fisica1-youn-e-freedman-exercicios-resolvidos

2.20: a) The bumper’s velocity and acceleration are given as functions of time by
dx
2
6 5
v x
= = ( 9.60 m s ) t − (0.600 m s ) t
dt
dv
2
6 4
a x
= = (9.60 m s ) − (3.000 m s ) t .
dt
There are two times at which v = 0 (three if negative times are considered), given by t =
0 and t 4 = 16 s 4 . At t = 0, x = 2.17 m and a x = 9.60 m s
2 . When t 4 = 16 s 4 ,
b)
x = (2.17 m) + (4.80 m s
2 ) (16 s
4 ) – (0.100)
a x = (9.60
m s
2 ) – (3.000
6
m s )(16 s 4 ) 3/2 = 14.97 m,
6
m s )(16 s 4 ) = –38.4 m s
2 .
2.21: a) Equating Equations (2.9) and (2.10) and solving for v 0 ,
2(
x − x0
) 2(70 m)
v0 x
= − vx
= −15.0 m s = 5.00 m s.
t 7.
00 s
b) The above result for v 0x may be used to find
vx
− v 15.0 m s 5.00 m s
0 x
−
2
a = =
1.43 m s ,
t
7.00 s
=
x
or the intermediate calculation can be avoided by combining Eqs. (2.8) and (2.12) to
eliminate v 0x and solving for a x ,
⎛ vx
x − x0 ⎞ ⎛15.0
m s 70.0 m ⎞
2
ax
= 2⎜
− 2
= 1.43 m s .
2
⎟ =
⎜ −
2
7.00 s (7 00 s)
⎟
⎝ t t ⎠ ⎝
. ⎠