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?

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) (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 =
:
Change 5 hr 20 min to hrs: 5 or hrs
:
Time upstream + time downstream = 5 hrs
+ =
Multiply equation by 3(s-3)(s+3)
3(s-3)(s+3)* + 3(s-3)(s+3)* = 3(s-3)(s+3)*
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 5 hrs at 6 mph?
6 * = 32 miles
:
:
Check solution by find the times
12/9 = 1.33
12/3 = 4.0
Total= 5.33 hrs
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