SOLUTION: On a canoe trip, Rita paddled upstream (aginst the current) at an avarage speed of 2mph relative to the riverbank. On the return trip downstream (with the current), her avarage spe
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Question 589485: On a canoe trip, Rita paddled upstream (aginst the current) at an avarage speed of 2mph relative to the riverbank. On the return trip downstream (with the current), her avarage speed was 3mph. Find her speed in still water and the speed of the river's current.
I don't understand at all how to do these so help ASAP would be great. (ps, we're supposed to use a system of linear equations to solve) Thancks in advance :(
Found 2 solutions by richwmiller, ankor@dixie-net.com:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
current is .5 mph
speed in still water 2.5 mph
(2+3)/2=5/2=2.5
3-2.5=.5
2+.5=2.5
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
On a canoe trip, Rita paddled upstream (against the current) at an average speed of 2mph relative to the riverbank.
On the return trip downstream (with the current), her average speed was 3mph.
Find her speed in still water and the speed of the river's current.
:
Let s = speed in still water
let c = speed of the river current
then
(s-c) = effective speed against the current
(s+c) = effective speed with the current
:
We have two effective speed equations,
s - c = 2
s + c = 3
------------adding eliminates c, find s
2s + 0 = 5
s = 5/2
s = 2.5 mph is the speed in still water
:
then using the "with" equation
2.5 + c = 3
c = 3 - 2.5
c = .5 mph is the speed of the current
;
;
You can confirm this in the "against" equation
2.5 - .5 = 2
:
I tried to make this understandable to you. Did I succeed?
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