jon_j_rambo Posted December 25, 2005 Share Posted December 25, 2005 A column of soldiers one mile long is marching forward at a constant rate. The soldier at the front of the column has to deliver a message to the soldier at the rear. He breaks rank and begins marching toward the rear at a constant rate while the column continues forward. The soldier reaches the rear, delivers the message and immediately turns to march forward at a constant rate. When he reaches the front of the column and drops back in rank, the column has moved one mile. Question- How far did the soldier delivering the message march? [ December 27, 2005, 09:54 PM: Message edited by: jon_j_rambo ] Link to comment Share on other sites More sharing options...
Ottosmops Posted December 25, 2005 Share Posted December 25, 2005 I will try it: a = speed of the column b = speed of the soldier c = distance d = time c = 2 * b * b / ( b * b - a * a ) d = 2 * b / ( b * b - a * a ) Example: a = 3 mph b = 5 mph c = 3.125 miles d = 0.625 hours ( 37.5 minutes ) The column marched 1.875 miles in this time. I hope that I am not completely wrong, because I drank a glass of wine, and I do this rarely. Link to comment Share on other sites More sharing options...
Ottosmops Posted December 25, 2005 Share Posted December 25, 2005 Sorry for my previous post. The solution is wrong. I shouldn't try such things when I drink alcohol. Link to comment Share on other sites More sharing options...
Kuniworth Posted December 25, 2005 Share Posted December 25, 2005 Simple. 2.4142 miles. Link to comment Share on other sites More sharing options...
J P Wagner Posted December 25, 2005 Share Posted December 25, 2005 If the soldier was marching at the same constant rate as the line, he would never return to the front of the line but would keep pace at the rear.....my answer 1/2 mile one way [ December 27, 2005, 05:23 PM: Message edited by: J P Wagner ] Link to comment Share on other sites More sharing options...
Codename Condor Posted December 25, 2005 Share Posted December 25, 2005 thats what i thought too JP Wagner, not enough data in this problem, if both constant rates are the same he will never come back to the front line. Link to comment Share on other sites More sharing options...
jon_j_rambo Posted December 26, 2005 Author Share Posted December 26, 2005 Ah, we have some takers on this question, that is good. Link to comment Share on other sites More sharing options...
zappsweden Posted December 26, 2005 Share Posted December 26, 2005 . [ December 26, 2005, 07:39 AM: Message edited by: zappsweden ] Link to comment Share on other sites More sharing options...
beginner's luck Posted December 27, 2005 Share Posted December 27, 2005 This may be totally wrong and hate to waste first post on this but my answer is he didnt move any. The column was going in circle so he waited for column to complete circle and then turned around again. Hope you old guys dont beat this new guy up too bad. lol Link to comment Share on other sites More sharing options...
Kuniworth Posted December 27, 2005 Share Posted December 27, 2005 Ok the solution for you chumps; Distance = rate * time so the guy walks back and forth in one hour. That means the meeting point when he meets the person in the rear is A(soldiers rate or speed) divided with A+1(+1 for the columns rate/hour) A/A+1 Meanwhile his distance will be 2*A/A+1 so that means we have the following; 1(1 hour) + square out of 2(back and forth) = 2.4142135623 etc etc THe soultion is that the soldiers have to walk about 2,4 miles. Link to comment Share on other sites More sharing options...
JerseyJohn Posted December 27, 2005 Share Posted December 27, 2005 Excuse me if I'm wrong but we don't know what speed a "Constant Rate" is. Nor do we know how long the column is ... Etc & so on. Assuming that photograph shows the front of the smallest unit represented in SC, a corps, and they're marching (quite leisurely it appears) in column of fours, approximately 50,000 troops, and all that. -- ? I say the poor bastard never gets back to the front and dies of exhaustion in the attempt. Anyway, the question is: "Question- How far did the soldier delivering the message march?" My answer is, All the way there and back again. [ December 27, 2005, 04:35 PM: Message edited by: JerseyJohn ] Link to comment Share on other sites More sharing options...
Mr.Dozer Posted December 27, 2005 Share Posted December 27, 2005 the question isnt how long it took how long he traveled. he traveled 1.5 miles because they moved 1 mile and they are 1 mile long. so if at constant he would travel 0.5 mile to get to back because each side moved half to make one. then would speed up to get to front at a constant speed and the line is 1 mile long so he would travel 1 mile back. so 1 + 0.5 =1.5 miles Link to comment Share on other sites More sharing options...
jon_j_rambo Posted December 28, 2005 Author Share Posted December 28, 2005 This is too funny! The truth is correctly stated in this thread by none other than Kuniworth! Hip Hip Hooray for Mr. Kuniworth! Education in Scando does pay! Now for the rest of you, go back to the end of the class to allow room for Kuniworth to march forward! [ December 27, 2005, 10:03 PM: Message edited by: jon_j_rambo ] Link to comment Share on other sites More sharing options...
JerseyJohn Posted December 28, 2005 Share Posted December 28, 2005 And with the rest of us, minus Kuniworth, at the back of the class ... Link to comment Share on other sites More sharing options...
beginner's luck Posted December 28, 2005 Share Posted December 28, 2005 I think this question was rigged so Kuniworth would win! :eek: How can he have the right answer? No where in the statement did it say he only traveled for one hour. So if Kuniworth would have assumed(you know what assuming makes both parties-_sses, )a different time say 1.5 hours than he would have come up with a different answer totally. This is all just double talk(no offense Kuniworth)-- "A(soldiers rate or speed) divided with A+1(+1 for the columns rate/hour) A/A+1 Meanwhile his distance will be 2*A/A+1 so that means we have the following; 1(1 hour) + square out of 2(back and forth) = 2.4142135623 etc etc" Cause how can you come up with this ****- A =Messegers rate(rate is defined correctly as Kuniworth stated - rate=distance/time ie 60 miles per hour is the rate of a car traveling. A +1 = columns rate(doesnt make sense that he figured there rate as "A+1" {distance/time/time}.That would be like saying 60 miles per hour per hour. Uhh doesnt work buddy. Then he states some great forumla- 2*A/A+1 to solve the distance(which was the question) but tries to solve for A but squaring 2,which he early said A was the rate of the soldier. So it is obvious that JJ rambo just posted this as a way to make Kuniworth "look" smarter than he really is. Do you guys even know if these two guys are the same person? :confused: Sounds like it to me. ------Man, I NEED!!! SC2 to come out quick!!! So I dont have so much time to write in this forum!! ---------- Link to comment Share on other sites More sharing options...
Kuniworth Posted December 28, 2005 Share Posted December 28, 2005 Look patethic chump are you for real? You need to assume 1 hour to know the rate cause distance is the same as pace multiplied with time. YOu could assume something else and that would alter the correct answer but the solution to the problem is the same. If I assume 1 hour I get this Distance=Messengers rate * 1 hour ok? This is important so you have something to work with. But to know the messengers rate this is not enough. You'll have to take into the formula the moving column. So then you get this A(messengers rate) / 1(1 hour for column to move 1 mile) + A(messsengers rate) This forumula above is the key to understand when the messenger meets th person at the rear. So thus A / 1+A = the meeting point the distance will be this plus the whole way back in which the column continue to move 1 mile/hour. That is 2 miles to travel plus and minus the different paces of messenger and columns. Thus we get this forumla; 2 * A / 1+A = The messengers total distance to travel What I need to do next is to take the square out of the times he travels back and forth(which is 2) plus 1 which will be the same as the distance to travel. Thus 2 square + 1 = 2.4 miles. That is for you 2.4 miles per HOUR he has to travel. [ December 28, 2005, 02:50 AM: Message edited by: Kuniworth ] Link to comment Share on other sites More sharing options...
BRO,JD Posted December 28, 2005 Share Posted December 28, 2005 I don't know the right answer, but I sure know that answer is wrong. Take two extreme hypotheticals: 1) The messenger is traveling at the speed of light and the column is moving extremely slow; or 2) The messenger is traveling 1% faster than the column. It's clear that in either case the messenger won't travel 2.4 miles. Link to comment Share on other sites More sharing options...
jon_j_rambo Posted December 28, 2005 Author Share Posted December 28, 2005 HaHaHaHaHaHaHaHa! This is too funny! Link to comment Share on other sites More sharing options...
BRO,JD Posted December 28, 2005 Share Posted December 28, 2005 Okay -- maybe it didn't read the hypo closely enough. This seems to solve the problem I saw: "...the column has moved one mile." Link to comment Share on other sites More sharing options...
zappsweden Posted December 28, 2005 Share Posted December 28, 2005 This problem seems quite strange to me. You ask "How far has the soldier travelled?" and the answer is "2.4 miles per HOUR he has to travel" ??? It is like asking how much a companis total sales was during a year and and answering "50 bucks per UNIT"... My Answer: "The problem has no solution". Rambo, will u give me a prize for being the first one with the correct answer? [ December 28, 2005, 07:05 AM: Message edited by: zappsweden ] Link to comment Share on other sites More sharing options...
Desert Dave Posted December 28, 2005 Share Posted December 28, 2005 Look patethic chump are you for real? You need to assume 1 hour to know the... etc, etc. Well gee, I remove me ball-hat and cringe. This is awful Harsh lingo Toward a Cat who has Entered-in and engaged the game place With really wry play On 'is stage name. See, he has deliberately MIS spelled Beginner's (... luck?) to be... beginer's! Now, THAT is cool laconic ironic, Should you ask me. A kind of... zonked Zen technique So to show He's no flame-for-fame Ego! Somewhere! In that clang lango Essayed thus far, Is a riddle, Should you be able To decipher through, Very-end to almost-middle. Can you do it - K-worth! So, la lo, and besides, The world ain't enny 2+2= 4? How improbably poor! Math formula. Rather, a kaleid O! scope! :cool: EVERYTHING changes a mighty TINY fraction Each & every nano instant, Including some straggling gaggle Of drummed into action soldiers, so To arrive at approximate answer, You might do - one of two things: 1) Buy a ticket, but! DO NOT board 'at milky-way train Ol' Einstein is YET riding around in, Here to Eternity, rather Stand beside the station tracks And wonder if it will EVER Come - parallax, again. OR, 2) Go ask Alice. I think she'll know. That LOGIC & PROPORTION Have long, long ago Fallen all sloppy dead... AND The White Knight's hop Scotching - awk!ward backward! So... remember - what the Dormouse said, Feed your head! A child's sing-song story, About not hot nor, too cold "porridge Of... perspective," instead. (... then you were a Child, you knew that ALL them adults were Cuckoo, you forgot this elemental school Truth... so soon? :confused: ) Link to comment Share on other sites More sharing options...
jon_j_rambo Posted December 28, 2005 Author Share Posted December 28, 2005 Last time a riddle such as this caused such a fuss was a man named Samson. @Zapp --- "You stupid or something?" -addressed to Forrest Gump. Link to comment Share on other sites More sharing options...
nemo Posted December 28, 2005 Share Posted December 28, 2005 For a more comprehensive solution, please consult the following link: http://mathforum.org/library/drmath/view/52915.html Link to comment Share on other sites More sharing options...
willebra Posted December 28, 2005 Share Posted December 28, 2005 The example at mathforum is good, but I think they make an error (that does not affect the result) in the solution that includes the hypothetical changing of the speed of the system. (If they impose a speed of -u on the system, the officer will be walking to the back of the column at v+u and returning at v-u, and not the opposite. But this doesn't affect the end result.) Link to comment Share on other sites More sharing options...
beginner's luck Posted December 28, 2005 Share Posted December 28, 2005 Either way Kuniworth reasoning/answer is incorrect.So given this new development JJ/KUNI what do you have to say. BTW So I am a " patethic chump " cause I dont agree with your answer Kuniworth. :eek: Or is it that I found out the connection between you and JJ Rambo? :confused: Amazingly others have stated a different solution or disagree with yours but they aren't chumps! So that leaves me to concluded that the real reason is the KUNI-JJ thing. Please dont get to upset about being found out. I'm sure it was going to come out anyway when SC2 came out. When JJ would lose to Kuniworth just so KUNI "looked" like he was a master. Sadly I have seen this many man times before when a new game came out. It makes my heart heavy knowing the stress of waiting for SC2 has driven someone to this extreme. Lets all try to help this poor soul and bring peace to his chaotic world. Link to comment Share on other sites More sharing options...
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