Question 381451
A boat made trip of 9 miles with the tide and back against the tide in 3hrs.
 If the tide had been half as strong the trip would had been made in 36min.
less time.
 find the rate of the boat in still water and the rate of the stronger tide.
:
Let s = boat speed in still water
Let t = rate of the stronger tide
then
(s+t) = effective speed with the stronger tide 
(s-t) = effective speed against the stronger tide
:
Write a time equation for each scenario; Time = dist/speed
:
{{{9/((s+t))}}} + {{{9/((s-t))}}} = 3 hrs
9(s-t) + 9(s-t) = 3(s+t)(s-t)
9s - 9t + 9s + 9t = 3(s^2 - t^2)
18s = 3(s^2 - t^2)
Divide both sides by 3
6s = s^2 - t^2
:
At half the tide rate; (.5t)
{{{9/((s+.5t))}}} + {{{9/((s-.5t))}}} = 3 - {{{36/60}}}
{{{9/((s+.5t))}}} + {{{9/((s-.5t))}}} = 3 - .6
{{{9/((s+.5t))}}} + {{{9/((s-.5t))}}} = 2.4
9(s-.5t) + 9(s+.5t) = 2.4(s+.5t)(s-.5t)
9s - 4.5t + 9s + 4.5t = 2.4(s^2 - .25t^2)
18s = 2.4(s^2 - .25t^2)
divide both sides by 2.4
7.5s = s^2 - .25t^2
:
multiply the above equation by 4, subtract 1st Simplified equation
30s = 4s^2 - t^2
6s = s^2 - t^2
---------------- subtraction eliminates t^2
24s = 3s^2
simplify, divide by 3s
8 = s
Boat speed in still water: 8 mph
:
Find t using the 1st simplified equation
6s = s^2 - t^2
t^2 = s^2 - 6s
Replace s with 8
t^2 = 8^2 - 6(8)
t^2 = 64 - 48
t^2 = 16
t = {{{sqrt(16)}}}
t = 4 mph is rate of the strong tide
:
:
Check solutions in the original half speed tide equation
{{{9/((s+.5t))}}} + {{{9/((s-.5t))}}} = 2.4
{{{9/((8+.5(4)))}}} + {{{9/((8-.5(4)))}}} = 2.4
{{{9/10}}} + {{{9/6}}} = 2.4
.9 + 1.5 = 2.4; confirms our solutions of s=8; t=4