# Cut The Bullshit In Physics

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### #324 Dubbelosix

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Posted 05 June 2019 - 05:00 PM

On the contrary, you keep asking people to varify with explanations why you are wrong, while I left you wordless?

I think I left you worthless, to be honest.

Well, this is the last time I answer you, because you let me wordless.

### #325 ralfcis

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Posted 05 June 2019 - 06:37 PM

Marco I've already answered your question but since you are illiterate in math you can't understand the answer. However, I'll give it another shot on my relativity and algebra thread because I'd like to keep things in one place for the future.

### #326 sluggo

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Posted 08 June 2019 - 12:01 PM

rhertz#308;

Let's examine the chase scene. The morons rob a bank, and speed away at 50 mph on route 66. The police get a call of a sighting 1 mile away. They immediately speed after them at 60 mph. How long will it take to catch up with the robbers?
distance between cars = 1, robber car speed r = 50, cop car speed c = 60, d = distance.
For robbers, d=1 + rt.
For cops, d=ct.
If cops and robbers are coincident, then 1+rt = ct.
Rearranging, t(c-r)=1, and t=1/(c-r).
Note, nothing is moving at a speed of (c-r)!
There are two objects in motion, so both speeds are required to solve the problem.
Transfer this relationship to an SR example.
A stick 1 unit long with a light at the near end and a mirror at the far end, moves at v along x. A photon moves at c toward the mirror. Using the same setup for the 1st example, t1=1/(c-v). For the reflection, t2 = 1/(c+v).
Nothing moves at (c-v) or (c+v).
Those two values are referred to as 'closing speeds'. How long to form or close a spatial gap/interval.

### #327 GAHD

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