Inspired by DoctorHelvetica's cool puzzle, I thought I give you guys and gals another one:
Donnie is standing on a train track, while a train is coming towards him with a constant speed of 60 km/h, without the ability to stop. Unfortunately donnie lives in a 2D world, and doesnt have the leg power to jump over the train. So he has 2 options: 1) Run away from the train and get hit exactly at point A. 2) Just get it over with and run towards the train, and get hit at point B.
The little vertical beams on the track are of equal distance from eachother.
Question: what is the average running speed of Donnie?
You know that if Donnie runs to the left, he will be at one notch to the right of A then when the train is at B. You also know that in the time he runs one notch and reaches A, the train has moved 3 notches to meet him there. So the train is on average 3 times as fast as Donnie. 60kmph / 3 = 20 kmph.
You know that if Donnie runs to the left, he will be at one notch to the right of A then when the train is at B. You also know that in the time he runs one notch and reaches A, the train has moved 3 notches to meet him there. So the train is on average 3 times as fast as Donnie. 60kmph / 3 = 20 kmph.
edit: usually i'd clean up my solutions but i wanted to leave my thought process this time lol
ok let's see.. so the key here obviously is that from donnie's current position, A is twice as far as B wait we'd also have to know how far away the train is, no? if not.. let's assume it's 60km away from pt b, so donnie can run 1 bar in 1hr so he can make it to A in 2hr, so the train is 120km away from pt a?.. so each bar is 20km so donnie runs at 20km/h? that's pretty fast but not unreasonable if ur a world class runner. hmm now does the distance of the train matter.. if not.. let's assume it's Xkm away from pt b, so donnie can run 1 bar in 1time so he can make it to A in 2time, so the train is 2Xkm away from pt a?.. so donnie can travel 2 bars in the time it takes the train to travel 2X, and we also know X is 3 bars so donnie travels 2 bars in the time it takes the train to travel 6, aka 20kmh
On November 01 2009 01:49 Railxp wrote: this smells suspiciously like a homework thread.
The last homework assignment I turned in was quite a few years ago.
And I wish that I used to get homework like this.
true that homework rarely makes you actually *think* in the same way and i'm like graduating from uni soon sad
here's a random question, best asked in person so you can see people's thought process.. so if you answer it, write down your thoughts as you think about it =)
you have 2 houses on unequal ground connected by rope. The rope also has some "sag" in it, i.e. it's not tight. the vertical distance/sag height from the lowest point of the rope to the highest point is ten meters. the rope is attached at a fixed point on each house, points A and B, such that A and B are perfectly level with each other. the rope is 20m in length. how far apart are the houses?
Why even bother saying they're on unequal ground if A and B are level?
You can solve these problems by dividing the rope up into segments and solving the differential equation, not sure if it falls into Newton's concatenary(?) object, which would be high point to low point.
On November 01 2009 09:26 igotmyown wrote: Why even bother saying they're on unequal ground if A and B are level?
You can solve these problems by dividing the rope up into segments and solving the differential equation, not sure if it falls into Newton's concatenary(?) object, which would be high point to low point.
please continue writing your thoughts if you're intent on solving the problem =)
On November 01 2009 01:49 Railxp wrote: this smells suspiciously like a homework thread.
The last homework assignment I turned in was quite a few years ago.
And I wish that I used to get homework like this.
true that homework rarely makes you actually *think* in the same way and i'm like graduating from uni soon sad
here's a random question, best asked in person so you can see people's thought process.. so if you answer it, write down your thoughts as you think about it =)
you have 2 houses on unequal ground connected by rope. The rope also has some "sag" in it, i.e. it's not tight. the vertical distance/sag height from the lowest point of the rope to the highest point is ten meters. the rope is attached at a fixed point on each house, points A and B, such that A and B are perfectly level with each other. the rope is 20m in length. how far apart are the houses?
On November 01 2009 01:49 Railxp wrote: this smells suspiciously like a homework thread.
The last homework assignment I turned in was quite a few years ago.
And I wish that I used to get homework like this.
true that homework rarely makes you actually *think* in the same way and i'm like graduating from uni soon sad
here's a random question, best asked in person so you can see people's thought process.. so if you answer it, write down your thoughts as you think about it =)
you have 2 houses on unequal ground connected by rope. The rope also has some "sag" in it, i.e. it's not tight. the vertical distance/sag height from the lowest point of the rope to the highest point is ten meters. the rope is attached at a fixed point on each house, points A and B, such that A and B are perfectly level with each other. the rope is 20m in length. how far apart are the houses?
ding ding ding =) it was a (paraphrased) question from one cognitive science book; this particular chapter was talking about different approaches to problems/proofs.
i must say at first i fell into the same mindset as igotmyown.. i was asked this in person (as originally intended) so i was like "yeah, i can do this but i'd need a paper/pen" and just kinda got stuck at that mindset.. then after a brief chat i was forced to think about it some more and approach it differently "so you can't solve this?" "i can.. but i need a pen/paper" "so you can't right now?" "no" "are you sure?" "well..Oh."
it's a fun little exercise to see how people approach a problem.. which is why it doesn't work as well over the internet
The train needs to travel 8 "ticks" to point A. Donnie needs to run 2 ticks. Ego, the train would be 4 times as fast.....his average is 15 km/h.
If he ran to point B, he'd cover 1 tick in the time that the train would travel 5. Thus, the train is 5 times as fast. That'd be 12 km/h.
If he runs 1 notch left when the train reaches point B, he averages 12 km/h on that interval. The next interval he would have to speed up to 20 km/h to reach A (as people have noted). His average is less than 20 and is 15. He doesn't run at 20 km/h the whole time.
(un)luckily I'm more interested in how to solve problems than solving them, and paying attention to the numbers is too much unnecessary hard work (usually). That's why god invented engineers, to think about those boundary condition problems.
On November 01 2009 10:48 N.geNuity wrote: Forgive my ignorance, but for the train:
The train needs to travel 8 "ticks" to point A. Donnie needs to run 2 ticks. Ego, the train would be 4 times as fast.....his average is 15 km/h.
If he ran to point B, he'd cover 1 tick in the time that the train would travel 5. Thus, the train is 5 times as fast. That'd be 12 km/h.
If he runs 1 notch left when the train reaches point B, he averages 12 km/h on that interval. The next interval he would have to speed up to 20 km/h to reach A (as people have noted). His average is less than 20 and is 15. He doesn't run at 20 km/h the whole time.
You can't assume that the train takes 8 ticks to reach b though (if it did the problem would be inconsistent) x = distance in kilometers between 2 ticks, y distance in kilometers between train and B, t = time in hours it takes donnie to run x kilometers.
then
y/t = 60 (y+3x)/2t = 60 -> x/t = 20
at least that's how I did it ~~
For jeejes problem it felt as if having access to pen and paper was almost cheating, (draw picture and done), kind of nice trick otherwise though :p
On November 01 2009 10:48 N.geNuity wrote: Forgive my ignorance, but for the train:
The train needs to travel 8 "ticks" to point A. Donnie needs to run 2 ticks. Ego, the train would be 4 times as fast.....his average is 15 km/h.
If he ran to point B, he'd cover 1 tick in the time that the train would travel 5. Thus, the train is 5 times as fast. That'd be 12 km/h.
If he runs 1 notch left when the train reaches point B, he averages 12 km/h on that interval. The next interval he would have to speed up to 20 km/h to reach A (as people have noted). His average is less than 20 and is 15. He doesn't run at 20 km/h the whole time.
You can't assume that the train takes 8 ticks to reach b though (if it did the problem would be inconsistent) x = distance in kilometers between 2 ticks, y distance in kilometers between train and B, t = time in hours it takes donnie to run x kilometers.
then
y/t = 60 (y+3x)/2t = 60 -> x/t = 20
at least that's how I did it ~~
For jeejes problem it felt as if having access to pen and paper was almost cheating, (draw picture and done), kind of nice trick otherwise though :p
It's not 8 ticks to point B for the train. It is 5. And why separate "y" as a separate variable than x? y is five ticks, and every tick is the same length.....keep x as the distance between two ticks and t as the time donnie takes to run x.
y = 5x (5x + 3x)/2t = 60 4x/t=60 x/t=15
Let's say each tick is 60 km. The train is 8 ticks away, so it is 480 km away. It takes the train 8 hours to cover 480 km.
Donnie is 2 ticks away. He is 120 km away. In 8 hours, he must travel 120 km.
120km/8h = 15km/h
IF donnie averaged 20 km/h, he would not get hit at point A. He would have travel 160km, well past point A.
On November 01 2009 10:48 N.geNuity wrote: Forgive my ignorance, but for the train:
The train needs to travel 8 "ticks" to point A. Donnie needs to run 2 ticks. Ego, the train would be 4 times as fast.....his average is 15 km/h.
If he ran to point B, he'd cover 1 tick in the time that the train would travel 5. Thus, the train is 5 times as fast. That'd be 12 km/h.
If he runs 1 notch left when the train reaches point B, he averages 12 km/h on that interval. The next interval he would have to speed up to 20 km/h to reach A (as people have noted). His average is less than 20 and is 15. He doesn't run at 20 km/h the whole time.
You can't assume that the train takes 8 ticks to reach b though (if it did the problem would be inconsistent) x = distance in kilometers between 2 ticks, y distance in kilometers between train and B, t = time in hours it takes donnie to run x kilometers.
then
y/t = 60 (y+3x)/2t = 60 -> x/t = 20
at least that's how I did it ~~
For jeejes problem it felt as if having access to pen and paper was almost cheating, (draw picture and done), kind of nice trick otherwise though :p
It's not 8 ticks to point B for the train. It is 5. And why separate "y" as a separate variable than x? y is five ticks, and every tick is the same length.....keep x as the distance between two ticks and t as the time donnie takes to run x.
y = 5x (5x + 3x)/2t = 60 4x/t=60 x/t=15
Let's say each tick is 60 km. The train is 8 ticks away, so it is 480 km away. It takes the train 8 hours to cover 480 km.
Donnie is 2 ticks away. He is 120 km away. In 8 hours, he must travel 120 km.
120km/8h = 15km/h
IF donnie averaged 20 km/h, he would not get hit at point A. He would have travel 160km, well past point A.
Ok so say it's five ticks between train and b then y=5x
EPT![[image loading]](http://i37.tinypic.com/231u0g.jpg)
