This is my last response to you about this matter until you back up your wild and crazy assertions. Put up or shut up!
Post #3 of this thread has an image that appears to be identical to the [POST=3062623]post #159 on April 20 of this year[/POST] which appears to be a screen capture of [POST=2917299]post #25 on March 20-21 of last year[/POST], where Motor Daddy was learning the description of one-dimensional motion with constant acceleration. He would also post a screen capture in [POST=2917781]post #34[/POST] of that earliest thread entitled "Acceleration and Deceleration from known height". Please favor AlphaNumeric and James R's posts over those of Motor Daddy (too ignorant) and Tach (too shrill).
Tach, I think your posts make the most sense but also I found a chart in an old physics textbook and it seems to explain d=16t^2 completely. So the 16 is neither 'g' nor a constant 'a' but the distance the object covered in the first time interval (if one decides to make a chart). Time Interval :::::::: Ave V :::::: Distance Traveled :::::::::: Cumulative Distance ::::::::: 0-1 :::::::::::::::::::: 5 m/s :::::::::::::::::::::: 5 m ::::::::::::::::::::: 5 m 1-2 :::::::::::::::::::: 15 m/s :::::::::::::::::::: 15 m :::::::::::::::::::: 20m 2-3 :::::::::::::::::::: 25 m/s :::::::::::::::::::: 25 m :::::::::::::::::::: 45m 3-4 :::::::::::::::::::: 35 m/s :::::::::::::::::::: 35 m :::::::::::::::::::: 80m You can arrive at d=80m by d = 5t^2 where 5 is the distance covered in the first interval d = 5(4)^2 = 80m Another try, for cumulative d=45m d = 5t^2 d = 5(3)^2 = 45m My question now is, which of the kinematic equations covers this relationship and also what do you call a proportional relationship such as this where the numbers work but the units don't? Because the resulting units here would be m-s^2
where: \(d\) is the distance \(u\) is the initial velocity \(v\) is the final velocity \(a\) is the acceleration \(t\) is the time of travel \(d=ut+\frac{1}{2}at^2\) The units m/s^2 are units of ACCELERATION. Acceleration is the rate of change of velocity. Your d=5t^2 is saying that the distance the object travels is equal to 5 times the time squared. If the time is one second then the equation is d=5(1^2), or d=5. So the distance the object traveled in one second is 5 meters. It started with an initial velocity of 0 m/s, and one second later it had traveled 5 meters. That is an acceleration of 10m/s^2. \(a=\frac{2(d-ut)}{t^2}\) For every second the object travels, the object's velocity increases 10 m/s. So at t=1 the object has a velocity of 10 m/s. At t=2 the object has a velocity of 20 m/s, at t=3 the object has a velocity of 30 m/s. So the DISTANCE that the object traveled is \(d=ut+\frac{1}{2}at^2\), where \(u\) is the initial velocity (0 m/s), \(t\) is the time of travel, and \(a\) is the acceleration (10 m/s^2).
[/QUOTE] \(d=\frac{g}{2}t^2\) If \(g\) is expressed in \(\frac{ft}{s^2}\) then \(g=32\) and \(\frac{g}{2}=16\). If \(g\) is expressed in \(\frac{m}{s^2}\) then \(g=9.8\) and \(\frac{g}{2}=4.9\).
There is a beach ball at rest on the beach. You claim g=9.8 m/s^2 at sea level. I start my stopwatch. I stop the stop watch 10 seconds later at t=10. Are you trying to tell me the distance the ball traveled is \(d=\frac{g}{2}t^2\)? So you're saying the ball traveled 490 meters in that 10 seconds? Is that what you're saying, Tach??
I asked you a question. I'll break it down for you into simpler terms, since you have comprehension problems. Do you claim g=~9.8 m/s^2 at sea level?
Again no answer to my question. You leave me no choice but to conclude that you're scared to answer my direct questions, because you know they make you look ridiculous, as is your equation. Answer my direct question: Approximately what is g equal to at sea level, 9.8 m/s^2?
Nope, the distance traveled in the FIRST second is 5m. The distance traveled in one second is a DIFFERENT number.
