Respuestas: Problemas Prácticos

1)Datos:
vi = 0 (m/s)
a = 8 (m/s2)
vf = vi + at = 0 (m/s) + 8 (m/s2) x 5 (s) = 40 (m/s)
d = vit + at2/2 = 0 (m/s) x 5 (s) + 8 (m/s2) x (5 (s))2 / 2 = 100 (m) 

2)Datos:
vi = 15 (km/h) = 4,167 (m/s)
vf = 60 (km/h) = 16,67 (m/s)
t = 20 (s)
a = (vf - vi)/t = (16,67 (m/s) - 4,167 (m/s))/20 (s) = 0,625 (m/s2)
d = vit + at2/2 = 4,167 (m/s) x 20 (s) + 0,625 (m/s2) x (20 (s))2/2 = 208,34 (m)

3)Datos:
vi = 15 (m/s)
a = 1 (m/s2) m/s2) x (6 (s))2/2 = 108 (m)

a) d = vit + at2/2 = 15 (m/s) x 6 (s) + 1 (m/s2) x (6 (s))2/2 = 108 (m)
b) d = vit + at2/2 = 15 (m/s) x 6 (s) + 1 (m/s2) x (-6 (s))2/2 = 72 (m)
t = (vf - vi)/a = (0 (m/s) - 15 (m/s))/(-1 (m/s2)) = 15 (s) 

4)Datos:
vi = 45 (km/h) = 12,5 (m/s)
vf = 15 (km/h) = 4,167 (m/s)
t = 5 (s)
a = (vf - vi)/t = (4,167 (m/s) - 12,5 (m/s))/5 (s) = -1,67 (m/s2)
d = vit + at2/2 = 12,5 (m/s) x 5 (s) + (-1,67 (m/s2)) x (5 (s))2/2 = 41,625 (m)

5)Datos:
vi = 12 (m/s)
vf = 5 (m/s)
d = 100 (m)
a) a = (vf2 - vi2)/2d = ((5(m/s))2 - (12 (m/s))2/(2 x 100 (m)) = - 0,595 (m/s2)
b) d = (vf2 - vi2)/2a = ((0(m/s))2 - (12 (m/s))2/(2 x (-0,595 (m/s2))) = 121 (m)

6)Datos:
vi = 10 (m/s)
a = 2 (m/s2
a) vf - vi = at = 2 (m/s2) x 60 (s) = 120 (m/s)
b) vf = vi + at = 10 (m/s) + 2 (m/s2) x 60 (s) = 130 (m/s)
c) v = (vf + vi)/2 = (130 (m/s) + 10 (m/s))/2 = 70 (m/s)
d) d = vit + at2/2 = 10 (m/s) x 60 (s) + 2 (m/s2) x (60 (s))2/2 = 4.200 (m)

7)Datos:
vi = 8 (m/s)
d = 640 (m)
t = 40 (s)
a) v = d/t = 640 (m)/40 (s= = 16 (m/s)
b) v = (vf + vi)/2, entonces vf = 2v - vi = 2 x 16 (m/s) - 8 (m/s) = 24 (m/s)
c) vf - vi = 24 (m/s) - 8 (m/s) = 16 (m/s)
d) a = (vf - vi)/t = (24 (m/s) - 8 (m/s))/40 (s) = 0,4 m/s2) 

8)Datos:
vi = 0 (m/s)
a = 5 (m/s2)
t = 4 (s)
vf = 0 (m/s) + 5 (m/s2) x 4 (s) = 20 (m/s)
d = vit + at2/2 = 0 (m/s) x 4 (s) + 5 (m/s2) x (4 (s))2/2 = 40 (m)

9)Datos:
vi = 0 (m/s)
t1 = 3 (s)
vf = 27 (m/s)
a = (vf - vi)/t = (27 (m/s) - 0 (m/s))/3 (s) = 9 (m/s2)
t2 = 6 (s)
vf = vi + at = 0 (m/s) + 9 (m/s2) x 6 (s) = 54 (m)

d = vit + at2/2 = 0 (m/s) x 6 (s) + 9 (m/s2) x (6 (s))2/2 = 162 (m)

10)Datos:
vi = 0 (m/s)
d = 250 (m)
vf = 80 (m/s)
a = (vf2 - vi2)/2d = ((80 (m/s))2 - (0 (m/s))2)/(2 x 250 (m)) = 12,8 (m/s2)

11)Datos:
vf = 600 (m/s)
d = 150 (cm) = 1,5 (m)
vi = 0 (m/s) El proyectil, antes de ser disparado está en reposo.
a = (vf2 - vi2)/2d = ((600 (m/s))2 - (0 (m/s))2)/(2 x 1,5 (m)) = 120.000 (m/s2)
Se habla de aceleración media debido a que en el interior del cañón, cuando se dispara el proyectil, la fuerza que lo impulsa no es constante, por lo que la aceleración tampoco lo es.

12)Datos:
vi = 20 (m/s)
vf = 60 (m/s)
d = 200 (m)
a = (vf2 - vi2)/2d = ((60 (m/s))2 - (20 (m/s))2)/(2 x 200 (m)) = 8 (m/s2)
t = (vf - vi)/a = (60 (m/s) - 20 (m/s))/8 (m/s2) = 5 (s)

13)Datos:
d = 1.800 (m)
t = 12 (s)
Suponiendo que parte del reposo:

d = vit + at2/2, despejando se tiene:
a = 2(d - vit)/t2 = 2x(1.800 (m) - 0 (m/s) x 12 (s))/(12 (s))2 = 25 (m/s)
b) vf = vi + at = 0 (m/s) + 25 (m/s2) x 12 (s) = 300 (m/s)
c) posición al primer segundo:
d = vit + at2/2 = 0 (m/s) x 1 (s) + 25 (m/s2) x (1 (s))2/ 2 = 12,5 (m)
posición al doceavo segundo:
d = vit + at2/2 = 0 (m/s) x 12 (s) + 25 (m/s2) x (12 (s))2/ 2 = 1.800 (m)
distancia entre el primero y el doceavo segundo:
d = 1.800 (m) - 12,5 (m) = 1.787,5 (m)

14)Datos:
vi = 60 (km/h) = 16,67 (m/s)
t = 44 (s)
vf = 0 (m/s)
a = (vf - vi)/t = (0 (m/s) - 16,67 (m/s))/44 (s) = -0,379 (m/s2)
d = vit + at2/2 = 16,67 (m/s) x 44 (s) + -0,379 (m/s2) x (44 (s))2/2 = 366,6 (m)

15)Datos:
vi = 40 (m/s)
a = -5 (m/s2)
t = 6 (s)
vf = vi + at = 40 (m/s) + -5 (m/s2) x 6 (s) = 10 (m/s)
v = (vi + vf)/2 = (40 (m/s) + 10 (m/s)/2 = 25 (m/s)
d = vt = 25 (m/s) x 6 (s) = 150 (m)

16)Datos:
vi = 0 (m/s)
d = 0,61 (m)
vf = 61 (m/s)
a = (vf2 - vi2)/2d = ((61 (m/s))2 - (0 (m/s))2)/(2 x 0,61 (m)) = 3.050 (m/s2)

17)Datos:
a = 9,8 (m/s2)
vi = 0 (m/s)
vf = vluz/10 = 3x108 (m/s) / 10 = 3x107 (m/s)
t = (vf - vi)/a = (3x107 (m/s) - 0 (m/s))/9,8 (m/s2) = 3.061.224,49 (s) = 35 dias 10 h 20 min 24,49 s
d = vit + at2/2 = 0 (m/s) x 3.061.224,49 (s) + 9,8 (m/s2) x (3.061.224,49 (s))2/2 = 4,59x1013 (m)

18)Datos:
vi = 100 (m/s)
a = -5 (m/s2)
vf = 0 (m/s)
t = (vf - vi)/a = (0 (m/s) - 100 (m/s))/(-5 (m/s2)) = 20 (s)
Para saber si puede aterrizar en una pista de 0,8 (km) = 800 (m) hay que calcular qué distancia recorre con la información que hay y luego se compara con esos 800 (m).
d = vit + at2/2 = 100 (m/s) x 20 (s) + -5 (m/s2) x (20 (s))2/2 = 1.000 (m)
Se observa que frenando a razón de -5 (m/s2) necesita 1.000 (m) de pista, por lo tanto no alcanza a aterrizar en una pista de 800 (m).