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

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Tim paddled his kayak 12 km upstream against a 3 km/h current and back again in 5 h 20 min. In that time how far could he have paddled in still water?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 171646This question is from textbook Algebra and Trigonometry Structure and Method Book 2
: Tim paddled his kayak 12 km upstream against a 3 km/h current and back again in 5 h 20 min. In that time how far could he have paddled in still water? This question is from textbook Algebra and Trigonometry Structure and Method Book 2

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Tim paddled his kayak 12 km upstream against a 3 km/h current and back again
in 5 h 20 min. In that time how far could he have paddled in still water?
:
Let s = his speed in still water
then
(s-3) = his speed upstream
and
(s+3) = his speed downstream
;
Write a time equation: time = distance%2Fspeed
:
Change 5 hr 20 min to hrs: 51%2F3 or 16%2F3 hrs
:
Time upstream + time downstream = 51%2F3 hrs
12%2F%28%28s-3%29%29 + 12%2F%28%28s%2B3%29%29 = 16%2F3
Multiply equation by 3(s-3)(s+3)
3(s-3)(s+3)*12%2F%28%28s-3%29%29 + 3(s-3)(s+3)*12%2F%28%28s%2B3%29%29 = 3(s-3)(s+3)*16%2F3
Results:
36(s+3) + 36(s-3) = 16(s-3)(s+3)
:
36s + 108 + 36s - 108 = 16(s^2 - 9)
:
72s = 16s^2 - 144
Arrange as a quadratic equation:
16s^2 - 72s - 144 = 0
Simplify divide equation by 8
2s^2 - 9s - 18 = 0
Factor
(2s + 3)(s - 6) = 0
Positive solution:
s = 6 mph; his still water speed
:
How far would he travel in 51%2F3 hrs at 6 mph?
6 * 16%2F3 = 32 miles
:
:
Check solution by find the times
12/9 = 1.33
12/3 = 4.0
Total= 5.33 hrs