Okay, so this is my first time organizing a challenge, I named it as SteemYourBrain Challenge.
The type of challenges appearing in the series are targeted to mind twist fellow steemians :)
Enjoy~
SteemYourBrain #1
You are running on a 1km track, you can run as slow as you like on the first 500m with v1.
At the moment you passed the 500m point, suddenly you decided to increase your average speed to finish the whole 1km track to 2v1.
Now the question is, how fast should you run now (v2):)?
Hint #1: For anyone getting weird answers, is this a mathematically possible method? How about physically?^^
The first one who came up with the correct answer will be awarded for 2 SBD!
However the first one who can explain the whole question/answer/the tricky part can get 3 SBD!
Do not edit your answers, comment again for modification
If you like my post, please upvote, resteem it and follow me and I will follow you back ^^"
There are also other interesting posts in my blog, in which you may find them interesting as well:)!!
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Related tags: @math-trail
I guess I should be running the next 500 m with a speed 3V1 so as to achieve 2V1 average speed in the entire 1km
Thx for answering!! The correct answer will be disclosed after payout :)
@kenchung @doughtaker @rycharde @dreamzchm
Seems like time is running short and no one actually discovered the tricky part, I have included new hints in the post.
Please check :)
I've got a guess for the tricky part. But first I need to compose the math portion of the answer. Look for another comment from me in the next few hours.
Ok, here's my 2nd attempt, I quickly gave up on trying to do a fallacious 1=2 proof so I'm just going throw my creative idea out here and see what sticks...
Let t1 = the time required to run the first 500m at velocity v1.
Let t2 = the time required to run the second 500m.
Thus the total time to run the track is t1 + t2.
v1 * t1 = 500m
2v1 * (t1 + t2) = 1000m
Therefore, 2v1 * t1 = 1000m = 2v1 * (t1 + t2)
t1 = t1 + t2 --> t2 = 0
So it takes zero time to run the second 500m, which seems to imply one must run at infinite speed. After all, the runner needs to be at the 500m mark and the 1000m mark at the same time... wait a second!
Is there a place on Earth (the physical hint) where it is possible for the 500m mark and 1000m mark to be in the same spot (or apart by less than the width of a runner's feet)? Indeed there is. There are actually two different regions on Earth where this can happen: near the North Pole, and near the South Pole.
The runner and the track should both start at a point that is [500/(2*pi)] m away from either the North Pole or South Pole. Then runner and track must both go either straight due west or straight due east. The track itself is just a simple line or a designated thin strip and as long as the runner's foot touches the track it should be considered on-track.
Once the runner has run straight due west or due east for 500m, he/she will have also completed a full circle of radius [500/(2* pi)]m and arrived back at their starting point. This starting point also happens to be where the 1000m mark (and end of the track) is. Which means that once the runner has covered 500m, they have also reached the 1000m mark at that very same moment. That satisfies the t2 = 0 requirement.
Therefore, the velocity the runner needs to run at is simply v1 the whole time. So, v2 = v1, and the runner is going both straight due west (or due east) and running a circle near one of Earth's poles at the same time.
wow I'm impressed! here is a 100% upvote :)
anyway, I do not agree that running straight west/east is running "straight". By straight, I mean a straight line define on the 3D Cartesian space, but not a straight line defined on a sphere's surface ^^
I'm completely happy with things getting deeper :)
Nice try!
Thanks. Now I'm really curious to find out what your unconventional, "outside the box" solution is.
:) go find out!
Thanks, but let me get this straight:
v2 is the average speed for the whole 1km, or is it the speed in the 2nd 500m?
Thanks!
oops, made some mistakes in the question, edited!
I just found your contest today.
So I'm a bit late to win.
But I still had fun playing with your little puzzle.
You STEEMED my brain. This is a fun challenge.
Intuitively, my guess would be something like v2 = 4v1
But to my surprise, when I played with the math,
here's what I came up with.
D = 1 km
T1 = Time to run the 1st half
T2 = Time to run the 2nd half
T = T1 + T2 = Total Time
V1 = (D/2)(1/T1) = Velocity in the 1st half
2V1 = 2(D/2)(1/T1) = D/T = Total average Velocity
divide both sides by D --> 2/2(1/T1) = 1/T
T = T1
But T = T1 + T2
Thus T2 = 0
V2 = (D/2)(1/T2) = D/0 = infinite velocity
Haha surprised?
Thanks for playing btw ^_^
answers will be disclosed after payout :)
btw you can go deeper :)
That was the Newtonian view, since we can expect that our runner can never run at relativistic velocities.
Considering Einstein, we need to consider that the speed of light is the ultimate speed limit, as far as we have experimentally detected. (Tachyons are still undetected.)
So the runner can never run at the speed of light due to the infinite gain in mass. But even if she could run that fast it wouldn't be enough. To a stationary observer, if she ran at the speed of light, it would take her 1/600,000 sec which is too slow and she would cause a black hole and suck inside the observer and the track and the earth, etc.
At very high velocities, time does some weird things, it slows down. Even our GPS satellites run at a slower time than the earth, in agreement with Einstein's equations.
If she could run faster than the speed of light, time might even go in reverse, which causes some really weird paradoxes.
Serious shit physicist here man ^^!!
Lets wait for 2 more days and I will disclose my answer here :)!!
You haven't yet disclosed it, have you? This is also my answer...
Boy, I love gentlebot
he should run with v2=2v1.
You have the answer in the question just LOOK FOR IT. :)
Nice trick though.
not quite :)
v2 is the speed you need to run from the 500m point but 2v1 is the average speed for the whole track
v2 = infinity
1000 / (2v1) = 500/v1 + 500/v2
"Do not edit your answers, comment again for modification"
anyway, nice try :)
answer will be disclosed upon payout ^^
oh sor9ry lor :(
but I didn't change my final answer ^^ I just added the steps
According to last edit timestamp, you are still the first one to give v2=Inf <-- this answer:)
Explanation: divide the track 500+500, first is V1, second is V2, entire is 2v1, so average 2v1. Now (V1+V2)/2=2V1, solving this, V2=2*2v1-v1=3v1. That's it.
But would it be little bit too easy for 5 SBD ^^?
yay hopefully it's correct :)
Keep in mind the SteemYourBrian series is a tricky mind twisting series :)
oh but I can't get the tricky part 🤔 am I doing the question wrongly...
<the assignmentThanks for your question, but I dun really understand it h
What is your (x,y) plane referring to? The Argand plane?
If so, the answer should be a vector pointing to (0,1), since on the Argand plane x-axis refers to the real part, y-axis refers to the imaginary part.>
haha, I used tricky jedi mind twists (!) to check the same book :-o
Yes, that's what I got. Didn't post coz thought was another error. v2 is infinite.
Hey @happychau123 you're much better at solving problems! hahaha
Thanks ^^
But I wanted the ability to mathematically solving problems more than practically like I did in your recent challenges :(
you are a scientist right ^^?
there is something more than this :)
you and ken are both very close to the real tricky part ^^
Any thoughts ^^? @kenchung @doughtaker @rycharde @dreamzchm @point @doughtaker
Hints #2: It is physics related as well:)
The point is: Is it possible to mathematically perform this action? And physically:)?
Upvoted, followed, and resteemed.
I think I know how to solve this but I have one question and it relates to the track. Is the 1 km track a defined straight-line track like in your drawing, or can I choose to "make my own 1 km track"?
Thanks for your question :)
It is a straight track as I drawn
The tricky point is not on the track for thua question^^
I will add this to the question thank you ~
Thanks for clarifying. I'll have to rework this, as I thought the trick was the track. Based on that assumption I was originally going to say
complete with work to mathematically back up that claim. But I guess that's not the solution you wanted.
hahaha XDDD
lemme clarify again, the track is a super-normal-non-moving-straight track :)
Now, let me ask this. Why would I run if I am imagining everything?
hmmm... lemme delete the word...
you guys are more creative then me :)
this is not a word game :(
the tricky part is more on the scientific side!
Oh. I'm too late.
Oops, you can still try:)
I will disclose the answer in 10hours;
Well, here's my work:
if I take 1 hour to travel 500m, then I'm travelling 500m/hour
In order to travel 1k/hour, I'll have to have travelled the last 500m already as well
Since that's impossible, I'll instead take an infinite amount of time to travel the first 500m. Which means it should be pretty simple to go the last 500m in the same amount of time, since infinity divided by 2 is still infinity.