{360/((s-6))}}} - (s-6)(s+6)*{{{360/((s+6))}}} = 5(s-6)(s+6)
Cancel the denominators, FOIL the right side
360(s+6) - 360(s-6) = 5(s^2 - 36)
:
360s + 2160 - 360s + 2160 = 5s^2 - 180
combine like terms
4320 + 180 = 5s^2
4500 = 5s^2
Simplify, divide both sides by 5
900 = s^2
s = {{{sqrt(900)}}}
s = 30 mph his speed in still water
:
:
see if this checks out, using the effective speeds
{{{360/24}}} - {{{360/36}}} = 5
15 - 10 = 5 hrs