Maths question.

In acceleration

2Pi*<m/s>/2.88=mph

 

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There are exactly 3600 seconds in 1 hour and 0.3048 meters in a foot and 5280 feet in a mile.

Thus we have exactly:
\( 1 \frac{\textrm{m}}{\textrm{s}} = 1 \frac{\textrm{m}}{\textrm{s}} \times 3600 \frac{\textrm{s}}{\textrm{hour}} \times \frac{10000}{3048} \frac{\textrm{feet}}{\textrm{m}} \times \frac{1}{5280} \frac{\textrm{mile}}{\textrm{feet}} = \frac{3125}{1397} \, \textrm{mph}\)

\(\frac{3125}{1397} \approx 2.236936 \\ \frac{2 \pi}{2.88} = \left( \frac{5}{6} \right)^2 \pi \approx 2.18166\)
So this isn't a close approximation if that was your question. \(\frac{85}{38} \approx 2.236842 \) is a better approximation.

It's also dimensionally inconsistent which is at least as big a problem to your bald assertion of an approximation using pi.

 

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Questions have question marks at the end of them.

 

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Acceleration is the rate of change of velocity.

If the initial velocity of a car waiting at a red light is 0 m/s at t=0 seconds, and the car starts accelerating, the velocity starts increasing, and time elapses.

There is a relationship with time and distance with acceleration. If the acceleration rate of the car is 27 m/s^2, that means that the velocity is increasing by 27 m/s every second. So if the car started a constant acceleration when the light turned green, at the 1 second point in time the car would have a 27 m/s velocity. At the 2 second point in time it would have a 54 m/s velocity, and so on. At t=289.4 seconds of constant acceleration the car would have a 289.4*27=7813.8 m/s velocity.

Now, if the car was at the red light and the light turned green and the car started accelerating, the car would have traveled a distance of 13.5 meters in 1 second. The average velocity of 0 m/s at the start, and 27 m/s at the t=1 second point in time is 13.5 m/s, and since it was 1 second, then the car traveled 13.5 meters from the start line while it was accelerating at the rate of 27 m/s^2 for 1 second.

You can imagine that if at t=289.4 seconds the car stopped accelerating and just cruised, the car would maintain that 7813.8 m/s velocity. That means that if the car maintained the 7813.8 m/s velocity for 1 second the car traveled 7813.8 meters in that 1 second. So from t=289.4 seconds to t=290.4 seconds the car traveled 7813.8 meters. The question is, what is the total distance that the car traveled in 290.4 seconds?? Gets a little complicated, so I made you a cheat sheet!!

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Motor Daddy, no need to spam the thread with a screen capture of Post #25 from the thread "Acceleration and Deceleration from known height".

That's a little larger than 2.75 g, a ridiculous acceleration for a car even on a downhill slope.

\(a_0 = 27 \, \textrm{m} / \textrm{s}^2 , \; a_1 = 0, \; t_0 = 289.4 \, \textrm{s}, \; t_1 = 290.4\, \textrm{s} \\ d = \frac{1}{2} a_0 t_0^2 + a_0 t_0 (t_1 - t_0) = a_0 t_0 \left( t_1 - \frac{1}{2} t_0 \right) = 1138.47066 \, \textrm{km} \)

At about 700 miles, that is a ridiculous distance which is longer than the distance between Berlin and Nice.

 

So you're saying once you post an equation it's "spam" to post it again? Does that hold true for you and everyone else too?

The great thing about my cheat sheet is that it holds true for ducks, cars, planes, boats, and swallow.

See above.

BTW, You're correct on your answer.

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...and if a light sphere was emitted at t=0 at the same start point as the car, the light sphere would have a radius of 299,792,458*290.4=87059729803.2 meters.

87059729803.2 meters - 1138470.66 meters = 87058591332.54 meters. So there is a distance of 87058591332.54 meters between the light sphere and the car along the x axis at t=290.4 seconds. See?

 

That's true in one inertial frame of reference, but just like your acceleration of 27 m/s^2 is not based on any actual experience with automobiles, your insistence of only one standard of simultaneity has no basis in physics education.

 

You complaining about my unrealistic 70 GTO acceleration with a stock Pontiac 400 in it, with an 068 cam and exhaust manifolds is nonsense. It's simply a learning tool. You want realistic go to a drag strip. Get the time for 1320 feet and report back.

BTW, there was acceleration, rest, and constant velocity, all measured, and you even posted the correct numbers. Now I add a light sphere to it and you complain? Not what Einstein says, is it?

 

It is what Einstein says. However Einstein also says that your "distance of 87058591332.54 meters between the light sphere and the car" has no universal validity because there are other ways to describe the physics of the situation without your particular choice of inertial frame.

 

What I posted is the results as measured. Does Einstein have different distances and times, as measured?

 

...and if my way has no universal validity, is that saying the laws of physics are not the same in all reference frames?

What does the driver of the car measure the speed of light to be???????? 290.4 seconds for sure, right??

 

rpenner, I think the driver measures the speed of light to be 299,788,537.6464876 m/s along the x axis. What do you think the driver measures the speed of light to be??

 

4,999?

 

5,000??

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unbelievable.
(shakes head)

 

At 5252 I equal 1 lb-ft of torque!

 

2Pi*<m/s> / 2.88 = mph

3.14159265359+3.14159265359=6.28318530718


6.28318530718*44.704=280.883515972

280.883515972/2.88=97.5289986014 mph


Is this not close, and if I adjust the value of Pi I can get exact conversion?

 

If I define my use of Pi has 3.22118826056 pi


Is my formula correct?

 

2pi*<m/s>/2.88=mph


3.22118826056+3.22118826056=6.44237652112


6.44237652112*1=6.44237652112

6.44237652112/2.88=2.23693629206 mph