jueves, 15 de octubre de 2020

cinematica: tren de (-b) a b con parada en zero

d_{t}[x] = a·( 1/(2b^{2}) )·x( x^{2}+(-1)·b^{2})

( ( 1/(x+(-b)) )+( 1/(x+b) )+(-2)·( 1/x ) )·d_{t}[x] = a

( ( 2x/(x+(-b))(x+b) )+(-2)·( 1/x ) )·d_{t}[x] = a

ln(x+(-b))+ln(x+b)+(-1)·ln(x^{2}) = at

1+(-1)·(b^{2}/x^{2}) = e^{at}

(b^{2}/x^{2}) = 1+(-1)·e^{at}

x(t) = b·( (1+(-1)·e^{at}) )^{(-1)·(1/2)}

d_{t}[x] = (ba)·(1/2)·( (1+(-1)·e^{at}) )^{(-1)·(3/2)}·e^{at}

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