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Chapter 2 Problem Solutions (Easy)
2.3
v
=
!
r
!
t
=
(
26
+
385
=
1760
)
mi
(
2
+
756
=
3600
)
h
=
11
.
9
mi
h
"
5
.
30 m/s
,
2.14
(a)
v
=
!
r
=
!
t
=
4800 km
=
11 h
=
436 km/h.
(b) The actual flight time is
!
t
=
6700 km
=
(960 km/h)
=
6.98 h,
so
11 h
!
6.98 h
=
4.02 h
is spent on the ground.
2.22
From the given function,
x
(
t
)
=
bt
+
ct
3
, and the definition of average velocity (Equation 2-1),
v
=
!
x
=
!
t
, one obtains:
(a)
v
a
=
[
x
(3 s)
!
x
(1 s)]
=
(3 s
!
1 s)
=
[(1.5 m/s)(3 s
!
1 s)
+
(0.64 m/s
2
)(9 s
3
!
1 s
3
]
=
2 s
=
9.82 m/s;
(b)
v
b
=
[
x
(2.5 s)
!
x
(1.5 s)]
=
(2.5 s
!
1.5 s)
=
9.34 m/s;
(c)
v
c
=
[
x
(2.05 s)
!
x
(1.95 s)]
=
(2.05 s
!
1.95 s)
=
9.18 m/s.
(d) The instantaneous velocity
(Equation 2-2b) is
v
(
t
)
=
dx
=
dt
=
b
+
3
ct
2
.
At
t
=
2.00 s,
v
(2 s)
=
1.5 m/s
+
3(0.64 m/s
3
)(2 s)
2
=
9.18 m/s. As the interval surrounding 2 s gets smaller, the average and instantaneous velocities
agree better; the values in parts (c) and (d) differ by less than 0.02%.
2.30
The velocity (positive downward) changes from
v
1
=
0
to
v
2
=
11.0
m/s in 1.12 s
while falling with average acceleration
a
=
!
v=
!
t
=
(11.0 m/s)
=
(1.12 s)
=
9.82
m/s
2
.
When stopping, the egg’s velocity changes from
v
2
=
11.0
m/s to
v
3
=
0
in 0.131 s,
with average acceleration
a
=
(0
!
11.0 m/s
=
0.131 s
=
!
84.0 m/s
2
. (A negative sign
means that the acceleration is upward; when the speed is decreasing, this is
called a deceleration.)
2.36
(a)
v
(
t
)
=
dx
=
dt
=
3
bt
2
,
and
v
(2.5 s)
=
3(1.5 m/s
3
)(2.5 s)
2
=
28.1 m/s
.
(b)
a
(
t
)
=
d
dt
=
6
bt
,
and
a
(2.5 s)
=
6(1.5 m/s
3
)
(2.5 s)
=
22.5 m/s
2
.
(c)
v
=
[
x
(2.5 s)
!
x
(0)]
=
2.5 s
=
(1.5 m/s
3
)(2.5 s)
2
=
9.38 m/s.
(d)
a
=
[
v
(2.5 s)
!
v
(0)]
=
2.5 s
=
3
)(2.5 s)
=
11.3 m/s
2
.
2.51
From Equation 2-9 with
v
=
0,
we find
x
!
x
0
=
1
2
v
0
t
=
1
2
(220 km/h)(29 h/3600)
=
886 m
(over half a mile).

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