Where's the solutions goddammit.

My guesses:

1) a pendulum.
2) 3x faster (this one seems too simple)
3) in this case forwards, but depends on the weight of the bike, and friction between the ground and the tyres?
4) Q2 and 3 of every wheel.

Are you sure about that? Let's say the track is 100m for simplicity's sake and you run it at 1m/s (so 100s).
If you ran the second loop 3 times faster, you run it at 3m/s and it would take 33.3s.
So, total time and distance is 200m in 133.3s at an average speed of 1.5m/s.

I think it's impossible.

Even if you ran the second loop 100000 times faster, your total time would be 100.001s and average speed would be 1.99998m/s.

It would only ever approach 2V1.

Unless there's something about rotational velocity (as opposed to speed) that I'm missing. Quite possible, given it's been a long time since I did any physics.

1) Some sort of viscous liquid, like honey.
2) Yeah... I thought v1=1 and v2=3, but then I realised that was actually more like t1=3, t2=1 which is not quite what the question asked.
3) Most of the bike will go forwards.
4) First thought was the wheels, but it would only ever be the very outside edge of the wheel in contact with the ground, only on driven wheels, and only if the wheel was slipping.

edit. I knew there was a reason it had to be a train and not a car. The rims of the train wheels extend past the contact point with the rails, so they will be moving backwards (well, part of them) even if the wheels don't slip.

1) Honey or similar liquid
2) Three times as fast.
3) Nowhere. If you pull hard, the wheel will spin, until the pedal is horizontal, then the bike will still not move (the back wheel will not turn).
4) Lower half of the wheels.

1) < 1/2 full of honey or other slime
2) bike will go backwards 1/4 pedal turn (forcing the pedal to rotate forward but move backwards) regardless of gearing because the wheel rotation/travel is much longer than the pedal rotation/travel

3) (T1+T2)/2 can never equal or be less than (T1)/2...so impossible unless you can finish lap 2 before you start it

4) a small portion of the lip of the wheels, farther out than the part that rides on the rail

Side note, how did you all get the numbers mixed up? #2 is the bike, not the track.

I think we all just copied @L0cky.

I just realised. I'm wrong about the lap question.

The trick is that it's a LAP of a track... the distance can change between lap 1 and 2, depending on how far from the centre you run.

The track question seems really straightforward, so I must be missing something.

The question is how fast do you have to run the 2nd lap such that the average of the two laps (Vavg) is twice the velocity of the 1st lap (2V1); so Vavg = 2V1 (says right in the video). Unless I'm missing something, V2 has to equal 3V1:

Since the problem states that Vavg must be 2V1, we can substitute that in the average calculation below:

So, Vavg = (V1+V2)/2 becomes 2V1 = (V1+V2)/2

Now solve for V2:

V2 = 4V1-V1
of
V2 = 3V1

i.e. your 2nd lap must always be 3x faster than your 1st lap so that the average of the two laps is twice the velocity of the 1st lap.

No?

This line is incorrect Vavg = (V1+V2)/2. That only applies if you run at V1 and V2 FOR THE SAME AMOUNT OF TIME.

Speed is Distance divided by Time, so the formula for calculating average speed is Dtotal / Ttotal.

The problem is that that only works if your second lap can be longer than the first lap.

If they are the same distance, the maths are undefined.

V1 = D1/T1
V2 = D2/T2

Vavg = (D1+D2)/(T1+T2)
if (D1 = D2) then
Vavg = 2D1/(T1+T2)

if Vavg = 2V1 then
2D1/T1 = 2D1/(T1+T2)
then T2 = 0

therefore V2 = D1/0 .... cannot divide by 0 (and no, it's not infinity )

I don't like the train one, as all of the train is moving forward, and the flange, at only one small part of the rotation is moving backward, otherwise it's mostly moving forward faster than the train. It's messy.

Yep, I realized my mistake while reading a comment over on YT.

Nice explanation here:
https://youtu.be/aFGEe5d70qY?t=6m53s

For 4, could it be the coupling rod? I guess alternating sides.

And for the bike one, you've got to remember the wheels are able to spin freely in that position, so he could drag it backwards as far as he wants. It's only the force of your body pressing down on the bike that creates enough friction that the gears drive it forward.

I guess the hint for these is the rotational test that they show at the first.

1) A sticky object that would let go like a wall crawler that climbs down a wall would create this effect. (see below)
2) You can't. As you approach infinite speed it would get very close. (see below)
3) The bike will move forward. (see below)
4) The outside parts of the wheels that overlap the rail. Also if the train has a flywheel that is larger than the wheel size the bottom of the flywheel would also always move backwards faster than the train was moving.

1) He says "what object is inside?" so I'm not sure a liquid would count. Also a viscous liquid would flow a slow rate and would probably not stop and start. You might be able to get a viscous liquid to stop and start if you had fins, but that still might just move slowly or gain enough momentum to roll fast without any flow.

2) A little excel calculation shows that the average velocity approaches twice the initial but will never hit it.

attempted m/s - total time - average m/s
1 100 1
2 50 1.333333333
3 33.33333333 1.5
...
200 0.5 1.990049751
201 0.497512438 1.99009901

3) I'm not sure if the parameters of this experiment are explained sufficiently.

If it is allowed to slip then no matter the mechanical advantage a hard pull should always be able to get the bike to skid back and defeat friction.

If the bike is not allowed to slip on the ground then I don't understand how it could ever move backwards, the only options would be that it doesn't move at all or it moves forward.

If it can't slip then the ratio of the pedal to the wheel is what is in question. Bikes only have gear ratios higher than 1 and the crank is smaller than the tire so the tire will always rotate more than the crank thus the bike should move forward.

3) it moves backwards because the force is applied to the entire system/bicycle. The wheel going backwards turns the gears and chain, rotating the pedal 'forwards' (rotating it in reverse). Because the wheel is much larger it travels much farther back than the pedal rotates forwards, even with 1:1 gearing the pedal actually travels backwards slightly compared to the ground, but in higher gears it gets much easier.
It's because the tire is so much larger than the crank, that makes it easier to rotate the tire than the crank even at 1:1, so it goes backwards.
It's counter intuitive, but I actually checked my work and yep, my bike went backwards in all gears....no sliding needed. Try it.

Yep, I was wrong on that one. Afterward I went and looked at solutions other people had and saw my mistake.

Ok, so every video I've watched on this still seems to take the view that both laps are the same distance. If that's the case then it's impossible (T2 = 0, etc).

But a lap is just a revolution of the track, you don't have to run at a fixed radius (and therefore a fixed circumference).

If you increase the distance (even by a tiny amount), it is possible.
If the 1st lap is 100m and you run it in 100s, then you get 1m/s. If the second lap is 200m, then you would need to run that in 50s (V1 = 1m/s, Vavg = 2m/s, 300m @ 2m/s = 150s, 150s - 100s (for the lap you've already run).

So if you double the distance, you need to run 4 times faster.
If you triple the distance, you need to run 3 times faster.
If you run 10 times the distance, you need to run 2.2222m/s

Basically the bigger your lap the slower you can go approaching 2m/s.

Conversely, the smaller the difference the faster you need to go (e.g. for 101m, you would need to do the 2nd lap at 202m/s or around 450mph )

Oh, and the bike CAN go forwards, you just need a ridiculously low gear.

What if you ran it backwards? Velocity is directional... No?

So how'd you do?

*related=https://videosift.com/video/4-Revolutionary-Riddles-Resolved

4 Revolutionary Riddles Resolved! has been added as a related post - related requested by PlayhousePals.

In that case, averaging your speeds would lower the average speed, not raise it, since lap two would be a negative distance, but not a negative time, so a negative speed in the equation.

What if you ran it backwards... IN TIME!