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"WHITE HOUSE PRESS BRIEFING" — A Bad Lip Reading

Sometime a building needs a waterfall

How Avengers Infinity War Should Have Ended

kir_mokum says...

imo, infinity war had the best ending of any marvel movie and they should keep it as the actual ending of the current MCU. the main story telling problem with most superhero movies is there is no real risk or threat to the main characters, only to the people around them. this makes all the action have much less impact as consequences to the violence and action are minimized. ending the series with real deaths from a foe they could not defeat finally gives the series the gravitas that it failed to have up until this point.

also, thanos was by far the most interesting and complex character in the whole MCU. the reframing of his motives vastly improved on his character.

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Avengers:Infinity War Trailer Breakdown

Marvel Studio's Avengers: Infinity War Trailer

Avengers:Infinity War Trailer Breakdown

Avengers:Infinity War Trailer Breakdown

2 Guys from the 80s React to Avengers: Infinity War Trailer

Marvel Studio's Avengers: Infinity War Trailer

2 Guys from the 80s React to Avengers: Infinity War Trailer

Vexus (Member Profile)

4 Revolutionary Riddles

ChaosEngine says...

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 )

Digitalfiend said:

The track question seems really straightforward. 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?

For example:

V1 = 1 m/s
V2 = 3 m/s
Vavg = 2 m/s

2m/s = 2V1

V1 = 5m/s
V2 = 15m/s
Vavg = 10m/s or 2V1.



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