# Evidencia 1 fisica TERMINADA - Posicion 10 8 6 4 2 0-2 1 2...

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Posición: x=-8m Velocidad: V=4m/2 Aceleración: a=0m/s^2 Velocidad positiva cons F(x)=X0+V0t+1/2at2 a= F(x) = X0+V0t Tiempo 0 1 2 3 4 F(t) =V0+at a=0 F(t) = V0 Gráfica de posición 1 2 3 4 5 -10 -8 -6 -4 -2 0 2 4 6 8 10 Posicion Posicion 1 2 3 4 5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 Velocidad

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Tiempo 0 1 2 3 4 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Aceleracion
Tiempo Posicion 0 -8 1 -3.83 2 -0.16 3 4.16 4 8.16 m=tan =co/ca θ =75.96375653 θ m=-4 12 stante Con tiempo antes de chocar con el muro =0 Área del rectangulo = 36 Posicion final = X= X0+Área Posicion final = -10 Posición: X= 8m Velocidad: V= -11m/s Aceleración: a= 4m/s^2 Velocidad Aceleración constante positiva 4 4 F(x)=X0+V0t+1/2at2 4 4 V= V0+at 4 Gráfica de aceleración Área del triangulo = 60.5 Área del rectangulo = 121 Velocidad final = V= V0+Área Velocidad final = 49.5 Posicion final = X= X0+Área

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