Question 1166744: On a canoe and camping trip, a group of Scouts decided to canoe downriver, set up camp, then canoe back to their starting point. The current stayed constant the whole time.
The Scouts canoed downstream (with the current) for 33.25 miles before stopping to camp. This took them 3.5 hours.
After camping, the Scouts canoed upstream (against the current) for the same 33.25 miles. This took them 9.5 hours.
Let x be the rate the Scouts travel in still water (with no current).
Let y be the rate of the current, such that:
Distance Rate Time
Downstream (x+y)
Upstream (x-y)
Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 3.5 hours for 33.25 miles is 9.5 mph
9.5 hours for 33.25 miles is 3.5 mph
x+y=9.5 mph
x-y=3.5 mph
2x=13 mph
x=6.5 mph speed in still water
y=3 mph current.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Did it not occur to you to ask a question about this...?
It's obvious what we/you are supposed to do with the given information... but you should make that part of your post.
The downstream and upstream rates are distance divided by time:
rate downstream: D = 33.25/3.5 ( = x+y)
rate upstream: U = 33.25/9.5 ( = x-y)
Algebraically, the canoe speed x is found by adding the upstream rate and the downstream rate and dividing by 2:
((x+y)+(x-y))/2 = 2x/2 = x
Without algebra, logical reasoning tells us the canoe speed is the average of the upstream and downstream speeds.
rate of canoe: x = (D+U)/2
The speed of the current is then the difference between the downstream speed and the canoe speed ((x+y)-x = y), or the difference between the canoe speed and the upstream speed (x-(x-y) = y).
rate of river current: y = D-x, or y = x-U
You can do the calculations.
Note the rate of the current is also equal to half the difference of the upstream and downstream speeds: y = ((x+y)-(x-y))/2
But it's hard to see logically why that calculation gives the speed of the current. It makes much more sense to find the rate of the current as the difference between the canoe speed and either the upstream or downstream speed.
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