4 Revolutionary Riddles
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- čas přidán 4. 05. 2024
- Can you solve these four rotation-related riddles?
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I came across these four physics puzzles over the years in discussions with Neil deGrasse Tyson (riddle 4: which part(s) of a moving train are going backwards with respect to the ground?), Simon Pampena (riddle 2: run around a track twice, the first time slowly, the second time much faster so that the average for the two laps is twice the speed of the first lap). Someone tweeted me a video of the mystery cylinder rolling down the ramp in riddle 1 (sorry I'm not sure who it was). Riddle three about a bicycle going forward or backward when it's bottom peddle is pulled back was brought to me by a number of people and I appreciate all of their help!
Filmed by Raquel Nuno.
Thanks to everyone at the Palais de la Decouverte! I've had this footage for five years and am only finally releasing it now. I wanted to talk about the way grass grows on a spinning turntable but I couldn't locate the footage...
1. Theres a hamster in the cylinder
Hahaha
The poor hamsuke!
Well, thanks for spoiling it fude what the hell
It is honey and a metal ball outrageous
Great one 🤪
'I want you to think about running two laps of this track'. No.
Harvey Fuller 😂
You can walk the first lap, as long as you run at the speed of light the second one.
LukeSumIpsePatremTe or I could do neither and not run at all
Yeah that's too close to exercise
That would work if he was talking about velocity and not speed.
Riddle 4: Lowest point of the inner edge of the wheel.
Any train wheel have an inner edge reaching further out than the surface of which the train rides on the track. Since the surface is connected to the track it will only move forward being rotated forwards until it meets the track again. Here it stands put while in contact with the track. But this inner edge is larger and therefore, the lowest point of that edge moves backwards when the surface stands still, as it rotates with the wheel.
Now, that's only a part of the wheel, a part of the rotation. So I'm not sure it can be said that it's always moving backwards. But a part of the inner edge always does.
The train wheel is a classic problem. Known mathematically as far as I remember as an epicycloid; the wheel flanges exhibit a backward rotation about the cusp of the enclosed cycloid described by the fixed point on the part of the wheel immediately inboard of the cusp. The bike problem is easy if you remember that pulling back on the bike's lower crank in the vertical position propels the bike fwd only temporarily for no more than 1/4 turn of the crank.
Running problem super easy v2=3*v1.
The intermittent movement of the outer cylinder is a more abstruse problem because we are forced to speculate on the inner geometry of the device. So I suspect that there is more than one mechanism which would do this. In any case it must involve an inner moving mass. One principle at work is that the combined centre of mass will move under the action of the constant accelerating force (m1+m2)*g*sin(alpha) neglecting friction, where alpha is the angle of inclination of the plane above the horizontaal. Although it may appear to be the case, motion of the system up the slope is not possible, even though this may appear to the eye to be happening.
I thought that two, but that would only be true if train wasn't moving despite the wheels turning. But as train move forward the point should remain same or move forward with train. Now, I'm gonna watch his answers bcz I'm few years late.
@@crustyoldfart I'd agree the running problem is easy but I think your answer is wrong (for the usual interpretation of average speed). If the length of the track is d the time taken for your first lap is t then your initial speed is d/t. In order to get an average speed of 2d/t while running a distance of 2d then your total time needs to be t, so the time taken to run the second lap must be 0.
@@rhysbyt I confess that I let my 30 odd years experience as an engineer trump the methods my high school physics teacher would have taught. In engineering we are constantly searching for solutions to real world problems, invariably without having complete information. In physics, there is a tendency to believe that if you can invent elegant mathematical propositions then nature is likely to follow the predictions of the mathematics.
In the running track problem the physicist might argue as follows :
V1 = d/t1 ; V2 = d/t2 ; Va = 2*d/(t1+t2); Va = 2*V1 ; which -> 2*V1*t1/( t1 + t2 ) = 2*V1 -> t2 = 0
Whereas ( V1+n*V1 )/2 = 2* V1 -> n = 3.
An empiricist might be forgiven for believing that t2 = 0 is absurd, while n=3 seems eminently possible.
@@crustyoldfart "Running problem super easy v2=3*v1." - Wrong. The answer is NEVER or when v2 is infinity.
"The earth is after all a giant turn table". This is going to be some good feed for flat earth society :) Be careful, you just could become their prophet lol
Stop denying facts
@@5446isnotmynumber wdym
normal customer guy dude- Whereyoumacha's do you have the whatchumacalleit's at? not so normal employee- Wechamahasit's on isle 4 1/2 right between isles 5 and 4 havemachagood day duh eh have a ma cha good day oh yeah it would all be a time pair of socks
Or for DJ's. ~rolls eyes~
1. Non newtonian fluid like corn starch
2. Not move
3. 3 times the original speed
4. The wheels on the train
2. the bike moves backwards. There's a backwards net reaction force acting on the bike - it's going to move backwards
Sebastian Morris nope go test it
For Q3, Its not 3 times. It is infinitely fast.
For Q2, I feel like it will move front. Net force is not zero since the pedals have larger torque relatively to ground
Sebastian Morris in order for it to move backward the rear tire would have to lose traction. The harder you pull the string the harder the bike wants to go forward.
the lap would have to be done instantaneously
3. That girl REALLY toned up from just 1 lap of walking !!
That was a good pump if I've ever seen one
I didn’t see any girl when he started talking about number 3. did the video get edited?
Lol
Yeah doctors hate her
@@MrCaptainTea he meant the girl from 2:24 that ''transformed'' into the guy at 2:32 , so basically a bad joke xD
1. A long weight attracted to the end with elastic bands. When it rolls, the band's twist. The weight eventually flips due to the rotational force of the mystery cylinder and the process repeats making it pause once more.
1 : pendulum of some kind don't exactly know how but id say it's a pendulum on something that makes it stay straight and start waving once the cilinder rolls.
2 : forwards, for a very small distance, then immobilise, then backwards. at first you pull, it turns the pedal and the bike mooves forward, until the half the distance that was pulled for the pedal to go at 90° then it stops, then backwards as you continue pulling.
3 : I was annoyed at the precisions of the calculations but it's not simply "three times as fast" as what stays the same is the distance and not time.
4 : a part of the wheel always goes backwards compared to the ground so that the train can moove forwards.
Let's hope i was right :D
1.C
2.C
3.C
4.C
wait this doesn't feel right
Sen3.14 no ur good. Cs get degrees
Wait. Too many Cs in a row. 5 HAS to be A. Definitely.
well the answer for every question is b
Good luck running at the speed of light!
Sen3.14 see?
"the earth is after all just like a big turntable" Oh my Oh my, flat earthers incoming
Oof
The same got into my mind 😁
Same
My thoughts exactly... Science channel undone in one sentence.. lol
He misspelled turngloble
1. There's a hamster inside the cylinder who's running in the other direction.
2. The two forces will cancel each other, and the bike will simply implode.
3. Unfortunately, I don't have a joke for this one.
4. People falling of the train.
4 = smoke
3. u have to teleport urself
3. V1=0, therefore 2V1=0, so you just stand there for the rest of time.
(V1+V2)/2 = 2(V1)
V1+V2 = 4 V1
V2 = 4 V1 - V1
V2 = 3 V1
Thus, your speed on the 2nd lap should be thrice as fast as your speed on the first lap.
Actually, you have to find harmonic mean, not arithmetic one
1 A pendulum attached to the centre of the tube with a elastic and a friction bearing
2 The bike moves slightly forward, but if you put it in a really high gear you can beat the friction on the wheel and pull it back with a strong pull.
3 3V1
4 The bottom flange of the wheels below the contact point
You’ve got the third one wrong. It’s impossible.
Imagine a dude who runs 1 km in 1 hour. His average speed is 1 km/h. There’s another dude who runs 2 km in 1 hour. His average speed is 2 km/h. So basically to double the average speed of the first guy from 1 km/h to 2 km/h you would need to travel an additional kilometre in 0 seconds, which is obviously not possible.
@@yajaman no wait. To calculate the average speed of both laps combined, yo need to calculate their mean, wich is adding them together and divide them by 2 (bc they are 2 values). So, knowing the mean has to be 2V, we can go backwards to find the velocity of the sencond lap, that is 2V*2=4 and 4-1V=3V
@@yajaman he said 3*v1 not 2V1
first lap = 5km/h
second lap = 15km/h
average = 10km/h which is twice the first lap.
you need to run 3times your 1st lap speed, in order the avg to be twice.
@@yajaman Hi, that is true. I saw the second video afterwards but I honestly got it wrong so am leaving it up. It took me a while to think it through properly for some reason.
@@brucedeo1981 no... you can not measure an average speed this way: imagine the lap has a length of a kilometer, by your way the first lap would be run in 12min, the second one in 4min, which makes 16min in total, what makes an average speed of 7.5km/h, not 10km/h... the only possible way is infinite speed
1:00 Who else flinched about how the tape just rolled away now he has to go get it 403942 miles away lol
i dont think thats how you use the word flinch
Tay • 19 years ago Ahaha, someone else who got recommended this video. Good day to you sir
@@Yametay ok
@@decaymak4400 hello there
1) 42
2) 42
3) 42
4) 42
1) 69
2) 69
3) 69
4) 69
The universe has spoken.
1) 420
2) 420
3) 420
4) 420
DO NOT ADD ANY LIKES, THIS IS PERFECTION
1) 420
2) 420
3) 420
4) 420
Derek always uses big words for simple terms and I love it
1. A rubber band mechanism Inside the cylinder
2. Bicycle is not move
3. V2 is 3×V1
4. I am not sure, but it's wheel of train
when the answers are edited
3 is actually not 3xV1, it’s 0 seconds. You need to do it instantaneously
@@ihexrt l agree 👍
@@ihexrt would you mind explaining why this is
@@-TheShaun- To do a 400m lap at 2km/hr would be 12mins, to do another lap is 800m total distance and the average speed to achieve would be 4km/hr. 800m at 4km/hr would take 12mins, you have already taken 12mins to do the first lap, therefore you need to do the second lap instantaneously......
That is by far the most intellectual click bait I have ever fallen for
@Range Man , It is pretty normal not to see exhale in winter here. It just requires weather were air humidity is low enough that exhale doesn't have water molecules condensation when hot wetter air becomes colder. If you happen to life next to a sea then winter humidity is likely to be always too high so you might think exhale should be always visible in cold weather.
@Range Man , One common case of visible exhale is basically fog warming. Fog warms because air collects heat and water from body. When you exhale it temperature drops which leads to higher relative humidity. If humidity rises high enough you will get visible fog. But exhale isn't visible always so in some weather conditions humidity doesn't rise high enough to form fog.
Obviously very simple explanation is wrong as there happens more complex processes than just simple cooling. Warm and cold air also mix to each other changing absolute concentrations too.
@Range Man , What tree flocking? I was wondering what you were talking about so I watched the video in question but I could spot which door you were talking about.
But as I watched the video I noticed that you can see exhale water turning to fog effect in an early part of video. There is an early part where camera lens becomes foggy from exhale directly to it from close range.
I'm not connected to the channel. I was just trying to explain you need more than just freezing temperatures to see exhale. I happen to life in a cold climate so I see effects every winter. If you want to go on set conspiracy theory then you probably should blame everyone else using same set as there is many other videos from the same vault.
It sure is lol
1 - Some kind of counterweight, possibly liquid
2 - forward
3 - no velocity can do it. you would have to run it in 0 seconds exactly
4 - the bottom part of the "wheels"
Theodor Dornonville de la Cour no it’s the steam
Hearth what if it’s not a coal powered train
Theodor Dornonville de la Cour if you run three times faster than your first lap your total average speed would be twice the first lap. If you run 1 mph the first and 3 the second your avg would be 2 mph
@@shdbsduskdh7663 THATS ExACTLY WHAT I DID imgur.com/ScQKDd1
I would say the rail
I love how the titles, captions and sound effects on this show make me feel like im watching a science show from SBS or ABC in the early 90s
1. A very dense liquid
2. Come backwards because the force you apply is larger
3. You run at 3 times V1
4. Lower part of wheel of train.
3 is not right
2. It depends on the bike’s gear ratio
4. Lower part of wheel in relation to ground is stationary
1) I guess sand, but any fluid that will turn with the cylinder, until the center of mass is behind the center of rotation, then pour back slowly to the front of the cylinder will do.
2) I don't see why it wouldn't go forward, at least for 1/4 turn of the crankset.*
3) Infinite speed. The trick is that the average speed is not the average of the two speeds (which gives you 3V) but the total distance divided by the total time. If you want twice the average speed over twice the distance, you get a time for the second lap that is zero.
4) This one is actually pretty clever, if I got it right. When rolling without slipping, the slowest a wheel goes with respect to the ground is 0 at the contact point. But train wheels have a flange to maintain it on its rails. The flange is on a larger radius than the contact point, so its trajectory will be a slightly prolate cycloid. That is sneaky, Neil !
*EDIT:
Ok, I could have been wrong with number 2. I guess that depending on the gear and the ratio between the radius of the crankset and that of the wheel, you could be unable to make the bike move forward.
When you look at the pedal of a moving bike from the side, you see a cycloid. In a low gear (a lot of distance traveled per rotation of the crankset), the cycloid will be more curtate, and in a higher gear more prolate (less distance traveled per rotation of the crankset). So if the cycloid is prolate, the bike moves forward a bit, because you are in that "backwards" part of the motion of the pedal. If it is curtate, then it stays still or goes backwards if you pull strong enough. Phew!
Love your thinking for the last one
Thanks, mate !
You're definitely right on #4. The answer can be found on page 13 of a very good mathematics book, Why do Buses Come in Threes? by Rob Eastaway and Jeremy Wyndham. :)
3) Watch the vid again. 2:30 says "the total speeds of the two laps combined". So the answer is 3V1.
Not really. He said "average speed over two laps".
I really appreciate how you integrate scientific language into normal explanations of what's going on (e.g. when you said "the tape rolls, or accelerates, down the ramp"). It's a little thing, but you always make sure people know what's happening while it's happening.
Down the ramp? You mean the incline plane?
@@ryanpowell6033 Incline plane? You mean a plank at a height of 10.052761cm at an angle of 5.02451015 from the ground?
@@donothack To the ground or a tangent to the Earth's rotation at a moment in time? Seems to be many degrees this can be refined. How about where you position the inclined plane along the tangent? If you move the inclined plane far enough away from the central point of the tangent the inclined plane's surface will point directly to the center of the Earth or by moving the inclined plane in the opposite direction will become parallel to another tangent to the Earth's rotation... The tangents are probably gravitational tangents and the rotation doesn't matter. That thinking thing is taking up time again.
Cheers.
I was about to be so mad when he wasn’t gonna give us the answers
but what he didnt
@@olof1382 he did its in a separate video
@@apollochaoz where
@@will-jh7yi In this exact channel???
@@smallone2351 great let me just look through the hundreds of videos
2. I think it depends on the gear ratio.
Above a critical gear ratio, it will stay stationnary, or tilt backwards it we pull too hard. It could also slip on the ground if static coefficient is low enough.
But with a lower gear ratio, it will be able to pull the lower pedal backwards and therefore move the bicycle forward.
I know it seems physically impossible that a backward force pushes the bicycle forward, but the bicycle’s mechanism will act as a pulley, that transfers force through static friction with the ground.
When I pull a weight with the help of a pulley, the weight also moves in the opposite direction than the force I apply.
wrong
So true you blew combustion's mind.
@@GangsterGumbo nah the 5 likes shows hes wrong
@@EazyDuz18 that's only cus you can't see the dislikes
My first thought was, “Wouldn’t it just fall over?”
1. *A hamster*
A deppressed hamster
@@sorsax7226 or may be one made to run in the direction opposite to incline plane...
Drunk hamster
Lazy hamster
2. A hamster
3. A hamster
4. A hamster
1 - Viscous liquid, maybe sand
2 - Disregarding the friction of the tires... If the radius of the pedal wheel is smaller than that of the tires (as it is in this bicycle) then the bicycle goes backwards. If they are equal, the bicycle doesn't move. And if the radius of the pedal wheel is bigger, the bicycle goes forward.
3 - The question is about the average speed of both laps combined. From what I understand, it's the distance of the two laps over the time you took to complete BOTH. Now, imagine that the track is 100m long, if in the first lap you run at 2m/s, you'll finish it in 50 seconds. For your average speed to be v1*2 (4m/s in this example) you would have to finish the 2 laps (200m) in 50 seconds, but you already spent 50s in the first lap alone, which means you'd have to run infinitely fast.
4 - Bottom of the wheels (those lips that prevent the train from derailing). And don't think the coupling rod goes backwards relative to the ground.
Now, thinking over, I might be wrong about question 2. I could be the opposite
Good call on #4, that one had me stumped.
reread his numbers again. and think about it. he said you spend 50s running 100m at 2m/s. Double that for 4m/s, if you are to run a total of 200m at 4m/s it would take you 50s. So you run 100m in 50s, then you want to run the next 100m at a speed that would give you a total of 50s. You have already used up your 50s doing the first 100m! So you have no time left to run the second 100m
Everything is regard to friction. There are only 2 lateral forces that will make the bike move forwards,backwards or stationery.
1st is the pull force (tension)
2nd is the friction
If there is no friction then regardless the radius etc. the bike will NOT move
Hitendra Kishanchandani great point
1. There’s honey with a small metal ball, or maybe just some slow flowing liquid in the cylinder.
2. The bike will go backwards if you keep your balance and pull with enough force.
3. You would have to run the second lap at 3V1.
4. The bottom of one of the wheels on the train.
For Q3, I did the same mistake. Average Speed is not the mean of the two speeds. It is the total distance over the total time.
my guesses:
1. there's a smaller heavy cylinder inside of the mystery one
2. it will move forward until its stopped by the string
3. if my algebra is correct, the second lap has to be 3 times as fast as the first
4. the only thing I could think of is the bottom of the wheels because they don't have infinite traction and will probably slide backwards a little
I would counter that the cylinder just has many weighted beads inside-like metal BB's. Also 'at any given time' is his specification for the backwards-moving part on the train. The wheels on the train would not necessarily have to slip to be moving backwards relative to the ground: assuming the train has a forwards velocity, the bottom half of each wheel on the train is always moving backwards relative to both the ground and the train's velocity. If you think about it, if you divided one of the wheels' circumference into 1-inch sections, every 1-inch section has to first move forward and then, upon reaching the bottom half of the wheel's rotation, move backwards relative to the ground in order to come back up to the top of the wheel and complete a full rotation.
For 3, you did (V1+3V1)/2=2V2. The problem is that if you ran at 3V1, you would finish the lap in a third of the time it took you in lap 1, so your average speed would actually be: (V1+(3V1/3))/2=(V1+V1)/2=V1. Seems to be impossible, because if your speed increases by XV1, the time it takes to do the lap is 1/X the time it took to do lap1, so it would be (XV1/X)=V1 every time…
I didnt use algebra i averaged. I started with 4+1 and then divided by 2 to get an average and it equalled 2 as he said meaning it would be 4x faster
I think the answer to 3 is the second lap would have to be instantaneous taking 0.0 seconds. If the first lap takes 10 mins and the second lap takes 3.33 the total time is 13.33 and the average time is 6.66, if you run the second instantly the total time would remain 10 mins which would average out to 5 mins/lap or twice as fast as V1, I may be completely wrong though.
1.) Inside the Jar: A really tiny bicycle with a string attached the pedal 2.) String on Bicycle: If you pull it like you are starting a lawnmower its going to come right at you3.) Track: Math and running is hard 4.) Moving Backwards Train: The chugga chugga choo choo sound
Only if it goes at less than the speed of sound tho. ;)
orbitarc Doesn't matter if you are going slower or faster. Doppler Effect takes place. There are articles explaining it and I don't know that much about it.
Maybe look it up, then you will also understand what I mean. It's rather related.
orbitarc The real answers right here
Not a part of train.
1)Black magic
2)It starts flying
3)V2
4)Something on the bottom
teach us your ways
u should start Verytasium 69
4) I think is the rail road
Deniz Can Gezgin haha great
Here are some answers:
1. There are some variations how it could be created, but my guess is some polygonal rod is in it. First the rod is slowly pulled up by the rotation of the cylinder, and when the rod turn on its next side, the cylinder can go faster a bit because of the weight change. Then lifting the rod again slows the cylinder. (Or you could put a gyroscope in it with a small motor in it and a micro controller to release the cylinder periodically, but that's probably NOT the answer)
2. As others have mentioned, it really depends on the gears. The front is on the smaller one, but the rear is not on its biggest gear, so that's what I think the trick is. On the biggest rear gear, the bike would move forward (That's when you have to turn it a lot to move a little). In this configuration the trick answer is: For a short time, the bike would be stationary, while its rear wheel is rotating forward (sliding on the ground) and the pedal is moving backwards. It's because the pedal would go just as much backwards as the bike would like to go forwards. (On higher gear it would move forward without sliding, on smaller gear it would slide backwards)
3. You'd have to do the second lap under zero seconds. Average velocity = total distance / total time. If you do the first lap under 1 second, you'd have 1 lap/sec as average speed. Then the goal for the 2 lap together is 2 lap/sec. That would mean you have to the whole 2 lap under 1 second, but you already used that second for the first lap. (s/t_1) * 2 = (2*s)/(t_1 + X) -> s * 2 * (t_1 + X) = (2*s) * t_1 -> 2s*t_1 + 2s*X = 2s*t_1 -> 2s*X = 0 -> (if the lap's lenght is not 0) -> X = 0
4. The trains' wheels doesn't meet the rails at their lowest point because of stabilization. That's the part that makes sure that the train doesn't get derailed. The lower parts of the wheels move backwards. Imagine rotating an umbrella on the table. The lower parts of the umbrella go backwards.
for number 3:
jit just run it 3x faster. first lap= 2, means the second one needs to be 6. if the first lap= 3, the second one needs to be 9. if the first= 17, the second one needs to be 51
@@Heqnry But I believe that the avg velocity is the total distance by total time (as the first guy said). It shouldn't (V1 +V2)/2 but total distance (2D) divided by total time (T1 + T2) that should be equal to V1
@@Heqnrythat is incorrect. You are making the assumption that the time it takes to complete both laps is equal, but remember, as you speed up, it takes you less time to complete each lap, so it impacts the average velocity less. Velocity is the change in position over time. Let’s say for the first lap you run 100 meters in 25 seconds. So your average velocity over the interval would be 100 meters/25 seconds or 4 meters / second. Now to double this velocity you would need to get your average velocity to 8 meters/second. So let’s write a new equation to model your average velocity between the two laps. Because the length of the track is the same for both cases, the distance travelled is 100 for both laps, so change in position is equal to = 100+100. Now the change in time is given by the time it takes you to complete both laps, so (25+new time). Given the equation for velocity is change in position over change in time, we can see that the average velocity for the interval is 200/(25+new time). But there’s an issue here, the velocity you need to reach is 8 meters / second, or 200 meters in 25 seconds. Meaning if average velocity is equal to 200 meters / (25 + new time) and the target velocity is 200 / 25, new time has to be equal to 0. Meaning you would have to travel instantaneously, so it is impossible.
1) Cylender has a length wise partition. One out of two part is empty and anather is half filled with liquid.
2) Cycle will go back.(If you pull rope in sitted position, ie below the paddel.)
3) 3xV1
4) Periphery of the wheel.
3v1 is wrong. no matter how fast you go, you have to teleport. i used to know the math on this, but you'd have to google it for an explanation now, sorry.
3V1 isn't right.
You have to remember that average speed isn't (V1+V2)/2, it is actually
(D1+D2)/(T1+T2) that is the total distance traveled divided by the total time taken.
It turns out if you travel at V1 for one lap, that brings your entire average to really low so you can never reach average speed 2V1.
This is like how, if you travel at a slow speed, then you travel longer. Therefore that speed has higher "weight" in your average, and it prevents you from ever reaching a certain higher average speed.
You are wrong about the third one:
Both laps you run the same distance(the olympic track): D1 = D2 = 1d
First lap speed: v1 = D1/T1 = 1d/T1
Second lap speed: v2 = D2/T2 = 1d/T2
The average speed is the total distance divided by the total time: Vavg = (D1+D2)/(T1/T2)
The average speed has to be double the first lap speed: Vavg = 2*v1 = 2d/T1 = (D1+D2)/(T1/T2) = (1d+1d)/(T1+T2) => 2d/T1 = 2d/(T1+T2) => T1 = T1+T2 => T2 = 0 => v2 = 1d/0h
So basically the second lap would need to last literally 0 time, which is impossible.
@@JustDoIt12131 so what is the correct answer then?
@@MohsinExperiments You have to teleport lap 2. Assume you're aiming to go 60mph and it's a 1-mile lap. Your first lap you go 30mph. That takes 2 minutes. To average 60mph, you have to take 2 minutes for BOTH laps. So unless you straight up teleport lap 2, you can't reach 2 minutes over both laps for a 60mph average.
1. Partially filled with viscous fluid.
2. It should go backwards, here's how: Taking the length of paddle to be same as the radius of the wheel, and assuming the wheels to be perfect rings, it wouldn't go anywhere if the crank and the cassette were equal but as the cassette is much smaller than the crank it reduces the moment of force on the wheel.
3. Infinite. to make the average speed double of the initial speed v, one has to run such that the total distance 2xd covered over the total time taken equals 2v. However the distance that the runner can run is restricted to twice the circumference of the track 2d, so, to get the average speed equal to 2v one would have to cover the whole track again in NO TIME.
4.The base rim of the wheel stays at rest wrt the ground (except when sudden breaks are applied). However, the wheels of a train are flanged meaning the have an external and internal ridge, or rim (lip) that extend beyond the point of contact of the wheel and the rail which would momentarily seem to go backward when below the point of contact of the rail and the wheel.
Ironically, these are the fastest moving part in a train too (the portion of flange when above the wheel.)
Actually, wheels only have one lip on the inside. Having two would make it impossible to turn
on 4th
I'd say part of the wheels go backwards, like the rotor blade of a helicopter does, when it circles around / is in rotation).
All the talk of the size of pedals and wheels is irrelevent as soon as you take gearing into consideration. Certain gearing ratios change the effective advantage the pedals have over the wheel. You can manipulate them so the rim of the wheel moves more, the pedal moves more, or they move the exact same amount. This also translates into the ratio of the force on the pedal and the force on the wheel
Thank you! Finally someone who thinks the same as I do
3)
(v1+v2)/2=2v1
v1+v2=4v1
v2=3v1
i think.. haha
You hate riddles.
+Physics Girl I did until you changed my perspective
But you do (Simone remembers !)
Guys just go out already...
Veritasium Just give us the answers already.
I wanna know!
Speaking of revolutionary riddles, I'd be very interested to see Veritasium's take on the Saturn hexagon, especially in view of additional research into the topic published last year! :)
Jews
I think the Saturn Hex was replicated with water on a soinning plate inside a cylinder. The speed made the water bounce off the walls even on all sides and at various speeds it would just change to more sides. If i remember correctly.
1. The cylinder contains H2O or D2O
2. It depends on the force, if it is balanced it will not move, if it is unbalanced it will move
3. No clu
4. The wheels are spinning backwards to the ground which moves the train forward
On 3, the answer is undefined because you have to run the same distance with 0 seconds as the total time would be the time of the first lap and the distance would double which consequently doubles the velocity. With this, the speed of the second lap would be the distance/0 which is undefined
surely its deuterium whats inside
has to be
not expensive af at all
I believe every point on a wheel moves forward with respect to the ground at all times.
@@NegatingSilence The wheels of trains have a sort of "rim" which reaches further down than the rail which the train is driving on. On this lowest part its moving backwards relative to the ground.
1> The tube contains another tube or a rolling weight.
As the tube rolls forward friction moves the weight relatively backwards inside the tube. Gravity pulling on the internal mass then makes it roll forward inside the tube and it overtakes dead bottom of the tube and rolls up the inner leading edge which accelerates the tube. The weight then rolls backwards again to the trailing edge of the tube which decelerates the tube until it stops rolling down the ramp. The internal weight then rolls forward again and the cycle repeats.
2> The bicycle will depend on the gearing. If the gearing of the pedal is such that the linear velocity of the pedal is slower than that of the wheel, then the bike will not move or the wheel will skid depending on the friction of the tyre to the ground.
If the gearing is low enough so that the linear velocity of the pedal is faster than the wheel, then the bike will move forward.
3> Vav=2(V1)
The average speed is (v1+v2)/2
(V1+V2)/2 = 2(V1)
V1+V2=4(V1)
V2=3(V1)
V2= 3 x V1
4> The part of a train that is always moving backwards is the flange. In particular the part that is below the top surface of the rail.
The answer for number 3 is impossible, because to reach the average of 2 times the speed of first lap you need to finnish both lap at the time you take on the first lap.
wrong wrong and again WRONG!
Vav=total distance/total time
Lets say v1=4m/s. Since 1 lap is 400m, that means you would finish 1 lap in 100 seconds. Make that t1. Now write an equation with the variable t2.
2V1= 8
Total distance = 800
Total time = 100+x
8=800/100+x
This means 8x+800=800
So 8x=0.
Therefore the solution to this riddle is that it's impossible.
Btw 2v1=8
And total distance of two laps=800
@@apextopweek1654 Are you an anti-vaxxer or why are you just saying "WRONGWRONGWRONG" with no input as to why what he's saying is incorrect
1) Honey?
2) It would depend on the gear that the bike is in. In low gear it will move forward, in high gear it will move back.
3) Let's plug in numbers and see how it works. Let's say you are running on a 1 mile track at 1 mile per hour. This means that you want to average 2 MPH. To average 2 MPH in two laps, the total time lapsed must be 1 hour. However your first lap takes 1 hour by itself. leaving you with no time. Perhaps if your moved at the speed of light, then time would stop and you would arrive at the same time your left? I don't have a full understanding of relativity to know if that works or not, but it's close enough.
4) The bottom flange of the wheel is always moving backwards.
#1 - fluid would yield fluid motion. I think there's more to it, but that's probably a part of it.
#4 - Agreed on the flange part, but I'm guessing it's only the bottom rear part that's moving backwards.
Only the bottom flange below the track is moving backwards but otherwise I think you're spot on
shouldnt v2=0? the second lap would have to be completed instantaneously. Vavg = Delta d/ Delta t, such that increasing the distance by a factor of 2 but keeping the time constant would result in an average velocity of 2*v1
The bottom flang of the wheel is not moving backwards it's stationary because it's always touching the ground. The part that is moving backwards is the inner border of the wheel when it's bellow the rail
train wheels are like a top hat on its side so there is a small part below the contact area. think of that part as the end of a levar and the contact area as the pivot.
”Earth is after all one one giant turntable.”
All flat earthers: "We knew it!"
1. Some sort of mechanism that keeps a weight on the rear part of the cylinder, so that it's centre of gravity is always at the back, countering it's inclination to roll.
2. If the gear ratio was 1:1, I'd expect the bike to not move, but because of the gears, the exertion required to move the pedal is less than that needed to hold back the bike, so I'd say it would move forward (briefly).
3. I already know the answer to this one.
4. Not sure about this one. Part of the wheel? Exhaust fumes (does that count as part of the train?) Sound waves? Something to do with the force applied (pushes backwards to make it move forwards)?
Going against the rotation of the earth maybe?
My guesses without reading any comments are:
1) Probably a liquid with high viscosity
2) Backwards. it is analogous to pulling on a wheel below the axis
3) infinite. if it took t seconds fininshing the lap with v1 and the second velocity is x*v1, the average speed is (v1*t + x*v1*t/x)/(t+t/x) = 2v1/(1+1/x). this equals 2v1 if x is infinite.
4) the part of the wheels that guides the train that is below the contact point. Actually, driving axis of trains don't move with a rolling contact because the friction is higher when they are sliding through, so also the part of the (driving) wheels that touch the rail move backwards.
He put in the lowest gear though so it might actually go forward.
i think that high viscosity liquid might me wrong, as it would simply rotate with the bottle.
I think a non-newtonian fluid may be more likely. whose viscosity will change based upon how much rotation it is experiencing.
Good job, for the train I was about to say the wheels accounting for imperfect friction.
4. The smoke
ChristofFreestyler NERD!
why don't you focus on more important questions. Like, how does the snowplough driver get to work?
LOL. At last, a sense of humour.
Piper at the Gates of Dawn
dog sled
I actually asked a snow plow driver that. The answer was quite interesting.
They have a tier system.
They drive a VW Beetle to work
Simple.Have the machine at home.We would not survive without it.
1. A partially liquid filled (using a medium viscosity liquid), sealed cylinder is inside the cylinder on the inclined plane.
2. 3(V1), because [V1 + 3(V1)] /2 = 2(V1) .
3. The bike should move forward as you pull the string backwards, because you are mimicking the actions of a functioning bike. The bike should not go backwards because the string is attached to the base of the pedal, which means the pedal will move backwards as you pull on it with the string.
4. The bottoms of the trains' wheels will always be moving backwards with respect to the ground, as they return to the top position in their rotation around their respective axles, after providing the necessary friction for forward movement.
Looks like some pretty good answers. Not sure about the fourth answer though, initially I thought the same thing but I don't think it's correct. Because if you were to make a dot on the edge of a wheel, and then follow that dot, at no point would the dot actually move backwards because of the forward motion of the wheels
The wheels would have to spin for them to move backwards with respect to the ground
For the second riddle, there is a trick.
If you consider time:
Running twice as fast means you ran the 2 laps in the same duration you ran one turn. This means you ran the second lap in ... 0s, as you already used all the possible time. To get my point, imagine two people running at V1 and 2V1. When the slower one ends its first lap, the faster one ends its second lap. So to reach an average speed equal to the faster one, the slower must run infinitely fast to make the lap instantly, and reach the same amount of distance in the same time.
Even if, mathematically i agree with your solution, i'm not sure it's true. (and because of the title of the video, I'd guess saying "infinite speed" is safer)
2. That is not the formula for avg speed
3) To double the average velocity while doubling the distance already covered, the overall time can't increase from where it is already. If any time gets added due to running the second lap, the average velocity will end up being less then double that of the first lap. Therefore the second lap has to be run in zero-time, or at an infinite velocity in order to double the average velocity from that of the first lap. This is independent of the speed of the first lap.
speed of time travel
Einstein is the key here
1. The silinder has a center of mass that moves
2. Forwards
3. 3*v1 coz
2*v1 = (v1+v2) /2
4*v1 = v1 + v2
3*v1 = v2
4. Wheels
Very smart
You need to think about 3 more logically.
Picture someone running at a certain speed and someone running double that speed take off at the same time.
At the same time the slow runner finishes the first lap, the fast runner finishes their second lap.
To run at the same average speed, he would need to already be done, so he has 0 seconds to complete lap 2
@@cerealdude890 correct, v2 is somehow coming out to be infinity for me. i really wonder what the answer is! owngamesgamer average velocity doesn't mean that. it's total distance upon total time => let length of a round be d and speed in second round be v2
2d/(d/v1 + d/v2) = 2v1 => on solving we get v1 + v2 = v2 => v1 = 0.... and that's a big LOL.
2 is backwards - you put a backwards force on the bike and a forwards force that is created through a mechanism therefore energy lost therefore greater force backwards
3. There is no possible speed.
V1=D/T1
V2=D/T2
2V1=D+D/(T1+T2)
2D/T1=2D/(T1+T2)
1/T1=1/(T1+T2)
T1=T1+T2
0=T2
V2=D/T2=D/0
The value of a number divided by zero is undefined.
1. Viscous liquid -
2. Forward
3. Infinite
4. The "near the bottom" flange on the train wheel. (entends below the rail)
Andrew Sachs +
Andrew Sachs 3 is 4 times faster, if you need an average of 2 and the first speed is 1 then running 4 times faster will leave you with 4x1/2 which is 2
+The deeper meaning
3 is wrong. We are finding the average of speed over a set distance, not the average
of two values.
Let's apply your example: A lap is 150m long. Your speed of the first
lap is 1m/staking 150seconds. But if you ran the second lap at 4m/s it
will take you 37.5 more sec. 300/(150+37.5)=1.6m/s, not even close to 2
times the speed of lap one.
So the reason why the answer is infinity is that you running 2 laps at the
speed of 2x taking you y seconds. Yet if you ran the first lap at the speed
of x, you will take y seconds, leaving no time for your second lap(unless you are a doge who can warp time).
The Doge oh alright, I understand now
3. is correct I think, so infinite. You have to travel twice the distance at the same time, but you already used up all your time. So you have to travel the remaining distance in no time.
2. There's a mistake I guess. The pedal will travel much less distance than the circumference of the bike wheel. So you could think of it as the pedal being connected to the ground too through some stick. And the part of this stick that's connected to the pedal will be pulled backwards while raising further from the ground. It's almost the same type of question as number 4.
1) A Liquid is in the mystery cylinder.
2) The bicycle will fall over to the opposite side from the string attached to the peddle.
3) You must run twice as fast for the velocity to = 2v1.
4) The exhaust fumes will be traveling backwards at 2x the speed of the train.
for 3 you would actually have to go 3 times faster i.e. if you go 10 miles an hour for your first and then 30 on your second the average is 20 as 10 + 30 is 40 divided by 2 is 20 which is 2 times as fast as the first one that was 10. if you don't believe me then do it with other numbers but I have already done it with other numbers.
riddle 2 answer
bike will stay where it is if u pull the string exactly horizontally
because pedal makes circle while moving and direction changes at every point in circular motion
so the pedal will not move
1. Oobleck inside the cylinder.
2. Depends on what gear the bike is in (and whether it is riding through oobleck).
3. Infinite speed. (Note: not physically possible to run this fast through oobleck.)
4. Oobleck..... brb going to go make some oobleck.
Damnit, make a new account named oobleck, and repost this comment. You did it wrong lol
Let's say you run 400 meters in 60 seconds. That's a velocity of 6.67 meters per second. Double that is 13.33 meters per second. So your second run has to be 400 meters at a speed that makes your run 13.33 meters per second on average.
If you run 400 meters in 20 seconds, that's 20 meters per second.
If you're just looking at velocity then you have (6.67 + 20) / 2 which is 13.33 which is double your first run's velocity on average.
If you're looking at distance over time then you ran 800 meters in 80 seconds which is 10 meters per second. The only way to do 800 meters in 60 seconds to get 13.33 after you ran 400 meters in 60 seconds is to do 400 meters in 0 seconds which is infinity.
So the answer depends on what exactly the problem is asking for. One is a nonsense answer, the other is 3 times V1.
For number 3) V1- 1mph.. V2-3mph. Its average is 2, twice the value of 1
XD oobleck so much oobleck
you love Oobleck
The first one is simple, inside the cylinder there is another heavier cylinder and in between the two cylinders there is a high viscosity fluid such as honey, as the cylinder rolls the inner cylinder is first put behind the point of contact stopping motion, then the fluid slowly let's it return to past the point of contact and the process repeats.
- - this person knows whats up
1 - the container has another round object inside that rolls differently from the can, causing it to lag and accelerate awkwardly as the weight favors rolling at times and favors staying other times.
2 - The bike will move forward a little bit. The pedal will move backward.
3 - You must run the lap 3 times as fast as the first. 3 + 1 is 4, divide by 2 is 2 times v1. ...or is it? speed is distance over time... speed is faster as it increases, but also faster as the time decrease... so I would be right I am sure.
4 - It's the flanges of the wheels. They protrude underneath the surface of contact, like the paddle of a rower.
Lets see how I did!
I've seen number 4 before.
Hey! You tricked me! I ran the first lap really slowly, like you said, then when I got to the end of it I found I had no time left to run the second lap! It was a trick! If only I'd run it a tiny bit faster... ;)
it's ok once you do manage to achieve 2V then you will be infinitely ripped
"So as always, thanks for watching. ...well that's... that's Michael's line."
Made my day :-)
1: some sort of liquid with a very high viscosity
2: slightly forward, the rotation of the pedal axel at the chain going backwards is smaller than the rotation of the wheel to which the movement is being transferred via the chain
3: V1 = 1, Vavg = 2, V2 = 3
4: the piston on the wheels goes backward and forward, all the rest should only move forward
I'v tried to come up with my most intuitive thoughts on each riddle, rather than look it up or try it out, so I have no idea how far or close to the truth I am, looking forward to the next video!
EDIT: and I feel like an absolute idiot, only now realizing that youtube suggested a video from 4 years ago...
3 is wrong
But the piston wouldn't move backwards in respect to the ground, only the train itself. It might at one point sit still above the ground but it wont move backwards.
@@CryoRaid wouldnt that depend on the speed? as the train just departs, surely the piston is pushing faster, unsure how it operates at higher speed though, we need a new video exploring this
It's not 3, it's infinity
4. smoke
1. Liquid inside baffles with a small hole to slowly equalise.
2. Bike will move forward until the sprocket has rotated 90° then stop.
3. Run 4 times V1.
4. The flange of the wheel moves backwards.
1. Some liquid inside the tube making it move this way
2. It'll stay stationarry if you don't pull to hard, but if you'd pull as hard as you can you'd just tilt up the back weel and pull it backwards
3. Your second lap would have to be 3xV1
4. The underside of the weels that touch the tracks are always moving backwards
the liquid is probably honey
@@creator6497 or mercuy also
I think no 3 is wrong. Average speed is not equal to two speeds combined and divided by two
I think no 2 is wrong: according to my intuition, it depends on the gear ratio.
Above a determined gear ratio, it will stay stationnary, or tilt backwards (or even slip on the ground according to static coefficient) it we pull too hard.
But with a lower gear ratio, it will be able to pull the lower pedal backwards and therefore move the bicycle forward.
I know it seems physically impossible that a backward force pushes the bicycle forward, but the bicycle’s mechanism will act as a pulley, that transfers force through static friction with the ground.
When I pull a weight with the help of a pulley, the weight also moves in the opposite direction than the force I apply.
@@creator6497 Honey would be too viscous. By the time the honey shifted enough to make meaningful difference, the momentum would carry it through.
1. Cylinder with liquid inside and partitions
The start/stop motion means there's some sort of oscillation happening inside the cylinder, where what's happening inside cycles between some slow and fast mode. The simplest I can think of is if the cylinder is partially filled with liquid (or something similar; say sand), and has some slowly permeable partition(s) dividing the cylinder into two or more semicirclular (or wedge) chambers. It could even be a single chamber with one radial partition, but the cylinder would roll one full turn between each time it rests, which doesn't appear to be the case in the video.
Assume all the liquid is in one chamber to begin with, and that chamber is at the bottom of the cylinder. The cylinder trying to roll downhill will try to raise the chamber along the uphill side of the cylinder. There'll be some angle where the weight of the cylinder trying to roll down balances against the weight of the chamber not wanting to go up. This angle shifts slowly as liquid works its way through the partition from one chamber to the next. As this shift happens, the weight of the bottom chamber is increasing and is pushing against an uphill chamber that's getting lighter and lighter. This eventually reaches a literal tipping point where the uphill chamber no longer has enough weight to matter, and the cylinder rolls until the liquid gets to the next partition and the cycle starts over again.
2. See after 4:
3. Infinitely fast.
For a given speed, it takes a certain time to run one lap. Running at twice the (average) speed, one covers two laps in the same amount of time. This means that after running the first lap, to double the average speed, there is zero time available to run the second lap - the second lap must be run infinitely fast.
The intuitive answer of running at three times the speed is correct if one is running for fixed times rather than fixed distances: if you run a certain time for one lap, then run at three times the speed for the same amount of time, you will have covered three more laps for a total of four laps in twice the original time.
4. Wheel flange
Train wheels have a conical section which rests on the rail, and a flange that runs along side of the rail to keep the wheel from slipping off the rail. The portion of the flange that is below where the wheel rests on the rail, will move in the opposite direction from the train. For small movements, you can think of the axle of the wheel, the point of contact, and the bottom of the flange as a lever pivoting on the point of contact: the axle, hence the train, going in one direction, levers the bottom of the flange in the other direction.
2. It depends...
This is the train wheel problem run in reverse; basically if you tie a string to the wheel and pull on it, does the axle move in the same or opposite direction? The answer depends on where the string is tied to the wheel - if it's at the flange, where it's at a larger radius than the rolling radius of the wheel, then it'll go in the opposite direction. If it's at a smaller radius than the rolling radius, then it'll go the same direction.
The question for the bicycle is what these two radii are. This depends on the gear ratio of the bicycle. If the bicycle has a very low gear where it looks like the crank is turning a very small wheel, then string may be acting on a crank longer than the wheel radius, and the bicycle will go forwards. On the other hand, if the bicycle is in a large gear, the bicycle will go backwards.
On a typical adult bicycle, the crank is 17cm long while the wheel has a radius of about 35cm. The smallest typical front chainwheel is 22 teeth, while the rear sprocket can have as many as 34, so the bicycle can roll with an effective wheel radius of 35cm*22/34 ~=23cm. This is longer than the crank, so the bicycle will roll backwards. This is only an empirical answer; theoretically, one could have a smaller wheel, smaller chainring, larger rear sprocket, or longer crank. Stretching these factors enough can result in the bicycle moving forwards.
Practically speaking, a typical bike would never stretch things this far; it'd make more sense to get off the bike and walk instead of riding with such a low gear ratio. You might however find such low gear ratios in in a cargo bike or rickshaw, but not in a road bike like shown in the video.
I love it when I see a fellow human that is much smarter than I am. Based on what I've read from the other comments, this answer makes a lot more sense than most of what I've seen. I commend you.
I too subscribe to /r/totallynotrobots, fellow human!
If posting in this thread is popular amongst fellow humans then I too shall post to this thread!
**SYNTAX ERROR**
**CANNOT RUN SCRIPT 42**
**PLEASE UPDATE DRIVERS**
**SHUTTING DOWN FOR MAINTENANCE**
08G0HFG85R-84-048047-8897890RHFH
Luns Tee #3 is much simpler than you're making it. You just need to travel three times faster the second time, and the average will be twice the starting speed.
1: The cylinder contains something similar to an hourglass. The motion is explained by the the movement of a fluid from one compartment to the next.
2: The bike wouldn't move. If he pulled hard enough to overcome the coefficient of static friction of the bike tires then it would move backward.
3: Impossible. 1 lap in 3 min plus 1 lap in 1 min gives a total time of 4 min. If you want the average to be equal to 2V, you'd have to complete two laps in the same time it took to complete the first lap. This gives 0 time to complete the second lap.
4: Train wheels have a lip on the inside to keep it on the track. On the bottom of the wheel, this lip is traveling backwards in relation to the ground because the part of the wheel which contacts the rail has a smaller radius.
Done and done.
Nailed it!
why would the bike move backwards? if you pull the bottom part on the padle the bottom chain would move back moving the Wheel forward
I'm not completely convinced with #2. Theoretically if wheels are small enough, crank arms are long enough and front-to-rear gear ratio is small enough it is possible for the pedal to move backwards relative to the ground while the bike is moving forwards, in which case I think the bike would move forwards (until the pedal was directly between the crank axle and Derek's hand of course). If this isn't a theoretical question though then there's not enough data about said parameters to solve this problem.
except 3 is speed not time
@grx70 I think you can disregard most of any gearing and whatnot. You're trying to lift a bucket you're standing in.
The laps one was so easy for me. It's been like a trick question in Physical Science (Physics) Exams
(1+v)/2=2 v=3
(2+v)/2=4 v=6
(3+v)/2=6 v=9
If u want the adverage speed to be twice the first u need to run three times as fast.
1. A precessing gyro is inside.
2. The bicycle won't move.
3. No speed can do it.
4. The parts of the wheel flanges that are below the wheel-rail interface.
Agree with you on all, except I think there is just some liquid in the cylinder of riddle 1. Did anybody officially post the results?
@@RainerSchrom very viscous glob
4th quarter of the wheel only which is leaving the ground
No. 4 is absolute right!!
I think the bike might depend on the length of the pedal crank arm. If it's long enough, it seems like it would travel backwards or remain stationary with respect to the ground, so the bike would move forward.
[1]: the cylinder has sand in it and i think i could hear it shaking while it was being handled,
[2]: the bike won't move -- unless it's in super granny-gear (in which case it will roll forward while the pedal moves back wrt the ground),
[3]: you'd have to run infinitely fast because you will need to do the 2 laps in the same amount of time you just did the first one,
[4]: it's the flange of the train wheel that moves backwards while it's below the point of contact, tracing out an epicycloid.
why 3 is not 3 X V1 ? if V1 is 1mph, and v2 is 3mpg, then their average speed is v1+v2/2
+Mazim Ginsburg
That was my first thought as well but doing the calculations shows that it is impossible. Here I'll show you:
average speed = time / distance
i.e: 100km/h = 1 hour / 100km
so if we put in our numbers where V(1) is first lap average speed, and V(2) is second lap average speed and 2V(1) is total average speed. And D is distance traveled in one lap we get the formula:
2V(1) = (V(1)*D + V(2)*D) / 2D
If you simplify this statement you get:
V(1) = V(1) + V(2)
The only number which fit this statement is if V(2)=0, 0 = 0 + 0, 1=1+0 and so on. Because of the fact that if you run at zero speed you will never complete a lap, we sadly won't fulfill the criteria of completing two laps.
So oppositely of what +Gabor Revesz you actually have to run at an infinitely slow speed to get as close as possible to V(1) = V(1) + V(2)
Is it? Your assumption that avg speed = time / distance is then contradicted by the example you gave. 100km/h = 100km / 1 hour, not 1 hour / 100km. Avg speed = distance / time. That's why your calculations state that you would have to run infinitely slow: you are calculating 1/avg speed
Nomulus Norwegicus: The simplification is wrong. The correct simplification gives 4V(1)=V(1)+V(2), so V(2)=3V(1).
1. Water in the cylinder
2. Forwards because of gear ratios
3. Three times V1
4. The lowest tip on the inside of the wheels
1-copper tube/coil with magnet inside.
2-if the string is non-elastic the bike cannot gain any net forward movement and would (depending on gear ratio) either the bike would remain stationary until the peddle is in the rearmost position then the bike would be dragged backwards or the bike would be dragged back from the begging.
3- (V1+V2)/2 = Vave and Vave=2V1 thus (V1 +V2)/2=2V1
Reduces to v1+ v2=4v1 then to v2=3v1 so if you run 2mph for lap 1 and 6mph (2mph x 3) for 2nd lap your your Vave =(2mph +6mph)/2 which = 8mph/2 =4mph which is 2xV1
4- bottom edge of the wheels
1. Sand
2. It's not moving if you pull smooth, but if you pull like a gorilla, it will definetly go backwards.
3. 3V1
4. A thing that just goes backwards with respect to the train would be the smoke, but with respect to the ground, if the train goes in the opposite direction of the Earth's spin, then it is actually the train.
I think the first answer is Honey or something, since it takes long for the liquid inside thé cylinder to catch up
I think you have 2 and 3 right. 1 I'm not sure about.
For 4 I think we want to consider the Earth as our stationary reference frame, so we ignore the rotation or the direction of the train. The point of a wheel that touches the ground will always be stationary with respect to the ground, assuming, the wheel doesn't slip, so I think the correct answer would be a part of the train what moves backwards with respect to the wheels. I would say probably the pistons, assuming he's talking about a steam train with horizontal pistons like you see in movies. The would go through cycles where for half the cycle they are moving much faster than the train in the same direction, and half the time are moving backwards. Not sure if you can prove that they always move backwards faster than the train goes forwards, though.
@@mattlawyer3245 or just tge top of the wheel
@@xavier9480 etiher that or the thingey that connects more wheels together, since it's set outcentered so at some point it's backwards.
Thought of this then realized the smoke tends to travel in the same direction as the train due to momentum. I thought it was a portion of the spinning wheels, but an experiment proved me wrong. Then I thought about gears of a different ratio that would rotate faster than the wheels. There would be regions on such gears that move backward as the gears spin.
1. The cylinder has another smaller cylinder which is oscillating.
2. Backwards, as the radius of the pedal is smaller so will rotate more as compared to the wheel with the same force.
3. given :
V1, V2, Vav
Vav = 2V1
since 1/V1 + 1/V2 = 2(1/Vav)
1/V1 + 1/V2 = 2(1/2V1)
1/V1 + 1/V2 = 1/V1
1/V2 = 0
V2 = Infinte !!
4. The part of the wheel which goes below the contact point of the track.
1. Something very viscous liquid. It will take time to readjust to level after it's spun a bit, so it will stay towards the left making it unable to roll for a sec.
2. It will go left.
3. 3 times as fast.
4. A part of the wheel that is beyond the contact point with the ground. The contact point will stay still, anything closer to the wheel center will move forward, but anything outside the contact point (below it) will go backwards.
1 - Mercury
2 - backwards
3 - infinite velocity
4 - a point just behind the bottom point of each wheel, where it touches the track
All correct. But not every wheels bottom only the engine vagon
1) a heavy cilindrical weight immersed in a viscous substance, like honey
2) Goes backwards, as any gear ratio that is not 1:1 will never accelerate it forward (plus it's not entirely energy conservative)
3) 3*v1, as AVG = (3*v1 + v1)/2 = 4*v1/2 = 2*v1
4) the universe, from the point of reference of the train itself
1) I have two hypotheses. The first is that there might be some gyroscope (or something inside that rotates on the same axes) inside and it might cause the cylinder to stop due to conservation of angular momentum. The second is that there's a really thick fluid inside, which is why it's movement is very "sloshy". When the can rolls forward, the fluid moves to the back end of the can and slows it down. Then when the can comes to a stand still, the fluid moves down the can to the front end of the can, causing it to move again.
2) the bike will move forward, and then back. By pulling the string, there is a torque on the pedal, causing it to turn, which will make the wheels turn. When the wheels turn there is friction between the tyre and the ground, and because the friction is greater than the tension, the bike will move forward. This continues until the tension has no lever arm on the pedal and the sum of the internal torques equals 0. Then the tension will pull the bike backwards.
3) The velocity is infinite (time taken is 0). If you travel the first lap at 1lap/t, and you want the average velocity to be 2laps/t. If the average velocity is 2laps/t and the distance travelled is 2laps, it takes time t. But if it took time t to travel the first lap (at 1lap/t), then you need to travel the remaining one lap in the remaining time, 1lap/0t. So Te velocity is infinite
4) I'm not too sure about this one, but if we set the ground level to be the rails, the part of the train that goes backward would be the part of the wheels that go lower than the track (the part that goes around the track so it doesn't derail). That's because the wheels are going at 2pi/T. The rolling motion of the wheel gives the lowest point the velocity (-)R*2pi/T, where R is the radius from the center of the wheel to the edge of the rim. And the translation motion of the train gives the lowest point of the wheel r*2pi/T, where r is the radius from the center of the wheel to the rail. So the linear velocity of the part of the wheel that goes below the track would be the sum of these r*2pi/T - R*2pi/T. And because R>r, it's linear velocity will be negative
Cool I agree
completely agree with these hypotheses.
3rd riddle answer
3 times fast
(v1+v2)/2=2v1
put v2=3v1
and equation is satisfied
so 2nd lap must be 3 times faster
3rd question reminds me of the calculating one directional speed of light video. In it also if light is c/2 one way it had to be instantaneous the other way to get average as c.
1 I thought maybe a spring and another cylinder
2 depends on what gear it's in, but go forward
3 impossible, you are out of time.
4 part of the lower half of the wheels
Couldn't you just go and make V2=3V1 (aka run 3 times faster on lap 2) to average your total speed to be 2V?
@@boxby6580 No, that doesn't work unfortunately.
Speed is distance over time, so to get the speed of lap one you divide the distance around the track by the time it takes you to complete a lap:
V1 = d/t (ex. 1m/s = 100m/100s)
Lap two is the same distance as lap one, so your combined distance for the two laps is 2d. (ex. 200m)
Your goal is to have an average speed of 2V1. (ex. 2m/s)
You can use the speed equals distance over time equation to determine the total time you have to complete both laps in order to reach that speed:
2V1 = 2d/Xt
Solving for X here results in 1, which means that you need to complete the two laps in the same amount of time you completed the first lap:
2V1 = 2d/1t (ex. 2m/s = 200m/100s)
This means that you have no time to complete the second lap.
4 - no, you're wrong. The lower half of wheel is also moving forward. Only the point, that touches the rail, have relative V = 0.
But train's wheels have small rim, that extends below rails and doesn't allows train to move left/rigth on rails.
@@SandroSmith And you just proved yourself wrong... because the rims on the inside that prevent the train from derailing have a greater diameter than the outer part of the wheel touching the rail, yes, where the velocity is v=0. At a spinning wheel the velocity is bigger with a greater radius.
@@jamielennard3388 Wtf is that conclusion. He never gave a time limit, you can run the second lap as fast as needed, which is actually 3*v1.
FYI the equation is (v1+v2)/2=2*v1, you can solve that after v2 and you got your answer.
1: a viscous liquid and a weighted ball
2: it will go forwards. the only point of friction is the point the wheels touch the ground, so any force on the pedals will cause a rotations independent of the ground
3: v2 = 3*v1
4: the light from the caboose facing backwards or the point on the bottom of the wheel
2. I think it depends on the height you are pulling it at. Pulling it slightly from above you may life the back tire a bit and it just spins in one place.
Pulling it from level or beneath I’d expect it to move forward. I like to think that the gears would make it do that
Actually when you ride a bike the bottom pedal moves backwards.. so pulling backwards on it won't pull the whole bike backwards, only the pedal.
Since the pedal is traveling backwards with respect to the ground, the bike will move forward
1. The cylinder's center of mass is not at the center of the cylinder and the cylinder can use gravity to travel up, stay put, or travel down the declined plane depending on where it's center of mass is.
2. The bike will initially begin to move forward as the pedal is being pulled back, but once the string has pulled the pedal to its back most position, all you could possibly do is pull the bike backwards.
3. The calculation of your average lap could be determined by (v1+v2)/2 = 2v1. This simplifies to v2=3v1 meaning your second lap would have to be 3 times as fast as the first.
4. A part of a train wheel that comes out further than the inner wheel to hold the wheels aligned with the tracks could travel the opposite direction of the train with respect to the ground from the lowest point in its rotation until a quarter turn after.
I'm looking forward to more riddles like these.
By the "many worlds" theory of quantum physics which says everything that can occur does occur in some universe... then it is possible, but very unlikely, we are living in the universe in which the answer to riddles is always "pink unicorns".
By "Al Gore" logic I can't be proven wrong...therefore I am right and the rest of you are wrong.
Pink Unicorns is the answer to all 4 questions.
1. Something with a high viscosity that needs to flow and catch up to the cylinder's position, and provides a counterbalance until it does so.
2. High gear: Bike goes back. (no matter the rotational speed, the crank is instantaneously travelling in the same translational direction as the bike relative to ground) Low gear: Bike goes forward (crank travels in opposite direction as bike relative to ground)
3. Well... can you run your second lap in zero seconds? :)
4. The inner portion of a train wheel that hugs the track and is wider than the part of the wheel in direct contact with the track. It has a larger radius than the part of the wheel in direct contact with the track, thus while the bottom of the contact is at velocity zero relative to the track, the bottom of the largest part of the wheel is at a negative velocity relative to the track.
#4 is wrong. So is number 3, but I've posted a dozen comments about it already, and you math geeks can't leave an equation unsolved even if it invalidates the question as a riddle by doing so....
4 is wrong because the ground does not have a direction as a stationary object. You have to still the train and rotate the ground for it to have a direction to move backwards relative to. The whole wheel opposes the rotation of the ground, not just the flange. Therefore, the wheels are the correct answer.
So a video like this is what got my subscription i look forward to seeing the answers
2. Depending on the friction of the tire it should go forward? You are applying a backwards force which will initially pull it back a small amount due to deformation of the frame backlash in the drive assembly, and the torsional deformation of the tire, but depending on the ratio of mechanical advantage of bike the backwards pull should propel the bike forward. Conversely if it is in a speed reduction range that reduces the mechanical advantage to the ground it will go backwards.
There is a hampster in the cylinder 😱
Salinox *hamster
Manuel Liriano Oobleck wouldn't have been able to stay on the ramp without the constant force being acted on it, hamsters have a random chance of going up, down, etc
Lole Damb train the hamster to act like oobleck.
100% completly agree
@@TheReligiousAtheists Hamper.
0mn0mn0m picnic.
1. Processing gyroscope
2. Backwards
3. 3 times the speed of the first lap
4. The bottom of the wheel that goes below the top of the track
Edit: I’ve changed my mind on #3. You would need to run infinitely fast in the second lap to double the total distance traveled without increasing the time elapsed.
Explination : for every action there is an equal and opposite reaction....physics simplified
I agree
#3 would have to be instantaneous in order to produce the desired result. It is impossible.
Let's demonstrate with numbers:
IF you ran a track of 60 meters in 60 seconds, your initial speed would be 1 meter per second (i.e. 60meters /60 seconds) and the desired average speed would be 2 meters per second. there is no way to make the next 60 meters in zero seconds in order to make the resulting division 120 meters per 60 seconds or 2 meters per second.
The bicycle will go backwards if the peddle is above the ground (which is the usual place). If the peddle was exactly on the ground the bicycle (at the instance of applying the force) would not move. If you were to place the bicycle on a platform that was above the ground and made the peddles long so that they went through a gap cut into this platform and then you pulled on a string connected to the bottom of this peddle, the bicycle would initially go forwards since in this case there is actually a rotation of the whole structure around a point at the level of the plank.
@@PIC18F That's if the drivetrain is set to 1:1, which it probably isn't.
1. Some type of square or hexagon type shape
2. The wheel will spin and when the string gets straight the bike will go back
1)A solid metal cylinder, 2) forward by wheel radius x the gearing ratio / r of the pedal stem sin (180-theta) for theta = [270, 180], 3) infinity, 4) the bottom of the wheel, which is going 0 - slippage.
Reading people's stupid explanations is more fun than I'd expected.
yes, one can really get a sense of the depth of the individual, their likes and dislikes, the way their hair turns or not, or which path they will choose when coming on a single choice, it is all there, explicitly for the reading, as your comment shows
T Hunter But I don't see you giving any answers...
Yeah, what's your answers?
He just said it was fun.
seeing stupid people criticizing at people's explanations as stupid is more fun that i anticipated.
Veritasium
1) a loose object that is carried to the back end of the cylinder due to fricton on the wall but then slides back forward
2)most of the cases it will travel backwards, since although you are turning the gears it still is a fixed Position relativ to your hand. Which means that it cannot travel forward unless you let go of the rope altogether.
3)if you see velocity on a skalar (I dont know how it's written in english but a unit that has no direction like Mass, Temperature)
then its not possible because you would have to run it instantly to double your average Speed.
If you think of speed as a vector anything is fine, because v_1 is 0, since you ran in a "circle" and 2*0 is still 0.
4)On the wheels of a train is a slight bit that extends below the height of the rail to prevent falling of the rails. Since the part of the wheel on the rail is exactly stationary relative to the ground the part at the side of the rail must be travelling backwards.
Had to give them sone thought but quite fun in general.
For three you would have to run three times as fast. It's not velocity it's speed he asked to calculate (V1+V2)/2=2*V1 V2= 3*V1
Jason Last
I know thats the intuitive solution but it has a flaw.
Since you complete the second lap faster you also run less time with that speed
so it's:
v_ avg = (1*v_1+(1/3)*3v_1)/(4/3)
=2v_1/(4/3) = (6/4)* v_1 =1.5
the better way to solve it is:
v_avg = s_total / t_total
so:
v = 2laps/(1lap/v_1 + 1lap/v_2)
2v_1 = 2laps/(1lap/v_1 + 1lap/v_2) |/2laps
v_1/1lap = 1/(1lap/v_1 + 1lap/v_2)
1lap/v_1 = 1lap/v_1 + 1lap /v_2
0 = 1lap / v_2
which shows that v_2 would have to be infinite
hope it can help you ^^
Yeah I was wrong lol, I saw the explanation in the next video
@@mmunier947 That is wrong. A riddle has a singular answer that is a valid solution. "Impossible" is not a solution. The solution is 3 times v1. This equation is a trap. Many riddles feature a trap that appears to solve the riddle, but is not the solution. The video creator is such a dolt, though, he thinks your solution is correct. SMH I wish he never called these riddles. None of them are if his answers are to be taken as correct, which none of his answers are.
1) The cylinder contains water and a filter at one point. So that it stops at each revolution until enough water has filtered through to shift the weight?
2) The bike would fall sideways?
3) At first I was thinking you'd have to multiply the speed by three, to average out at double.
BUT
Say it takes 60 seconds to run lap 1. That's 1 lap/minute. For you to then double your total average would mean running 2 laps/minute... But you've already wasted that minute in your first lap. So it's not possible.
4) I can only guess at this one, and the guesses are so simple that they're probably wrong. My immediate response was 'steam'. Not all trains run on steam. Sound? Air? The light reflecting from the back?
For 3 triple it does work, he just wanted speed as an average so if you speed up from 1 lap/minute to 3 laps/minute it would average to 2 laps/minute, even if you're only running two laps, as the speed is independent. Traveling the 5 miles in a car at 5 mph or traveling 5 miles in a car at 15 mph would get you an average speed of 10 mph, regardless of time.
3 is wrong. Your math is way off.
1. Magnet on a metal track.
2. Gear reduction. The bike moves backwards.
3. 3x
This is easy. 1mph (V1) plus 3mph (V2) equals 4mph. 4mph/2 equals 2mph or 2(V1). Overcomplicating something gets you into trouble. The way you thought it is impossible. The way you worded it is simple.
4. Sound.
I think you are right with 2,4 and maybe 1, if the magnet isn’t too strong, but 3 is wrong because when you run 1 mph with run V1 and 3 Mph with lap V2 your average speed isn’t 2 Mph. You would have to teleport if you wanted twice the average speed as V1.
To clarify that:
Someone is running 10sek/100m
double the velocity is 10sek/ 200m or 5sek/100m because velocity equals time devided by distance.
So it is acually impossible to run to double up your average speed, you would have to teleport.
4 the light emitted by the rear light
Out-of-the-box answer, but it's not part of the train at that point.
Ha, clever!
e why not? the total momentum of the photon plus the one of the train after the photon emittion is equal to the total momentum before the emittion, so we can consider it all like an isolated system
Sebastiano Buson its the upper part of the trains wheel .
Naveen Rawat i haven't said yet the light is the only part which goes the opposite direction
These questions are almost all the same. It has to do with points on a rotating object.
1) Viscous liquid
2) It depends on the gearing of the bicycle. If the rear gear to rear wheel ratio is higher than the front gear to pedal ratio then the bicycle will move forwards and vice versa. I think I said that right.
(Update: Search Prolate Cycloid)
3) The second lap would need to be completed in zero time. Draw a position vs time graph and you will find that in order to double the area under the line you will need double the distance but without adding any more time.
(Update 2: If you don't likes using graphs I have done the math in a comment below)
4) The part of the train wheel that sits below the track. This is why pulling on this part of the wheel would push the train forward. See how this ties to question 2?
Mr12mic12 You're the only commenter that had the same thought on the bicycle as I did.
Mr12mic12 surely moments or torque is still conserved so that doesn't work
The moving points on the wheels form a cycloid which does not go back. Unless you consider outer parts of the wheels not touching the rail.
Mr12mic12 The Hurricane part ar the beginning of the video actually explains all four of them at the same time.
Exactly as I thought about the bike, it'll depend on 2 things: THE WEIGHT OF THE BIKE(amount of friction) & THE GEARING (wheel size/pedals/chain length)
But of course generally speaking if you try this at home folks, the bike will go backward 😏
These videos make me feel let down with no conclusion.
The next video is the solution
@@gregg8721 how am I supposed to find that lmao. Scroll through 4 years of videos??? I don’t even know the title or thumbnail. If I was on a PC sure. But on a phone it’s not the easiest.
@@Badoopboop he doesn’t have a lot of videos and it has almost the same thumbnail and title
@@gregg8721 crazily… it got suggested later last night lol.
Or just search the same title and add solutions. And CZcams will do the rest
1. Sand but separated by quadrants so that it's like a rotating hourglass?
2. That bike is going forward.
3. Impossible. In order to achieve double the speed, the second lap needs to happen instantaneously.
4. Sound emitted from the train?
You're wront for 3 : v2 need to be equal to 3.v1
Because if you take the average of the two (v1 + v2)/2, you can replace v2 by 3.V1
So you get (v1 + 3.v1)/2 = 4.v1/2 = 2.v1
If you run the second lap at 3 time the speed of the first one, your average speed will be 2 times the one of the first lap !
I solved it by posing :
(v1 + v2)/2 = 2.v1
v1/2 + v2/2 = 2.v1
v2/2 = 2.v1 - v1/2
v2 = 4.v1 - v1
v2 = 3.v1
@@elinatural2058 Nah, you'd need to run second lap at infinite speed. Watch his leading video. You can't compare velocities independently of the distance required to travel for the two laps. You need to add distances together too.
1- viscose liquidi
2-forward until reaches 90degrees relative to the start point if is in small gear and backards if in high gear
3-infinite becose the total overage speed is (S1+S2)/(t1+t2)
4-the point near the ground if the wheel have the format of the H
It could also be shaped like >
wait sorry i lied train tracks never do that it looks like
Romeu Capela sa You can undershoot it and overshoot it, therefore it must exist.
1) probably a certain amount of sand or water
2) move forward slightly then stay in its place because when you pull the string the pedal will come back a little, but will stop when it has an equal plane with the line.
3) impossibly fast
4) the wheel
Nope nope yep unspecific
1. Liquid in the tube
2. Bike goes backwards (due to mechanical advantage in the gears)
3. Infinite speed so that it’s instantaneous (impossible as it would exceed the speed of light)
4. Wheel part below the track (on the inside of the track).
3. You need to walk 3 times the speed of lap one, because
V(avg) = 2V(1)
V(avg) = (V(2) + V(1))/2
This gives us
(V(2) + V(1))/2 = 2V(1)
Both sides times 2 gives us
V(2) + V(1) = 4V(1)
If you do -V(1) you get
V(2) = 3V(1)
So you need to walk/run your second lap three times as fast as your first lap to have an average of 2V(1)
1. Probably some viscous liquid like syrup. So as it rolls down, it takes time for a center of mass to go to the right of the contact point of the cylinder with the ground. It might also contain a spinning gyroscope (but i doubt it).
2. Move backwards (not completely sure)
3. Not possible (infinite speed). Twice the average speed means covering twice the distance of the lap in the same time, but that time is already up, leaving zero time for the next lap.
4. The part of the wheel below the track head. Logically, point of contact has zero forward velocity. Anything below must have reverse velocity.
I'm pretty sure the bicycle would move backwards slightly, because although theoretically the force applied to the pedal should be counteracted by the wheels perfectly, some energy is probably lost somewhere along the gears, string or chain, making the pull backwards slightly greater than the wheel's forward motion.
it think it's the metal bar thing on the train's wheel
Robbie Keith
as might know, the track is fixed to the ground.
my opinion on the bicycle question:
It depends entirely on the gear ratio.
high hear ratio (like driving fast) - won't move.
low gear ratio (like driving very slow) - moves forward a little bit.
I agree with your answers. Bicycle goes backward (confirmed with my own bicycle).
Except mountain-bikes with a very low gear ratio.
For #2: I'm pretty sure it depends on the gearing ratio between the pedals and the rear wheel. On a low gear, it would move forwards, but on a high gear it would move backwards. It's an imbalance of forces and the direction the bike moves depends on whether the force on the string or the force from the wheel on the ground is greater. In both cases, the pedal has to move backwards with respect to the ground, though, else you're creating energy.
For #3: That depends on whether you're asking for the average speed over distance, or the average speed over time. If you're asking for average speed over time, it'd be 3x the speed of the first lap.
1. A thick fluid (because when cylinder rolls the fluid goes a little far up with respect to platform that creates an un balance which stops the cylinder from rolling and as soon as the liquid flows down in cylinder, the cylinder rolls further...)
2. Bike will be going backwards as there is no weight on the bike I think the friction won't allow the system function plus when the bike's ridden the force is applied downwards in this case it's being applied backwords hence I think it would go back
3. (I know there definitely a trick behind it being this easy but m gonna try anyway) If u run 2 kmph in your first lap then ran 6 kmph in ur second the avg of them would be 4 which is twice the first lap
4. Photons emitted by the tail light of the train they go backwards with respect to ground no matter what
@veritasium
sufiyan shaikh sweat
Honestly for the 1st riddle I would say oobleck
xD
For the Bike, it depend on the gear ratio, depending on the selected gear, it stands still as the pedal will move the same speed to the back as the bike drives forward. Basically, the pedal would keep the same distance towards you so to actually make the bike move, you have to pull the cord up.
The train, it is his sound.
No. It will always be still there is a force balance between the pedal and the ground. Assuming no losses the force you impart on the pedal is equal and opposite to the force the tire imparts on the ground. The gear ratio is not relevant.
The train is its wheels
For the black cylinder, there should be plenty of possible mecanisms that use centrifugal force to trigger a break... but I have a feeling it's much simpler. Maybe a mecanism using water. For the bike, it shouldn't move unless you pull hard enough that you drag it back with the wheels losing adherence. For the laps, You would need to run infinitely fast. For the train, the part of the wheels that is below the contact point with the rails would go backward compared to the ground.
For the laps, wouldnt you need to run 3 times as fast? 1v + 3v = 4v. Divide by the amount of variables to get the average. 4v ÷ 2 = 2v. Or am I missing something?
@@collinkirk2940 yeah I'm lost too. so average speed is distance divided by time (v = d/t). the distance is always the same, only the time changes. let's say the distance is 100m and the first time I go I take 100s. So V=1, and we need to know how fast I'd have to go so V=2. at the 2nd time, the total distance is now 200m so the formula is 2 = 200 / (100 + x) 2(100 + x) = 200 200 + x = 200 x = 0??? I miss maths and now I'm dumb
@@ntonioproductions1591 I've tried to solve it too.
I used different numbers to have a different result.
Distance: 400m (1 lap)
Speed 1: 100s (4m/s)
Speed 2: X
Speed 3: 8m/s
(4m/s + X) / 2 = 8m/s | *2
(4m/s + X) = 16m/s | - 4m/s
X = 12m/s
Therefore you'd have to run 3 times as fast. But I could be wrong, didnt have math for a few months now.
@@ehrenamtlicherehrenmann4152
it's impossible because you are limited by the distance you can run.
let's keep it extremely basic.
The lap is 100m in distance. You finish in 100 seconds.
Now, if you want to get your average speed over both laps to be twice as high as lap 1, you would need both laps combined to be run in 100 seconds. but since you already needed 100 seconds to finish lap one, this achievement is impossible.
let's go with your way of thinkin.
lap 1: 100 seconds
lap 2: 33 seconds (three times as fast)
now you ran 200 meters in 133 seconds. That averages to 100 meters in 66,5 seconds. Which is only 50,3% faster.
@@BlackspaceLP Yep, I've read through some more comments. I don't think that my answer is that wrong, its just a different question that I've answered.