[Youtube Review][Veritasium] 4가지 혁명적 수수께끼 : 해답편!

(Recommended)Popular Videos : [Veritasium] 4가지 혁명적 수수께끼 : 해답편!

 

This time, I will review the popular YouTube videos.

These days, even if it's good to watch on YouTube, sometimes people skip it or don't watch it if it's too long.

When you watch Youtube, do you scroll and read the comments first?

To save your busy time, why don't you check out the fun contents, summary, and empathy comments of popular YouTube videos first and watch YouTube?

(Recommended)Popular Videos : [Veritasium] 4가지 혁명적 수수께끼 : 해답편!

https://www.youtube.com/watch?v=72DCj3BztG4

 

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Summary Comments : [Veritasium] 4가지 혁명적 수수께끼 : 해답편!

KW**:

For the people that are confused with the second riddle, I’ll show you why he is correct:

Let’s call the average speed of each lap: V1=D/T1, and V2=D/T2. Distance is unchanged.
The one many are using is Vmean=(V1+V2)/2, which is actually the MEAN SPEED for this case where you have two different speeds, independently of its value, you can find that Vmean=D/2(1/T1+1/T2), so you can actually solve that to find that if you want Vmean=2V1, then you must have V2=3V1.
This means that if you run the same distance 3 times faster than the first lap, the mean speed doubles the speed of the original lap.

The actual definition of average speed, involves the total distance which is 2D
in the total time which is T. Is easy to tell that T=T1+T2.
Now if you use this you have that
Vavg=2D/(T1+T2), and here is the tricky part. Since you want this to be 2V1 then by solving you have:
2D/(T1+T2)=2*(D/T1), So you’ll arrive to T1=T1+T2,which can only be achieved if T2=0.
This in short means that you run at an infinte speed in your second lap, which is physically imposible.

Remember don’t confuse MEAN SPEED with AVERAGE SPEED, they are only the same when the speed remains constant during the motion. I hope that this helps. Grettings.

 

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Playtime Comments : [Veritasium] 4가지 혁명적 수수께끼 : 해답편!

PH****:

1:36 when you use the wrong formula but somehow stun yourself and the teacher by getting the right answer.

In************:

4:08 "This may seem like a trick question, but actually, it's a trick question"

Dr*******:

3:56
If you run at the speed of light, according to special relativity, in the runner frame of reference, it will take 0 time to go anywhere. So the average speed is indeed twice.

sa********:

3:25 that implies that your average speed is measured over *time*. If you were to take the average speed over distance it is definitely possible.

Ry*******:

2:30
damn, he cleans up quick tho

Su********:

3:56 Rather than running twice the distance, the equation will hold true if same distance is covered in half the time; i.e. 2V = d/t/2

ve*********:

4:05
*laughs in teleportation

qw******:

I've been wondering for some time what direction do cars in front of me throw up debris against my beloved Miata. 5:03 looks like the answer! What simulator is that?

ew*******:

4:13: @Derek isn't that the definition of a trick question!?
Otherwise nobody would be tricked into believing it could be done in the first place.

Ch*******:

3:56 : "Even if you when the speed of light..."

Actually, from your reference frame going the speed of light, no time will have passed going the second half of the distance. It really depends on the reference frame of the measurement.

 

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Top Comments : [Veritasium] 4가지 혁명적 수수께끼 : 해답편!

ca*****:

When it takes you 2 years to actually get the joke.

Co************:

Oh look. It's Dirk from Veristablium!!

Da**************:

I actually really like these types of videos, please do more!

Vi*****:

Lots of surprising answers in this one... but nothing baffles me as much as the fact it took me since last week to realise why you called those riddles "revolutionary". Damn puns...

Co*******:

I definitely thought the running around the track one was just a verbal typo that didn't get caught in editing. It would have been a lot more clear to say "the value of" instead of directly saying to compare speed and time.

Sa**:

Am i the only one who just realized that revolutionary meant rotation related and not world changing?

Ja**********:

You actually can get your average to be 2V1 by running 3 times the speed but zig-zagging to reach the same time as the first lap

NM***:

one thing to consider is if your velocity is zero (you don't move). then next lap your velocity is zero and zero is double of zero. ;)

ma*********:

Wait I just realized the true solution to riddle number 2 you must not walk then the time it takes you is never now if you don't start the second lap it will also be never so the solution is 0m

CB*:

Now I Understand why The Title is "REVOLUTIONary Riddles"

ra********:

in 3:45. why not use instead of 2d/t , d/(t/2)? you can't increase the distance but decrease the time it's possible.

Ni***:

Pulling the bike forward with the rubber duck tied, and then replaying the video backwards is the coolest way I have seen someone prove a point.

Wi***********:

You got the wrong answer for 3. :D
The correct answer is 0. Lay down and have a nap. Don't go anywhere. Twice zero is still zero. You live at the starting line now.

Ma********:

Nice video, cool riddles. That said, I have an argument about the flange: the flange is going forwards with the speed of the train: the part of the flange that goes backwards is not the same at all times so there's no part of the traind that is always going backwards - otherwise, there would not be much train left after a while.

Br************:

Havaianas, todo mundo usa

OB***:

Finally i can feel smart for a change! Riddle 2 was easy with simple math and times of laps, not speeds. If you run a lap in 30 seconds, how fast should you run the second lap in order for the average of the 2 laps to be 15 seconds? Well... you should run the second lap in 0 seconds, so.. there...

Ar****************:

you could zig-zag during the second lap, thus doubling your distance traveled.

Kr********:

Theoretical answer to the Running Question-

Let the circumference of the ground be 'd'

average speed = total distance/total time

therefore

let the velocity for first lap be v1 and velocity for second lap be v2

time for first lap be t1 and time for second lap be t2

t1= d/v1
t2= d/v2

total distance = 2d (circled the ground 2 times)

average velocity = 2d/(d/v1 + d/v2)
=[2d(v1v2)]/[d(v1 + v2)] (d gets cancelled out)
=2v1v2/(v1+v1)

now this velocity should be equal to 2v1

therefore 2v1=2v1v2/(v1+v2)
= 2v1(v1+v2)=2v1v2 (2v1 gets cancelled from both sides)
= v1 + v2 = v2 (v2 gets cancelled from both sides)
= v1 = 0

but if v1 = 0 the time will become infinite, so practically this is not possible unless v2 = infinite

But going deep into the terminology used

He said "what will be the velocity"

velocity is a vector quantity and it is equal to displacement/time

while circling a ground we are coming back to our original position so displacement = 0 hence velocity in both the cases would be zero

the correct terminology here would have been speed or angular velocity (omega)

Thanks for reading.

Ar***********:

Riddle 2: since distance can’t be altered, run the same distance in half the time.

Hu**********:

The answer to the running riddle seems more predicated on language than actual science

ro******:

i really enjoyed it! :D one of the most valuable channel on youtube! Greets from Poland!

E*:

For the track one you achieve it every time because your average velocity around the track is always 0

We****:

The next question: How to ask the 2 lap question in the way that it can be answered as expected?

Bo******:

the second one you could just run the same distance in 1/2 the time to increase v.

Kh*******:

Awesome video as always. I've just recently started my own science channel and I'm happy to say I found inspiration watching your channel and many others. Thank you and keep up the great work!!

Mo***********:

Train: Any gear/motor/engine that is rotating faster than the wheels involved in speed reduction/torque increase...

Ba*****:

For the people who are confused by the second riddle, just use numbers!

Assume the following:
- The lap distance "d" is 10 meters.
d = 10 m
- Your speed in the first lap "V1" is 1 meter per second.
V1 = 1 m/s
- You have to run the 2 laps at an average speed "Vavg" of 2 meters per second
Vavg = 2V1 = 2×1 = 2 m/s

Remember that V = d/t
So:
t1 = d/V1 = 10×1 = 10 s
t(total) = 2d/Vavg = 2×10/2 = 10 s

Notice that the time needed for both laps (10 s) equals the time that you spent in the first lap alone (10 s)!!!

So basically, you have to complete the second lap in 0 seconds to be able to achieve Vavg = 2V1 !!!

Which is impossible of course.


Unless you can teleport..

Et********:

Loved the thought experiments. Failed #1 (thought it was a weight + rubberband) - maybe an extended video or knowledge that the cylinder made its way all the way down would have clarified. I had an incorrect assumption (woops!) it would stop at some point, maybe go back? Semi-got #2 - I thought it was impossible due to a math quirk, but I got that math quirk wrong... Got #3 and 4, woot, dead on; and it looks like I get a new video to stare at to boot!
Loved this exercise! I'd subscribe but I already am...

Ad******:

You're right that the flanges are to keep the train on the track; but only a last resort, they're not designed to contact it - they'd be moving relative to it so they would wear and waste energy. It's the fact that the wheels' rolling surfaces are slightly wider on the inside that keeps it on the track normally: so the train is lowest when it's central and gravity corrects any deviation.

Incidentally, it's the conical wheels which mean trains don't need differential gears. In a curve, the train is thrown out, so the outer wheel contacts the rail at a point where its diameter is greater (and vice versa)

 

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[Veritasium] We gathered comments about popular videos and looked at them in summary, including play time, and order of popularity.

It's a good video or channel, but if you're sad because it's too long, please leave a YouTube channel or video link and I'll post it on this blog.

 

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