Math Riddle, Insoluble?

Factual Questions
1

I’ve been banging my head against this for a few hours now, and I think that there is no real answer to it. I may just be missing something obvious though, so here it is:

"Biff is out canoeing. On his way upstream, he passes a log floating along with the current. He continues upstream for another hour, then turns and heads downstream. Two miles downstream from the point where he saw the log before, he passes it again.

What is the speed of the current, and what are Biff’s possible upstream and downstream rates?"

If anybody can get a real answer not involving negative numbers and quadratic formula answers (which I am told are incorrect), I am definitely impressed, and would love to see how.

2

First, consider the speeds of the canoe and log with respect to the flowing water, not with respect to the shore. This is like pretending that it’s a pond with no current, and pretending that the log is not moving. If he paddles one hour away from the log, he will paddle one more hour to get back to it. So the round trip from the log and back to the log again is two hours.

Now add in the speed of the water (and log) with respect to the shore. The log, moving with the current, has travelled 2 miles in 2 hours, so the current is one mile per hour.

As far as I can tell, his speed could be anything at all, except maybe less than 1 mph. He could, for example, paddle at 10 mph with respect to the water, so he goes upstream (with respect to the shore) for 10-1=9 mi. in the first hour and downstream for 10+1=11 mi in the second hour. Or he could paddle at 5 mph for 5-1=4 miles upstream and 5+1=6 miles downstream. The riddle seems to imply that he is actually moving upstream while paddling upstream, so I guess his speed in paddling (with respect to the water) must be at least 1 mph to keep up with the current. But even if it wasn’t, he would still meet the log 2 miles downstream after 2 hours. He could, for example, paddle at 0.1 mph with respect to the water so he goes downstream at 0.9 mi. in the first hour and downstream 1.1 mi. in the second hour. In either case, he would still meet the log after 2 h, 2 miles downstream.

3

Let S = Biff’s “paddling speed”
Let C = speed of the current

(The unwritten assumption is that both of these are constant.)

Then Biff’s speed upstream is S-C, downstream = S+C.
The log travels downstream at speed C.

Let T=time travelled in hours.
The log travels CT = 2.
Biff travels upstream -(1)(S-C), then down (T-1)(S+C).

Then CT=2=CT+ST-S-C-S+C, or
CT=2=CT+ST-2S.
Ergo ST-2S=0, T=2, and since CT=2, C=1.

Biff’s “paddling speed” can be anything. His travelling speed downstream will be 2 mph greater than up, due to the current.

4

The whole river (Biff, log and all) are moving with the current at the same speed. So relative to each other, current can be ignored. We know that Biff goes away from the log for an hour, and then turns and comes back to the log. Assuming Biff is moving through the water at a constant speed, he must therefore take an hour to get back to the log, and he has been away from the log for two hours total.

In two hours, the log has moved downstream two miles. Therefore the current is at one mile per hour.

You cannot say how fast Biff is going. He could head upstream at one mile an hour for an hour. His rate over land will be nil (the current will cancel out his speed). Then when he turns round his rate over land will be two miles an hour and he will meet the log two miles downstream as expected. He could head upstream at a million miles an hour giving him land speed of 999,999 mph. Over the hour he will move 999,999 miles upstream. Then he’ll turn around and come back at a land speed of one million and one mph resulting in him meeting the log 2 miles from where he first passed it.

Have I just done your homework for you?

5

This is what I figured out.

Let T = time from when Biff first passes the log until he passes it a second time.

d = distance Biff covers upstream.

W = velocity of the stream

V = velocity of Biff with respect to still water.

V_up = Biff’s upstream velocity with respect to the ground

V_down = Biff’s downstream velocity with respect to the ground.

in the time T, the log moves 2 miles downstream

2 miles = W*T

In traveling 1 hr upstream, Biff moves

d = V_up*(1 hr) = (V-W)*(1 hr)

Downstream Biff must move to pass the log again

d + 2 miles = V_down*(T-1 hr) = (V+W)*(T-1 hr)
You can see that we have 3 equations but 4 unknowns (T, d, W, V) so the problem cannot be solved completely.
I can substitued the first equation into the third to get

d + WT = (V+W)(T-1 hr)

d= VT +WT - V1hr - W1hr - WT = VT- V1hr - W1hr

setting this equation equal to the second one above

V1hr - W1hr = VT -V1hr -W*1hr

Solving this gives me

0 = V*(T-2hr)

so T = 2hrs

The velocity of the water is W = 2 miles / 2hours = 1 mph.

Now

d = V_up * 1hr and d + 2 miles = V_down * 1hr

This tells us that

V_down - V_up = 2 mph.

Any choices for V_up and V_down that satisfy this will work
GTPhD1996

6

I still don’t see it. The problem I run into is the three different times. He passes the log, then goes upstream for 1 hour at (Vu-Vc) where Vu is his paddling speed and Vc is the current. He then turns around, and the time it takes him to get back to the point where he originally passed the log is (Vu-Vc)/(Vu+Vc), going at a rate of (Vu+Vc). He then spends an additional (Vu+Vc)/2 in time, before he finally reaches the log again and passes it again. So, the log has traveled a total of 2 miles, at Vc. We can then say that Vc=2/(1+(Vu-Vc)/(Vu+Vc)+(Vu+Vc)/2 ). And thats about it. How can you resolve those three different times? You can’t really just say that the current can be ignored, because part of the time Biff must paddle against it and part of the time he must paddle with it at the same paddling rate. So, as I am very dense, if somebody has the patience could they please explain to me where the times go to get any sort of answer.

7

I get that either
[ul][li]Biff just drifts with the current, while the current has arbitrary nonzero velocity (this is the trivial solution), or[/li]Biff travels at an arbitrary (constant) velocity while the current travels at 1mph.[/ul]

8

You are tying yourself in knots by calculating all sorts of unnecessary things.

Break it down into two stages. The first is calculation of the time involved. The second is the calculation of the speed involved by applying time to a known distance.

The only time we are given is a time in relation to movement of the canoe away from the log (one hour). As we are working here with only the canoe and the log, we need only consider their movements relative to one another, and we can ignore the land for the moment.

Current can be ignored in this context because the log and canoe are both floating on the same water, and all the water is moving at the same speed downstream. As long as Biff is putting in the same amount of effort, his speed through the water is the same. Whether he is going up or down stream will only affect his speed relative to the land.

So, bearing that in mind, if you move away from an object at a fixed speed for a fixed time, then turn around and come back at the same speed, it will take you the same fixed time to get back to that object.

Therefore, Biff moves away from the log for an hour, it will take him an hour to get back.

You have now determined that the time involved is one hour each way.

Now, to arrive at current speed relative to land, you need to compare the time involved (two hours) to the land distance travelled. The land distance moved by the log is 2 miles. Therefore the current is one mile per hour.

I don’t think I can explain it any better.

9

"The only time we are given is a time in relation to movement of the canoe away from the log (one hour). As we are working here with only the canoe and the log, we need only consider their movements relative to one another, and we can ignore the land for the moment. "…"Whether he is going up or down stream will only affect his speed relative to the land. "

This is my problem right here. According to the wording of the problem, he paddles upstream for an hour, away from the point where he first passes the log. He then turns around, and takes a different amount of time to get back to the point where he passed the log, a point which must be referenced with respect to land. His speed relative to land is important to know to have the time he actually spent going back down the distance he traveled up for that one hour. You can’t just talk about the canoe and log in the water’s reference frame if the point at which you are measuring speed or time is outside of that reference frame.

10

He then turns around, and takes a different amount of time to get back to the point where he passed the log, a point which must be referenced with respect to land.

Yes, but not a point which must be referenced to solve the problem presented. The only reason the point where he first passed the log is mentioned is to establish that when he meets the log again, the log has moved two miles downstream.

The question is framed in terms of his meeting the log on the way up and again on the way down. You are not asked to solve anything relevant to when he passes that first meeting point while headed back downstream.

11

Here’s my solution:

C= speed of current
B= Biff’s speed(without current effect)
T= Time Biff goes downstream
1= Time Biff goes upsteam

So the distance Biff travel’s upstread and downstread has a difference of 2 , ergo:

(B-C)=T(B+C)-2 (Distance equals distance)
(T+1)C=2 (The log’s distance)

These are two equations in three unknowns. One can solve for B or C or T, but you’ll never get an equation that doesn’t have a variable on the right side of the equation.

12

Here’s my solution:

C= speed of current
B= Biff’s speed(without current effect)
T= Time Biff goes downstream
1= Time Biff goes upsteam

So the distance Biff travel’s upstream and downstream has a difference of 2 , ergo:

(B-C)=T(B+C)-2 (Distance equals distance)
(T+1)C=2 (The log’s distance)

These are two equations in three unknowns. One can solve for B or C or T, but you’ll never get an equation that doesn’t have a variable on the right side of the equation.

13

I worked it out somewhat similarly, but I get a different answer. Looking for holes in it, can’t see any.

Stating knowns:

A : position at which canoe meets log for the first time.
B : upstream turnaround point
C : downstream point at which canoe meets log for the first time.

Vc : velocity of canoe relative to river (positive upriver)
Vl : velocity of log relative to A (positive downriver)

Assumes Vl > 0 and Vc > Vl since the river must be moving, and ‘continues upstream for another hour’ implies some net upstream distance over the hour.

Dab : distance from A to B (and also subsequently the distance from B to A.
Dac = 2 : distance from A to C
Tab = 1 : time from A to B
Tab = Dab / (Vc - Vl)
Dab = Vc - Vl

Tba = Dab / (Vc + Vl)
Tba = (Vc - Vl) / (Vc + Vl)

Tac = Dac / (Vc + Vl)
Tac = 2 / (Vc+Vl)

From the moment at A that the log is first encountered, the log moves downstream at a rate of Vl. It continues moving downstream until the canoeist has gone upstream and all the way back downstream again, so the covered distance based on time is:

(Tab + Tba + Tac)Vl

But the covered distance is also 2 miles downriver of A, so:

(Tab + Tba + Tac)Vl = 2

So:

(1 + (Vc - Vl)/(Vc + Vl) + 2/(Vc + Vl))Vl = 2

Collecting the terms inside over a common (Vc+Vl) denominator:

((Vc+Vl) + (Vc - Vl) + 2)Vl/(Vc+Vl) = 2

Collapsing:

(2Vc + 2)Vl/(Vc+Vl) = 2

Expanding again:

(2VcVl + 2Vl)/(Vc+Vl) = 2

Multiplying by (Vc+Vl)

2VcVl + 2Vl = 2Vc+2Vl

Getting rid of the 2Vl from both sides, and dividing by 2

VcVl = Vc

Dividing by Vc

Vl = 1

I end up with the log floating downriver (with the current) at 1 mi / h.

I also end up with Vc being absolutely anything.

14

Well, okay, I end up with Vc being absolutely anything subject to the initial assumptions that Vc > Vl, so Vc is anything greater than 1 mi / h.

Where’s the hole?

15

This might help illustrate it. Imagine he’s paddling so slowly that in still water he would only move 10 feet per hour. He and the log are drifting downstream at nearly the same rates, but he’s going slightly slower. The log passes him, and he paddles upstream for an hour, at which point he’s 10 feet upstream from the log. He then turns around and starts paddling downstream at the same snail’s pace. An hour later, he’s caught up to the log, and they have both moved two miles.

Now imagine the same scenario, but with him going 100 miles per hour relative to the water. It all works out the same.

The way the problem is worded may disallow the rate of 10 feet per hour, since it does say that he actually travels upstream, rather than merely paddling in an upstream direction. If that’s the case, then his speed can be anything greater than 1mph.

16

I still think my formula works. Of course B>C and 0<C<2.

Can someone check it?

17

il31415li,

Try this:

Biff is standing with a log on a moving walkway (you know, like at the airport). He puts the log on the ground and starts walking backwards for an hour (OK, its a really really long walkway). Then, he starts walking forwards again. Lo and behold, he comes upon the log.

Does that image help at all?

18

We can’t both be right, aahala, except where your C (my Vl) is exactly 1.

(B-C)=T(B+C)-2 (Distance equals distance)
(T+1)C=2 (The log’s distance)

Getting rid of T:

TC + C = 2
TC = 2-C
T = (2-C)/C

(assuming C is not zero, of course)

(B-C)=(2-C)(B+C)/C - 2
(B-C)=(2B + 2C - BC - CC - 2C)/C
(B-C)=(2B - BC - CC)/C
(B-C)=2B/C - B - C
B = 2B/C - B
2B = 2B/C

Oh, of course, that results in C=1, and B being anything more than C according to the original stipulation. hey, we agree. :slight_smile:

19

(Bolding added)
Don’t be disheartened, il31415li. You are trying to solve the problem exactly the way I solved it, but you just made a little mistake. In the sentence I bolded above, you should have 2/(Vu+Vc), but you accidently wrote the reciprocal (Vu+Vc)/2. Putting in the correct expression, your equation becomes Vc=2/(1+(Vu-Vc)/(Vu+Vc)+2/(Vu+Vc) ), which you’ll find reduces algebraically to Vu = Vc * Vu. From this, it follows that if Vu is nonzero, then Vc = 1, and Vu may take on any arbitrary value.

20

Aahala, your equation is correct, but your statement that T is incalculable is incorrect.

We know that the distance travelled Biff travels through the water (ignoring his land speed and distance) away from the log, equals the distance Biff travels through the water towards the log. Therefore, assuming Biff’s speed through the water is constant, the time up and down stream must be the same.

Or if you want it in a formula:

d1 = distance relative to log travelled upstream
d2 = distance relative to log travelled downstream
1 = time travelling upstream
T = time travelling downstream
v = speed through water

In general, distance = velocity x time

d1 = 1v
d2 = Tv

d1 = d2

Therefore

v = Tv
T = 1

Have we done this one to death yet, or what?