fisica1-youn-e-freedman-exercicios-resolvidos

2.74: a) From Eq. (2.17), x(t) = αt –
β
t 3
3
(2.5), the acceleration is a(t) = –2βt = (– 4.00
= (4.00 m s )t – (0.667
3
m s )t.
3
m s )t 3 . From Eq.
α
b) The velocity is zero at t = ± (a = 0 at t = 0, but this is an inflection point, not an
β
extreme). The extreme values of x are then
The positive value is then
⎛
3
α β α ⎞
x = ± ⎜α − ⎟ = ±
⎜
3
β 3 β ⎟
⎝
⎠
2
3
3
2 ⎛ (4.00 m s) ⎞ 2 2
x = ⎜ ⎟ = 32 m
3
3
2.00 m s
⎝ ⎠ 3
1
2
3
α
.
β
= 3.77 m.
2.75: a) The particle's velocity and position as functions of time are
2
3
vx
( t)
= v0x
+ ∫ (( −2.00m s ) + (3.00m s ) t)
dt
x
= v
0x
t
0
2
⎛ 3.00 m s
− (2.00 m s ) t + ⎜
⎝ 2
t
2 2
3 3
( t)
= ∫ v
0
x
( t)
dt = v0xt
− (1.00 m s ) t + (0.50 m s ) t
3
⎞
⎟
t
⎠
2
3
= t(
v0x − (1.00 m s ) t + (0.50 m s ) t
where x 0 has been set to 0. Then, x(0) = 0, and to have x(4 s) = 0,
2
3
2
v
0 x
− (1.00m s )(4.00s) + (0.50m s )(4.00s) = 0,
which is solved for v
0x
= −4.0 m s. b) v
x
( 4s) = 12.0 m s.
2
,
2
),
2.76: The time needed for the egg to fall is
t
=
2∆h
9
=
2(46.0 m − 1.80 m)
(9.80 m s
2
=
)
3.00 s,
and so the professor should be a distance v y t = (1.20
m s )(3.00 s) = 3.60 m.