SOLUTION: Tim paddled his kayak 12km upstream against a 3km/h current and back again in 5 h 20 min. In that time how far could he have paddeld in still water?

Question 72137This question is from textbook Algebra and Trigonometry: Structure and Method
: Tim paddled his kayak 12km upstream against a 3km/h current and back again in 5 h 20 min. In that time how far could he have paddeld in still water? This question is from textbook Algebra and Trigonometry: Structure and Method

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Tim paddled his kayak 12km upstream against a 3km/h current, and back again in 5 h 20 min. In that time, how far could he have paddled in still water?
:
Here again we are given the total time so we will write a time equation:
:
Let s = speed in still water
So we have:
Speed with the current = (s+3)
Speed against current = (s-3)
:
The one way distance is given as 12 km
:
The total time is given as 5 hr 20 min, we have to convert that to hrs only
5 + 20/60 = 5 & 1/3 hrs, since we are dealing in fractions anyway let's call it
(16/3) hrs
:
remember Time = Distance/speed
upstream time + downstream time = 16/3 hrs
:
=
:
A complicated common denominator 3(s-3)(s+3); but we can tough it out!
:
3(s+3)(12) + 3(s-3)(12) = 16(s-3)(s+3)
:
12(3s+9) + 12(3s-9) = 16(s^2 - 9)
:
36s + 108 + 36s - 108 = 16s^2 - 144
:
72s = 16s^2 - 144
:
arrange as a quadratic equation:
16s^2 - 72s - 144 = 0
:
Simplify, divide equation by 8:
2s^2 - 9s - 18 = 0
:
Factors to:
(2s + 3)(s - 6) = 0
:
2s = -3; ignore this solution:
and
s = +6 km/hr the speed in still water
:
:
A quick check using the original equation:
12/3 + 12/9 =
4 + 4/3 = 5 & 1/3 hrs as given
:
But they said,"In that time, how far could he have paddled in still water?"
:
Remember: Distance = speed * time
We have: 6 * 5 & 1/3 hr
or
6 * 16/3 = 96/3 = 32 km in 5 hr 20 min
:
:
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