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Wednesday, February 10, 2016

On the deceleration or acceleration

Further, in simple terms under acceleration a galilean =s / T2 means '' A body moving bc with 4 M / 1 Sec2 '' (i.e. 1 Sec Squared). That means '' The body in 1 SEC IS MOVING AT 4 M / SEC '' So how '' At the end of a sec body cover 4 M IN 1 sec i.e. 0,4 m in 0,1 sec ''. However (since like we said, if you choose more galilaiikḗ slide would be undermined as a statistical the proportion of the block) Here-because the instant speed is fluid concept-could have said this: '' At the end of a sec body cover 0,4 m in 0,1 sec at the end of which 0,1 sec body cover 0,04 m in 0,01 SEC '' And Set the acceleration as a =s / t3 i.e. 4 M / sec sec3 (i.e. in the cube-and not on the block). It should be amended and other similar types. Bc the-Average-type of Galileo for '' Average speed '' A body of accelerating χ=(at2 1/2) not only for the acceleration of gravity '' G = 9,8 m / sec2 '' But every other acceleration however big, so I'll epidécheto and power in the cube instead of power on the block. (besides-even if we accept the block of Galileo-a falling body accelerating cannot be easily definition fixed medium speed bc steady speed on average 5 M / sec, because if the first sec has comes 5 M, in 2 O sec will have travelled far greater distance bc at least 20 m. If we put cube the acceleration would be even greater with obvious consequences in Newton '' F=Gmm / R2 ''. Basically it is a largely types these weekly meetings, autoanaphorikoús, by way of trigon not relating to much broader conditions.

*** ΑΒΟUT THE ACCELERATION AND GALILEO : further , simply put , Galileo's acceleration a =S/t2 means that '' a body moves with eg 4 m/1sec2 '' (ie 1 sec squared). This means that '' the body in 1 sec is moving at 4 m/sec '' or that '' in the end of 1 sec the body covers 4 m in 1 sec i.e. 0,4 m in 0,1 sec as momentum speed''. However (since as we said if we choose another Galileo ski slope the Galileo square analogy in time would not be exactly correct) -since the momentum ιnstantaneous speed is fluid , it could be said that : '' at the end of 1 sec the body covers 0 ,4 m in 0.1 sec at the end of which 0.1 sec the body covers 0,04 m in 0,01 sec '' and the acceleration could be defined as a =S/t3 ie 4m/1sec3 (ie sec cubed and not squared). Thus other equations should also be changed. Eg the equation of Galileo for the ''average speed' of an accelerated body X =(1/2)at2 has to do not only with the acceleration of gravity ''G = 9,8 m / sec2'' but also with any other acceleration even larger , so it also could be cubed instead of squared. (Even if we accept the Galileo square , the truth is that a falling accelarating body does not easily accept a definition of a stable ''average speed'' 5m / sec , because if in the first sec it has traveled 5 m, in the 2nd sec it will have traveled much greater distance , for example at least 20 m. And if we put the cube in the acceleration equation it would be even bigger with obvious consequences in the Newtonian ''F = GmM/r2''. So we are talking about rather fluid and calculus type equations that do not involve much wider and bigger conditions.

Trully if you accept ''average speed'' 5m/sec = in 1 sec 5 m , you cannot easilly accept ''average speed'' 20m/2 sec=10m/sec=in 1 sec 10 m , so it is all statistic and fluid.

The most famous guy for straight smooth movement is the u =S / T. The guy who gave himself Galileo to express the average speed of an accelerated college is x = (1/2) a t2, whom I have. As I mentioned the guy under the galilean concerns expression of the average speed of an accelerated college and no rectilinear smooth movement (U= s / T) neither of the acceleration (a =S / T2). If in a mean in acceleration put the g so the acceleration due to gravity 9,8 m / sec2, we get 5 M IN 1 SEC AND 20 M IN 2 sec so we have an expression of the average speed for A falling body as follows: average speed of 5 M IN 1 sec and average speed of 20 metres in 2 sec and average speed of 45 M IN 3 SEC, etc.
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