Posted by CINDY on Thursday, December 21, 2006 at 2:16am.
TIM PADDLED HIS KAYAK 12 KM UPSTREAM AGAINST A 3 KM/H CURRENT AND BACK AGAIN IN 5 HR 20 MINUTES. iN THAT TIME
HOW FAR COULD HE HAVE PADDLED IN STILL WATER
12/(v+3) + 12/(v3) = 5.33 hours
Solve for this speed in still water, v.
12(v3) + 12(v+3) = 5.33 (v^2 9)
24 v = 5.33 v^2  48
Use the quadratic equation and take the positive root for v.
v = 6.0 km/h
In the same time, the distance he could travel in still water is
D = 5.33 * v = 32 km

RATE, DISTANCE, TIME  Anonymous, Wednesday, June 19, 2013 at 10:38am
6
Answer This Question
Related Questions
 Algebra 2  Tim paddled his kayak 12km upstream against a 3km/h current and back...
 algebra  Abby Tanenbaum paddled her canoe 10 miles upstream, and then paddled ...
 MATH!!!!!!!  The river boat paddled upstream at 12km/h, stopped for 2 h of ...
 Help...More Math Help!!!...Please!  A kayaker paddled 2 hours with a 6 mph ...
 Math  Steve can kayak at 8 km/h in still water. On the river, Steve and the ...
 Algebra  on a canoe trip, Rita paddled upstream (against the current) at an ...
 math  a kayak travels 15/h in still water and if the current flows at a rate a ...
 Algebra  When Neil was 2 miles upstream from camp on a canoe trip, he passed a ...
 algbra  a paddle boat can move at a speed of 14 km/h in still water. the boat ...
 math  Seven people rowed 12 miles downstream with the current in 1.5hrs. Coming...
More Related Questions