Chapter 2
Problems:
14, 18, 22, 23, 30, 36, 51, 52, 59, 61, 78
14
. (a) The average speed is
v
= 4800
km=
11
h
= 436
:
4
km=h:
(b) The plane travels at an average speed of
v
air
= 960
km=hr;
the total
distance covered is
s
= 6700
km:
The total time in the air is
t
air
=
s=v
air
=
6700
=
960 = 6
:
98
h:
Hence the time on the ground is
t
g
’
4
h:
18
sec
v
avg
=
x
t
=
2
:
5
2
= 1
:
25
m=s:
(b) After 4
sec
v
avg
=
x
t
=
0
4
= 0
m=s:
(c) After 6
sec
v
avg
=
x
t
=
2
6
=
:
333
m=s:
(d) During the interval from
t
= 3 sec
and
t
= 4 sec
v
avg
=
x
t
=
0
3
1
=
3
m=s:
23
. (a) The velocity for the rocket as a function of time is
v
=
dy
dt
=
d
dt
bt
ct
2
±
=
b
2
bt
= (82
9
:
8
t
)
m=s
2
:
(b) The velocity is zero when
t
= 82
=
9
:
8 = 8
:
4
s:
30
. The average acceleration during the fall is
a
v=
t
= 11
=
1
:
12 = 9
:
82
m=s
2
:
The average acceleration on contact with the ground is
a
v=
t
=
11
=:
131 =
84
:
0
m=s
2
1
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View Full Document36
. Given that
x
=
bt
3
;
where
b
= 1
:
5
m=s
3
;
(a)
v
at
t
= 2
:
5 sec
;
v
=
dx
dt
= 3
bt
2
= 3 (1
:
5) (2
:
5)
2
= 28
:
125
m=s;
(b)
a
at
t
= 2
:
5 sec
;
a
=
dv
dt
= 6
bt
= 6 (1
:
5) 2
:
5 = 22
:
5
m=s
2
;
(c) the average velocity during this time is
h
v
i
x=
t
=
bt
3
=t
=
bt
2
= (1
:
5) (2
:
5)
2
= 9
:
375
m=s;
(d) the average acceleration during this time is
h
a
i
v=
t
= 3
bt
2
=t
= 3
bt
= 4
:
5
2
:
5 = 11
:
25
m=s
2
:
51
. The acceleration of the airliner is
a
=
±
v
t
=
±
220
10
3
3600
1
29
=
±
2
:
107
m=s
2
Hence the distance required to stop is
x
=
v
i
t
+
1
2
at
2
=
220
10
3
3600
29
±
1
2
(2
:
107) 29
2
= 886
m:
Since the acceleration is constant this distance can also be found from
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 Spring '08
 Hicks
 Velocity, 10Km, 1 3M, 18km, 13km, 14km

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