algebra 2 word problem using systems of equations.

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elizabethA.

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Here is the problem:
A canoeist paddles upstream from a starting point. At two miles, he passes a log floating downstream. He continues to paddle upstream for one more hour, turns around and paddles downstream to his starting point, where the log now is.
What is the speed of the current?

I understand I need a system with two unknowns, usually the speed of the paddler and the speed of the current, but I am unable to figure out how to represent time using only these variables, and the distance traveled in the one hour is also unknown! I also tried using only the two mile distance (from where the log started back to the starting point) but still can't get it down to two variables.
 
elizabethA. said:
I am unable to figure out how to represent time
Click to expand...

Time may be expressed as the ratio Distance/Rate.

What have you tried or thought about, thus far, in this exercise? Did you define any symbols, yet? Did you draw a diagram?

These boards are for tutoring; we need a better idea of where you're stuck.

I don't have time to work this exercise at the moment (for the purposes of trying to guess what you might have done already), but here are some quick thoughts. In the future, please be complete in your requests for help.

When a boat moves against water that's also moving, then the boat's rate is reduced by the water's rate. Likewise, when a boat moves with the flow, the boat's rate is increased by the water's rate.

c = canoeist's rate in STILL water

s = stream's rate

Hence, the rate of the canoe upstream is c - s, and it's rate downstread is c + s.

d = rt

The time upstream for the first two miles is 2/(c - s)

The remaining distance upstream is c - s

The time from turnaround to the 2-mile mark is (c - s)/(c + s)

The time during the last 2 miles is 2/(c + s)

The rate of the log is s

Try to see how these expressions all come from manipulating the relation d=rt and substitution.

Do these quick thoughts help at all? If not, then you need to be more specific about what you need.

Cheers :cool:

PS: Please confirm my expressions because I did not
 
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Yes, this is helpful. I have made a chart, with upstream and downstream. I know that because the distances are equal, I will eventually make my two expressions (r times t)for the upstream and downstream portions equal. I did originally have the rates as c-s for upstream and c + s for downstream.

I get why the first part of the time can be represented as 2/c-s for the upstream leg of the journey, but I don't understand how to represent TIME for that second part (after the log but still upstream). Time is distance over rate, so I know the denominator will be c-s, but what is the numerator (distance for that portion)? Once I have that I believe I will add those together to get the expression for the total time, multiply that by the rate (c-s) and set that equal to rate x time for the downstream portion.

Additionally, for the downstream portion, I have the rate as c+s, and the time as 2/c+s (for that two mile part) and you instructed me that the other portion is c-s/c+s. I know I would add those together to get the total time (as above) but I don't understand why that portion of the downstream is c-s/c+s.
Why is that distance c-s?

This is confusingly worded and I hope you can understand my question!
Thank you.
 
Ok, another question. I just tried setting these equal and that obviously doesn't work! I need two equations with two unknowns! So, do I do rate x time ==distance for the upstream portion and the same for the downstream? and if so, won't the two sides just equal each other because I have to represent distance as rate x time? It seems like the two sides (for both the upstream and downstream) would just cancel each other out.

rate time distance

upstream c-s 2/c-s + c-s/c-s ?


downstream c+s 2/c+s + c-s/c+s ?



Thank you!
 
Hi Denis,
Thanks for this, but I'm not sure how to approach it this way. I' m thinking that although their distances are different, the time they travel is the same? But that would include the part of the hour that the canoe travels upstream. Do I still use two unknowns (c for canoeist speed, s for stream current)? I still don't get how to represent the distance traveled by the canoe after if turns around and heads downstream and reaches the two mile point! We don't know what part of the hour is used downstream. Totally confused!
I have made a diagram with the following


<---------------log, rate=s, distance=2, time=2/s---------------------------

<-----------------------canoeist, rate=c+s, distance= 2+ ?, time=?--------------------------------------------------

Also not sure how to relate these into two different equations.
Thank you,
Liz
 
Oops, just re read problem and saw that the canoe turns around AFTER one hour. So, is the time downstream for the canoe the time of the log minus one?
 
Ok, thanks. I was getting that everything cancelled out because of time being the same. But, the original question asks for the speed of the current. I must be missing something because I still don't get how to find the answer to that!
 
elizabethA. said:
Here is the problem:
A canoeist paddles upstream from a starting point. At two miles, he passes a log floating downstream. He continues to paddle upstream for one more hour, turns around and paddles downstream to his starting point, where the log now is.
What is the speed of the current?

Nice going Denis. I remember this coming up once before but could not remember the logic. I am a fan of the old fashion way however. This first came to my attention over 15 years ago.

1-Let X = the speed of the canoe.
2-Let Y equal the speed of the stream
The time for the log to travel east, from the point at which the canoe and log were along side one another, to the starting point is defined by Tl = 10,560ft/Y.
The time for the canoe to continue traveling one hour further west and back to the starting point is defined by Tc = 60 + [60(X - Y) + 10,560]/(X + Y)
3-Let S1 equal the distance from the canoe passing point to the turn around point or (X-Y)60 feet.

4-Therefore, 60 + [60(X – Y) + 10,560]/(X + Y) = 10,560/Y
5- [60X + 60Y + 60X – 60Y + 10,560](X + Y) = 10,560/Y
6-[120X + 10,560]/(X + Y) = 10,560/Y
7-[120X + 10,560] = 10,560(X + Y)/Y
8-120X + 10,560 = 10,560(X/Y + 1)
9-120X + 10,560 = 10,560(X/Y) + 10,560
10-120X = 10,560(X/Y
11-120 = 10,560/Y
12- Y = 88ft/min = 1 MPH….Whew!
Click to expand...
 
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