Leaving The Lake: CR4 Challenge (05/27/08)

Hi jim35848,

I have given you a good answer too.

I have checked your method which, by the way, is "integration by summation".

Using time values of 0.01 sec your answer has a margin of error of only 0.0136%

Don't forget to add a constant of integration "C" when integrating. And then apply your initial conditions to solve for "C" (at t=0, s=20). So the integred equation should be (before applying your integration limits) s^2/2 = 100t + C. Applying your initial conditions at t=0, and s=20 to solve for C, you get 20^2/2 = 100*(0) + C, C=200. The equation then becomes s^2/2 = 100t + 200. Now apply your limits of integration; distance s from 20 to s and the time from t=0 to 20 which gives
1/2*(s^2 - 400) = 100*(20-0) + 200.
s^2 = 4400 + 400.
s^2 = 4800.
s = 4800^0.5
s=69.28 m

Sorry! I can't agree with your comment.

A constant of integration is only required for an "indefinite integration" (one without limits). When you form the "definite integration" any constant cancels out; since you evaluate the integral (with a constant of integration if you like) at the upper limit, then evaluate the integral (with the same constant of integration if you included one) at the lower limit and subtract the two values; at which point the constants cancel each other out.

Verification also comes from the posting #1 which is an excellent digital approach used in many situations, to approximate by adding small pieces together.

I stand by my answer.

I agree with SlideRuler. The constants drop out. If you substitue time zero into the Guest"s solution, you get a distance of 28.3 m and not 20 m.

I agree that V= 100/S at distance S from the lake.

V=ds/dt or 1/V= dt/ds.

Also, 1/V= S/100.

Therefore, dt/ds= S/100.

Integrating, t= S^2/2*100 = S^2/200.

t=20 seconds which would mean 20= S^2/200.

Therefore,S^2= 20*200=4000.

S= (4000)^0.5=63.24metres from the lake.

Sorry, you're wrong

You have integrated "indefinitely" and therefore require a Constant of integration

Your resulting eqation t= S^2/2*100 = S^2/2 is not valid at t=0. It gives s=0 instead of the correct s=20.

See also my response to post #16. (guest) who overestimates the answer by including a Constant of integration and then integrates with limits, resulting in an overestimate of the distance. By NOT including a constant of integration, you under-estimate the distance. I find "definite integration" muchmore convenient to use in cases like this, although "indefinite integration gives the correct answer of course when used correctly

15 feet

15 feet. That was when the runner tripped and fell after 3 seconds....

I live in an area with lots of man-made lakes, and a few natural ones. One thing I can tell you about all these lakes, is that the ground slopes up away from the lake. So, you will be running uphill and working against gravity. Just thought I'd mention that so you could figure it in. I haven't taken the time to cogitate on this much, but I'm fairly sure this will have to be factored in somehow.

And then you have places like New Orleans, where much of the city is below the level of Lake Pontchartrain....

Well, 3Doug, if one is 20 meters back from the shoreline, it is entirely possible that the ground has been leveled from that point on to make this hypothesis calculable. (Calulable?? Is that a word?) Or perhaps the runner has sufficient strength and the ability to run consistently stronger to overcome those forces. Sounds like a trick question to me.

Velocity is inversely proportional to the distance from the lake.

I don't agree at all.

At least when I'm relaxing I don't do it at 5 m/s. If I decide to exercise and start running, means that I wasn't running yet. So, initial velocity must be 0 m/s. At least I never could pass from 0 m/s (or a fraction of m/s if I were quietly walking) to 5 m/s in 0 seconds.

I must be mad, but I don't understand the question.

You are relaxing 20 m away from the shore of a lake ....

If this means floating 20 m from shore than any attempt at leaving the floating devise would put you in the lake.

My answer is that in 20 seconds you have re-surfaced and are mostly still in the lake with a distance of 0 m.

relaxing 20 m. away from the shore? yes but in the opposite direction to the lake!!!!

If I'm inside the lake at 20 m. from the shore I can't start running (I didn't attend the school when the teacher explain how to walk over the water).

So, sorry but you have to look for a different answer

Now that I'm thinking of it, the choice of a 2-dimensional reference (a lake) with unspecified shape as opposed to picking a single point to measure distance, smells tricky. Of course, as no other information is available, we can assume that there is no influence to the solution. The lake's shape may as well be convex (e.g. a circle), so we can accurately define the meaning of "running away from the lake". Moreover, the distance that is run, readily adds up to the distance from the lake, as the direction of the course is always perpendicular to the shore of the lake.

If on the other hand, the lake had a non-convex shape, then we couldn't exactly define the direction of the run, let alone, that we may have to constantly redefine the direction we have to place a hypothetical measuring tape to count the distance from the lake, at every moment. Imagine, for example, a PacMan-character shaped lake and a starting point somewhere inside the "mouth" of the beast...

I doubt there is any slight intention to pose a tricky question as such, but I cannot help thinking: "why a lake?"

Gentlemen, Fellow engineers,

Not sure how some of you come up with 100 (5 * 20 ?)

I see it this way. 20 m away from the shore (on land of course) you (or an object) starts running at 5 m/s away from the lake. The velocity is decreasing inversely proportional to the distance from the lake.

If I run 20 m, the distance from the lake is now 40 m and the velocity is 5 * 20 /40 = 2.5 m/s, after running a total of 60 meters, the distance is 80 m to the lake and the velocity is 5 * 20 /80 = 1.25 m/s and so on. (At 160 m distance, v is .625 m/s).

We have a case of non-linear deceleration. If it would be linear i could solve it with:

a = (Vf^2 - Vi^2) / (2*S). Plugging in above numbers in various cases show a non-linearity for a, -a really.

So perhaps someone else can show the way to solve that.

Well my friends I get 33Km and 42.4 metres

Can this be correct. If my interpretation is correct it may fit the question better to have a rocket launcher rather than running

Starting at 5m/s and increasing by 21/20 then 22/20 upto 40/20 gives a terminal velocity of 33 Km a second

And the distance travelled in the 20 seconds is 33Km and 42.4 metres

If the lake were frozen and you were in the middle of the lake 20 m from the shore and started running away from the lake... never mind.

If you were on the top of a frozen lake, the only direction you could go to get away from the lake is upwards, but generally, the first step would possibly put you flat on your face, leaving either a nice face plant, or a red smudge where you hit

For the awkward squad (as that's really asking for trouble, I hereby give notice that I will only respond to consequent comments if I find it amusing so to do):

It would appear that, for this case, the pedants are correct - i.e. the (technically excellent) solutions given by SlideRuler and TKOT don't answer the question as written.

If we take the wording of the challenge, initially you are in one of the following positions (unless I've missed something):
On or in some structure, and 20-metre above/below the shore,
In the lake (or some other body of water)**, or
20-metres from a shoreline that is on the other side of a narrow section of lake.

Unlike the other options, the interpretation that places you somewhere in the lake is uniquely defined as regards solving the problem - albeit the solution is grossly non-physical, and to make any sense you have to define time wrt your inertial frame of reference and distance wrt the frame of reference of the shore(!!)

You start at the speed of light, so (in your time-frame) it takes you no time at all to reach the shore (effectively infinite velocity?). As we've been forced to define infinite velocity in the inconsistent manner of the previous sentence, we continue to take time in your frame of reference, and distance in the frame of reference of the shore (though the difference between this and calculating everything subsequent to crossing onto the shoreline wrt the frame of reference of the shore will be very small). In that case, we can use whichever of SlideRule's or TKOT's formulations you prefer, to arrive at a distance of √(200.t), or ≈ 63.24 metres.

That means that the silly solution (that allows running on water and at the speed of light) isn't so very different...

**Maybe all I'm doing is explaining bhasker03's solution?

you mean s = √(200.t+400) which is 66.332m and not 63.25m

This also gives the correct value at t=0.

As stated above, that does not answer the question as written; it is the answer as you sensibly re-interpreted the question (I do try not to indulge in void repetition of others' contributions).

If you start 20-metres from the shore (i.e. in the lake), your initial distance from the lake is zero, and therefore your velocity is infinite** until you reach the shore. So the time is the same as if you started at the edge of the lake, so you should use s = √(200.t).

**For detail on how I cope with that without violating special relativity, please read my previous answer.

If you start 20-metres from the shore (i.e. in the lake) → False

I live in Madrid, 350 km away from the closest shore and I'm completely dry. Distances can be measured in a line in two opposite directions.

Please, see post nº 18 (It's mine, but I didn't realize I wasn't logged in)

Sorry, but you must be using a dictionary that I can't access. According to my dictionaries (I have now checked Webster, Oxford, Cambridge and Chambers) you are not "away from the shore" - you are ON the shore, albeit so far away from the lake that you would not naturally describe your position in that manner. You are 350km away from the shoreline, which is a completely different matter.

Okay, I think I finally get it. A strict definition of shore is the strip of land along a water body that is alternately exposed and covered by waves and/or tides. So the answer will depend on how wide the shore is when we start running. I (being not fully understanding of the english language) was thinking we were measuring from the water line (edge of the water). Therefore, I've reformulated my analysis to include the new parameter S = width of the shore. (My first answer was a first order approximation for S = 0).

My second, more enlightened, answer from distance from lake after 20 seconds is:

X = distance from lake = sqrt(s^2 + 200m*S + 4000m^2)

I got this by using SliderRulers approach but including a shore width in the analysis.

"A strict definition...": unusually for Wikipedia, that is the (strict and unique) definition of the "foreshore"**, which is equally correctly defined as the part of the shore that lies between the low and high water marks. So far as I know, shore in normal usage (and as related to water) has two valid definitions:

i) a strip of land bordering on the sea, lake, or other body of water
ii) land as opposed to sea

The use of shore to imply foreshore applies, so far as I know, only in legal documents where it relates specifically to rights of use of land that can be covered by tides. (It may also be used in some technical areas, but I am not familiar with these).

The legal definition would in any case need to be extended to apply to the shore of a lake (presumed non-tidal) - possibly the normal range that could regularly be covered or uncovered due to variable rainfall? To get a unique answer you would then need to assume that the lake is at its fullest.

But have we spent long enough beating Dobin's deceased relatives?

**Believed to be derived as "the region at the front of the shore". Backshore is also used to describe the part of the shore that is above the high water mark.

Since the challenge states the "shore of a lake", doesn't this imply it is a strip of land along the lake. Otherwise, wouldn't the challenge state the "shore near a lake"???

O.K. I get it! Checking your interpretation of the question however, are your first 20m in a direction "away from the lake?" Admittedly the are "towards the shoreline" but it sounds to me as if they are "along the lake"

I agree. But the only way you can run away from the lake is if you first leave it - so I reckon that running along the lake towards the edge is a necessary precursor to the exercise you decided on. I also stretched a point with the velocity, which is only strictly defined once you are running away from the lake; but you are running in/along the lake for zero time, so I initially thought that it was the closest I could get to answering the original question.

On consideration, however, it could be that the challenger's clock only starts ticking once you have started running away from the lake, so you could perhaps take as long as you like to reach the shore. (Fortunately, it doesn't affect the value of the answer).

Will 140.954 m away from the pond do for an answer?

When I find how to put equations and plots with my comments (See General Discussion Blog), I will be able to show how I almost made it to 141 m.

But, suppose I got up and started directly toward the pond and the pond water supported me or running at a sufficient speed I remained on the water's surface, I would end up 60.954 m from the shore in the water. Now I would hope I either touched bottom or could swim to shore.

soaralone1

No, it won't - unless you can explain what you interpret differently from jim35848 and SlideRuler. (66.33-m works)

Click on the blue Omega sign above your text entry for symbols in equations, or cut and past your text. If that's not what you want, you can make a jpeg file and use the camera to paste it (but the scale for text can be awful - it's probably best bite the bullet and type directly into the frame and accept the limitations of the CR4 editor.

Just about the only things in the question that are uniquely defined are the reference for the velocity and the direction of travel and the - the lake and away from the lake respectively.

No! It is not a good answer. That's what I get by working carelessly - too fast.

I went to work on the lake problem the next day -- but whew all the interruptions lead me to finish up a more thoroughly prepared answer nearly a week later - LATE. Tsk Tsk! I goofed by inadvertently including the 20 m distance from the lake 2 times!!!!

A better answer is 120.9545 m. It's about three pages of explanation and calc's I did in Mathcad 6Pro. Should I paste this in? How do equations respond that might be wider then the page allows? I'd like to compare with Jorrie's solution.

soaralone1

You might as well spare us from reading wrong answers and hypothetical thinking of running into the lake and over the water. You got to stay real.

You needed three pages of calculations??? That alone tells you it can't be right.

Floram,

Nice comment - and I appreciate that you didn't slam me too hard. I thought showing my calculations [with explanations] would be more fun for you, and all the friendly folks, to tear it apart to show me where I went out of step. But, if it turns out my solution is more "You got to stay real." and I did, I would appreciate at least a "Thank you" or "Gees!"

I have several comments at each stage of the calculations. The principal reason for me was, at 1st blush, I thought "k" may not be constant.

I set up the problem (here without comments other than the singularity), I paste in a simplified (and corrected from my earlier) calculations -- below

??? !! ?? Well, I have just spent from about 09:00 hrs to 12:40 hrs trying to put my work on this comment in this blog with no success, none what so ever! I am glad to have done the repeated calculations. I found a miscalculation so I have a different answer and I want so very much to show you and the other good folks how I arrived at them ????? Yes, "them."

Problems --!!?? I find no way around this pop-up flag:

[The MIME of the uploaded file "application/x-mathcad" was not accepted by the server.]

I do not find any means for me to put my calculations on this blog. I do not have a URL I can use and I have the document I have "Browsed" to (has been tried many times).

So, at least I can report my new-to-me answers

"forward" distance (away) from the shore 55.8268 m

"backward" distance (toward) the shore -35.8268 m. For this to happen, the runner really picks up speed (to negative infinity (-m/s) as the runner goes over the shore line (water's edge on the land of the beach), then immediately slows to a running speed capable of (and large-enough feet/shoes for) keeping the runner on the surface of the water, or the power to plow through the water without slowing).

The "complication" is that the velocity needed to find the distance the runner finds him-/her-self at the end of 20 seconds is

v*t = k/(v*t+a)

which yields a quadratic for the runner to choose running away from the shore, or running toward the shore and performing some spectacular physical gymnatics.

soaralone1

dear Soaralone1,

My comment was not meant to offend in any way. Regarding others analyzing were you went wrong, I think there have been plenty of correct answers and various ways to calculate it, including also some near-correct answers where there would be little room for anther wrong answer. Of course, the possibilities of going wrong are near endless.

An interesting solution was given by tkot, solving for s^2, where the graphic representation is a straight line? I found that also a very good answer, among the many other correct ones. You know when it is correct when the answer is 66.33 m.

Now, why did you want a 'Thank you' or a 'Gees'?

See again at the next challenge question.

Dear Floram,

Sorry if I clicked your comments when I replied to Physicist? His comments are what I was addressing, not yours: Yours, I appreciate very much.

I may need your personal help. How do I publish my work on these comments? What are the specific actions I must perform? Check out my most recent comments - I have spent - literally hours - trying to publish on these (and other) specialty blogs like CR4. I must tell you: If there is a "Murphy" around, my work is always on his list.

The answer, 66.33 m, is not supported when we consider the runner starts with a 5 m/s velocity and the decreasing runner-added velocity is also positive. It seems the distance must be well beyond 100 m, wouldn't it?

Here's the "kicker" omitted in the others' work: the momentary runner velocity (that we solve for to multiply by related momentary time) appearing in the left side of the equation of velocity AND in the denominator of the right side of that equation. This leads to a quadratic that has 2 solutions: 1) in the forward direction away from the lake AND 2) the reverse direction away from the lake.

If you can give me specific steps to take to put my calculations on this site in these comment threads, I will be greatly pleased.

soaralone1

First, I suggest you try a quick sanity check:
Your initial velocity is 5 m/s. If your velocity was constant, in 20 seconds you would move 100 metres. That would place you 120 metres away from the lake. We are told that your velocity varies inversely with distance - which means you get slower as you get further from the lake, so you must be less than 120 metres away.

Then I suggest you look at the official answer and at the solutions that were rated "good"; if you can identify errors, then you will have something to add.

Physicist?,

soaralone1

Physicist?,

My forgoing answers were to check on some of all you good folks' thinking. Hm-m-m?! The answers were only some calculations of the traveling from the starting position. incomplete, which had I been unable to put them paste them in --??

HERE, I announce my prior answers were not at all correct.

Try these, and I am open to reviewing my work once I can get them to you, Physicist?, all the others - good folks.

Forward (away from) the shore/lake End position 155.8318 m from the shore line.

Backward (toward the lake) End position 64.1782 m from the shore line (losing distance trying some pretty tough gymnastics to overcome (to the backward facing runner) initial backward going velocity, 5 m/sec.

Again, the two direction options are the mathematical fall-out of the quadratic generated by the velocity (to be multiplied by time 0...20 sec) appearing in the left-hand side of the velocity equation and in the denominator of the right-hand side of that equation.

soaralone1

Clearly, the questioner intended that 'you' run away from the lake - but that doesn't mean that the option of running towards it is without interest.

However, I think there is ambiguity as to what happens after the runner (now changed from 'you' to hopefully improve clarity) reaches the lake.
. If the velocity is specified as the distance from point on the lake that is closest to the runner, the direction once he reaches the lake is undefined - but the speed is infinite. So, once the runner reaches the lake, he can instantaneously be anywhere on it - but will never leave it**.
. Alternatively, we may define a distance from the runner to the edge of the lake. In that case I think we should assume that the velocity is also towards the edge of the lake. So he can go around the edge of the lake at infinite speed, but will stay at the lake's edge.
Whichever we choose, once he has reached the edge of the lake, other than being in the lake or on its edge, we have no idea where to find him.

**But- even though we can set up a one-to-one correspondence between the points on a line and the points on a surface, that does not mean that he will visit every point on the lake in a finite time...

Physicist?,

Yep, I'm with you. That is ,I believe my re-enactment of the runner's backward-going is explained another way. Hope this avoids some of the ambiguity plight.

I chuckled when I saw the two directions and thought I could have a little fun with it. Even though the runner turns his/her back to the starting post (20 m from the shore), that results in negating the runner's position starting at the pole, the pole remains 20 m from the shore, 20 m behind the runner placing the runner on the shore line. So, I added an additional 20 m putting runner's starting position at the pole. Then the 5 m/sec positive velocity "pushes" the runner backwards away from the pole and shore. The runner's toes "claw" the sand trying to go in the direction of a negative position toward the lake -- of course, without success. The runner finishes the 20 sec at 64.1782 m from the shore line. ERGO, the runner never gets any closer than 20 m from the shore line, if the runner starts from the 20 m pole/from the shore line.

Find my solution on my computer in Mcad 6PRO? I have not been successful converting it to Mcad13 SR2. Murphy owns my computer. He does what he wants to do and I cannot get around him to change some of my Mcad6 to Mcad13; the lake challenge being one of them he holds back. I have tried the conventional methods.

soaralone1

I think I see what you mean - the challenge didn't say where the other end of the distance from the lake was being measured. I (in common, I suspect, with most other readers) merely assumed it was the distance to you (as the runner).

On the other hand, the consequent reasoning appears too contorted (even for me)

I know that velocity is inverse proportional to distance, thus

(a) v = d/dt (x) = k/x
where d/dt (x) is the derivate in time of space, thus velocity.

I know that when x=20, then v = 5 m/s (as if the man would have been started sometime earlier, but do not worry about this now), therefore

(b) k=100

I am now looking for a function x(t) which solves the differential equation

(c) d/dt (x) - k/x = 0

We know that the square root function Sqrt (x) has exactly this carachteristic: the differetial is proportional to the reciprocal.

Therefore

(e) x(t) = A.sqrt (x)
where A is a constant to be found.

Substituting (e) in (c) and solving, gives

(f) A = sqrt (2k)

The definitive equation for distance in function of time is therefore:

(g) x(t) = sqrt (200*t)
where x=0 and t=0 references to the shoreline !!

Now: How many seconds would elapse to reach the 20 m? (the "starting" point): it is
t0 = sqr (x) / 200 = 400 / 200 = 2 seconds

And finally: which distance after additional 20 seconds, that is at t1= 22 seconds?

x(t1) = sqrt (200*22)=sqrt (4400) = 66.33 m

0 ft. I am not running away from the beer.

Hi ONE and ALL,

I have had to get a better computer -- the old one was not that very old -- 6 or 7 months. But like all computer installations I have personal acquaintance with, all have Murphy helping me !!!! My "new" computer crashes regularly -- this time, right now as I write this, it has taken 3 professionals, years well experienced, 3 weeks to get me up and running. They did not believe Murphy was in the beast, before they started -- BUT now they do!!!!!

So now I am back and have -- along with other ,many files -- I have to redo my files on the Lake problem.

I'll be back -- and hopefully be able to paste my work on it, here .

Soaralone1

Format your Hard Drive and install Linux, it'll do you for the next 10 years

Hi every one,

I will try again to paste in my 3 pages yes, xxx, it took 3 pages,; but I hope that is not too crazy. Please read it carefully to find and m-m-m goofs??

The second try to paste did not work. So again

The third try RTF did not work. So again

Murphy helped me the past few weeks -- I had a computer change, program re-installations, this project having to be redone -- all together much delay.

I repeat the challenge statement briefly so readers (and evaluators) can help me for any fixings needed.

A pole is located 20 m from the shore. From the pole , the runner starts with 5 m/sec velocity running way from the water and shor......!??!!?

NOTICE!!! This is my more than 7th attempt to get my computer together enough to get this reply to the "Leaving The Lake" Challenge. Yes, I have reformatted my computer drives several times by professional, well experienced computer folks. They have met and come to respect my brand of Murphy. I will be very brief in the comments but attempt completeness in the equations.

OK folks I have the file I want to down load and all attempts result in denials

I will need specific detailed step help -- I have already confessed that I am not a computer-phile; so no noises about my limitations here I thank you very much. What do I do??

soaralone1