fisica1-youn-e-freedman-exercicios-resolvidos

2.50: a) From Eq. (2.15), the velocity v 2 at time t
v
2
= v
1
+
∫
t
t1
αt
dt
α 2 2
= v1 + ( t − t1
)
2
α 2 α 2
= v1
− t1
+ t
2 2
= (5.0 m s ) – (0.6
= (4.40 m s ) + (0.6
3
m s )(1.0 s) 2 + (0.6
3
m s ) t 2 .
m s ) + (0.6
At t 2 = 2.0 s, the velocity is v 2 = (4.40
to two significant figures.
b) From Eq. (2.16), the position x 2 as a function of time is
x2 = x1
+ ∫
t
t 1
v
x
dt
3
m s ) t 2
t
3 2
= (6.0 m) + ∫
t
((4.40 m /s) (0.6 m /s ) t ) dt
1
+
(0.6 m s
= (6.0 m) + (4.40 m s)( t − t ) +
3
At t = 2.0 s, and with t 1 = 1.0 s,
c)
x = (6.0 m) + (4.40
= 11.8 m.
m s )((2.0 s) – (1.0 s)) + (0.20
3
m s )(2.0 s) 2 = 6.80 m/s, or 6.8 m s
3
) 3 3
1
( t − t1
).
3
m s )((2.0 s) 3 – (1.0 s) 3 )