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Pregunta a)
tramo de 0-10 s
aceleración? pendiente de la recta
v - 30 = (30/10) (t - 10)
a = 30/10
a = 3 m/s^2
distancia recorrida en ese tramo
v^2 = vi^2 + 2ax
x = v^2/(2a)
x = (30 m/s)^2 / (2*3m/s^2)
x = 150 m
tramo de 10-30 s. Movimiento uniforme
x = v*t
x = (30 m/s)(20s)
x = 600 m
tramo de 30-56 s. Movimiento retardado
v - 30 = (30/30 - 56) (t - 30)
a = -15/13 = -1,15 m/s^2
v^2 = vi^2 -2ax
-vi^2 / (-2a) = x
x = -(30m/s)^2 / (-2)(1,15 m/s^2)
x = 391,3 m
Distancia total recorrida = 150 + 600 + 391,3
= 1141,3 m
vii)
y = yi + vi(t) + (1/2)(g)(t)^2
y - yi = (1/2)(g)(t)^2
t^2 = 2 (20 m) / (9,8 m/s^2)
t = 2,02 s
viii)
v^2 = vi^2 + 2gy
y = -(5 m/s)^2 / (2)(-9,8 m/s^2)
y = 1,28 m
v = vi - gt
t = -vi/(-g)
t = - (5 m/s) / (-9,8 m/s^2)
t = 0,5 s