fisica1-youn-e-freedman-exercicios-resolvidos

2.18: a) The velocity at t = 0 is
and the velocity at t = 5.00 s is
(3.00 m s ) + (0.100
(3.00 m s ) + (0.100
3
m s ) (0) = 3.00 m s ,
3
m s ) (5.00 s) 2 = 5.50 m s ,
so Eq. (2.4) gives the average acceleration as
(5.50 m s) −(3.00
m s)
(5.00 s)
=
2
.50 m s
.
b) The instantaneous acceleration is obtained by using Eq. (2.5),
Then, i) at t = 0, a x = (0.2
ii) at t = 5.00 s, a x = (0.2
dv
a 2 (0.2 m s
3 x
= = β t = ) t.
dt
3
m s ) (0) = 0, and
3
m s ) (5.00 s) = 1.0
2
m s .