worksheets answer key
TRANSCRIPT
Motion In One Dimension
Study Guide Answers
1. Yes, from t1 to t4 and from t6 to t7.
2. Yes, from t4 to t5.
3. greater than
4. greater than
5. Yes, from 0 to t1 and from t5 to t6.
6. Yes, from t1 to t2, from t2 to t4, from t4 to t5, and fromt6 to t7.
7. −5.0 m (or 5.0 m to the west of where it started)
Displacement and Velocity, p. 7
Holt Physics Solution ManualIII–2
III
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ed.
1. vf = 0. The car is stopped.
2. vi = 2
∆∆t
x
3. a = −∆v
ti
4. a = −2∆
vi
x
2
5. vi = −a∆t ∆x = 12
vi∆t
Acceleration, p. 8
1. a. −g
b. initial speed = g(∆t/2)
c. elapsed time = ∆t/2
d. height = g∆t 2/8
2. a. −9.81 m/s2
b. 12 m/s
2. a. vf = a(∆t)
b. vf = vi + a(∆t); ∆x = 12
(vi + vf )∆t or ∆x = vi(∆t) +12
a(∆t)2
c. 1.2 s
Falling Objects, p. 9
1. a. t1 = d1/v1; t2 = d2/v2; t3 = d3/v3
b. total distance = d1 + d2 + d3
c. total time = t1 + t2 + t3
3.
Time interval Type of motion v (m/s) a(m/s2)
A speeding up + +B speeding up + +C constant velocity + 0
D slowing down + −E slowing down + −
4. a. b. 1 s
c. 2 sTime (s) Position (m) v (m/s) a(m/s2)
1 4.9 0 −9.81
2 0 −9.8 −9.81
3 −14.7 −19.6 −9.81
4 −39.2 −29.4 −9.81
Mixed Review, pp. 11–12
Section Two — Problem Workbook Solutions II Ch. 2–1
Motion In One Dimension
Problem Workbook Solutions
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olt,
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1. ∆x = 443 m
vavg = 0.60 m/s∆t =
v
∆
av
x
g =
0
4
.6
4
0
3
m
m
/s = 740 s = 12 min, 20 s
Additional Practice A
Givens Solutions
2. vavg = 72 km/h
∆x = 1.5 km∆t =
v
∆
av
x
g = = 75 s
1.5 km
�72 k
h
m��36
1
0
h
0 s�
3. ∆x = 5.50 × 102 m
vavg = 1.00 × 102 km/h
vavg = 85.0 km/h
a. ∆t = v
∆
av
x
g = =
b. ∆x = ∆vavg∆t
∆x = (85.0 km/h)�36
1
0
h
0 s� �
1
1
0
k
3
m
m�(19.8 s) = 468 m
19.8 s5.50 × 102 m
�1.00 × 102 k
h
m� �36
1
0
h
0 s� �11
00
k
0
m
m�
4. ∆x1 = 1.5 km
v1 = 85 km/h
∆x1 = 0.80 km
v2 = 67 km/h
a. ∆ttot = ∆t1 + ∆t2 = ∆v
x
1
1 + ∆v
x
2
2
∆ttot = + = 64 s + 43 s =
b. vavg = = = = 77 km/h2.3 km
(107 s)�36
1
0
h
0 s�
1.5 km + 0.80 km(64 s + 43 s)�
3
1
60
h
0�
∆x1 + ∆x2∆t1 + ∆t2
107 s0.80 km
�67 k
h
m��36
1
0
h
0 s�
1.5 km
�85 k
h
m��36
1
0
h
0 s�
5. r = 7.1 × 104 km
∆t = 9 h, 50 min
∆x = 2πr
vavg = = =
vavg =4.5 × 108 m
(590 min)�16
m
0
i
s
n�
4.5 × 108 m
(540 min + 50 min)�16
m
0
i
s
n�
2π(7.1 × 107 m)
�(9 h)�60
1
m
h
in� + 50 min��1
6
m
0
i
s
n�
∆x∆t
vavg =
Thus the average speed = 1.3 × 104 m/s.
On the other hand, the average velocity for this point is zero, because the point’s dis-placement is zero.
1.3 × 104 m/s
Holt Physics Solution ManualII Ch. 2–2
Givens Solutions
6. ∆x = –1.73 km
∆t = 25 s
7. vavg,1 = 18.0 km/h
∆t1 = 2.50 s
∆t2 = 12.0 s
a. ∆x1 = vavg,1∆t1 = (18.0 km/h)�36
1
0
h
0 s� �110k
3
m
m�(2.50s) = 12.5 m
∆x2 = –∆x1 = –12.5 m
vavg,2 = ∆∆
x
t2
2 = –
1
1
2
2
.
.
0
5
s
m =
b. vavg,tot = ∆∆x
t1
1
++
∆∆
x
t2
2 = 12.
2
5
.5
m
0
+s +
(−1
1
2
2
.0
.5
s
m) = =
c. total distance traveled = ∆x1 – ∆x2 = 12.5 m – (–12.5 m) = 25.0 m
total time of travel = ∆t1 + ∆t2 = 2.50 s + 12.0 s = 14.5 s
average speed = to
t
t
o
a
t
l
a
d
l
i
t
s
i
t
m
an
e
ce =
2
1
5
4
.
.
0
5
m
s = 1.72 m/s
0.0 m/s0.0 m14.5 s
–1.04 m/s
8. ∆x = 2.00 × 102 km
∆t = 5 h, 40 min, 37 s
vavg� = (1.05)vavg
∆x� = 12
∆x
a. vavg = ∆∆
x
t = =
2.0
2
0
0
×43
1
7
05
s
m
vavg =
b. ∆t = = = 9.73 × 103 s
�2.00 ×2
105 m�
(1.05)�9.79 m
s�
∆x�vavg�
9.79 m/s = 35.2 km/h
2.00 × 105 m
��5 h� �360
h
0 s� + �40 min� �
m
60
in
s� + 37 s�
vavg = = –1.73
2
×5
1
s
03 m = –69 m/s = –250 km/h
∆x∆t
∆t = (9.73 × 103 s)�36
1
0
h
0 s� = 2.70 h
(0.70 h)�60
1
m
h
in� = 42 min
∆t = 2 h, 42 min
II
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Section Two — Problem Workbook Solutions II Ch. 2–3
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olt,
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ed.
1. vi = 0 km/h = 0 m/s
aavg = 1.8 m/s2
∆t = 1.00 min
vf = aavg ∆t + vi = (1.80 m/s2)(1.00 min)�16
m
0
i
s
n� + 0 m/s =
vf = 108 m/s = (108 m/s) �36
1
0
h
0 s��110k
3m
m� = 389 km/h
108 m/s
Additional Practice B
Givens Solutions
2. ∆t = 2.0 min
aavg = 0.19 m/s2
vi = 0 m/s
vf = aavg ∆t + vi = (0.19 m/s2) (2.0 min)�16
m
0
i
s
n� + 0 m/s = 23 m/s
3. ∆t = 45.0 s
aavg = 2.29 m/s2
vi = 0 m/s
vf = aavg ∆t + vi = (2.29 m/s2)(45.0 s) + 0 m/s = 103 m/s
5. ∆x = (15 hops) �110
h
.0
o
m
p�
= 1.50 × 102 m
∆t = 60.0 s
∆t stop = 0.25 s
vf = 0 m/s
vi = vavg = +2.50 m/s
a. vavg = ∆∆
x
t =
1.50
60
×.0
10
s
2 m =
b. aavg = v
∆f
t
−
sto
v
p
i = 0 m/s
0
−.2
2
5
.5
s
0 m/s =
−0
2
.2
.5
5
0
m
m
/
/
s
s = −1.0 × 101 m/s2
+2.50 m/s
6. ∆x = 1.00 × 102 m, backward= −1.00 × 102 m
∆t = 13.6 s
∆t � = 2.00 s
vi = 0 m/s
vf = vavg
vavg = ∆∆
x
t =
−1.00
13
×.6
1
s
02 m = −7.35 m/s
aavg = vf
∆−t�
vi = = 3.68 m/s2−7.35 m/s − 0 m/s
2.00 s
7. ∆x = 150 m
vi = 0 m/s
vf = 6.0 m/s
vavg = 3.0 m/s
a. ∆t = v
∆
av
x
g =
3
1
.
5
0
0
m
m
/s =
b. aavg = vf
∆−t
vi = 6.0
5
m
.0
/s
×−1
0
01
m/s = 0.12 m/s2
5.0 × 101 s
4. ∆x = 29 752 m
∆t = 2.00 h
vi = 3.00 m/s
vf = 4.13 m/s
∆t = 30.0 s
a. vavg = ∆∆
x
t = =
b. aavg = ∆∆
v
t = =
1.
3
1
0
3
.0
m
s
/s = 3.77 × 10−2 m/s24.13 m/s − 3.00 m/s
30.0 s
4.13 m/s29 752 m
(2.00 h)�36
1
0
h
0 s�
Holt Physics Solution ManualII Ch. 2–4
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olt,
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8. vi = +245 km/h
aavg = −3.0 m/s2
vf = vi −(0.200) vi
Givens Solutions
9. ∆x = 3.00 km
∆t = 217.347 s
aavg = −1.72 m/s2
vf = 0 m/s
vi = vavg = ∆∆
x
t = = 13.8 m/s
tstop = vf
aa
−
vg
vi = = = 8.02 s−13.8 m/s−1.72 m/s2
0 m/s − 13.8 m/s
−1.72 m/s2
3.00 × 103 m
217.347 s
10. ∆x = +5.00 × 102 m
∆t = 35.76 s
vi = 0 m/s
∆t � = 4.00 s
vmax = vavg + (0.100) vavg
vf = vmax = (1.100)vavg = (1.100)�∆∆
x
t� = (1.100)�5.0
3
0
5
×.7
1
6
0
s
2 m� = +15.4 m/s
aavg = ∆∆t
v
� =
vf
∆−t �
vi = = + 3.85 m/s215.4 m/s − 0 m/s
4.00 s
1. ∆x = 115 m
vi = 4.20 m/s
vf = 5.00 m/s
∆t = vi
2∆+
x
vf =
4.20
(
m
2)
/
(
s
1
+15
5.
m
00
)
m/s =
(2
9
)
.
(
2
1
0
1
m
5
/
m
s
) = 25.0 s
Additional Practice C
2. ∆x = 180.0 km
vi = 3.00 km/s
vf = 0 km/s
∆t = vi
2∆+
x
vf = =
3
3
.
6
0
0
0
.0
km
km
/s = 1.2 × 102 s
(2)(180.0 km)3.00 km/s + 0 km/s
3. vi = 0 km/h
vf = 1.09 × 103 km/h
∆x = 20.0 km
∆x = 5.00 km
vi = 1.09 × 103 km/h
vf = 0 km/h
a. ∆t = vi
2
+∆x
vf =
∆t = =
b. ∆t = vi
2
+∆x
vf =
∆t = = 33.0 s10.0 × 103 m
(1.09 × 103 km/h)�36
1
0
h
0 s��11
00
k
0
m
m�
(2)(5.00 × 103 m)
(1.09 × 103 km/h + 0 km/h)�36
1
0
h
0 s��11
00
k
0
m
m�
132 s40.0 × 103 m
(1.09 × 103 km/h)�36
1
0
h
0 s��11
00
k
0
m
m�
(2)(20.0 × 103 m)
(1.09 × 103 km/h + 0 km/h)�36
1
0
h
0 s��11
00
k
0
m
m�
vi = �245 k
h
m��36
1
0
h
0 s� �110k
3
m
m� = +68.1 m/s
vf = (1.000 − 0.200) vi = (0.800)(68.1 m/s) = +54.5 m/s
∆t = vf
aa
−
vg
vi = 54.5 m
−3
/s
.0
−m
6
/
8
s
.21 m/s
= = 4.5 s−13.6 m/s− 3.0 m/s2
Section Two — Problem Workbook Solutions II Ch. 2–5
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olt,
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4. vi = vavg = 518 km/h
vf = (0.600) vavg
∆t = 2.00 min
vavg = �518 k
h
m��60
1
m
h
in��11
0
k
3
m
m� = 8.63 × 103 m/min
∆x = 12
(vi + vf)∆t = 12
[vavg + (0.600) vavg]∆t = 12
(1.600)(8.63 × 103 m/min)(2.00 min)
∆x = 13.8 × 103 m = 13.8 km
Givens Solutions
5. ∆t = 30.0 s
vi = 30.0 km/h
vf = 42.0 km/h
∆x = 12
(vi + vf)∆t = 12
(30.0 km/h + 42.0 km/h) �36
1
0
h
0 s�(30.0 s)
∆x = 12
�72.0 k
h
m��36
1
0
h
0 s�(30.0 s)
∆x = 3.00 × 10−1 km = 3.00 × 102 m
1. vi = 186 km/h
vf = 0 km/h = 0 m/s
a = −1.5 m/s2
∆t = vf −
a
vi = = −−
5
1
1
.5
.7
m
m
/s
/2s
= 34 s
0 m/s − (186 km/h) �36
1
0
h
0 s� �11
0
k
3
m
m�
−1.5 m/s2
Additional Practice D
6. vf = 96 km/h
vi = 0 km/h
∆t = 3.07 s
∆x = 12
(vi + vf)∆t = 12
(0 km/h + 96 km/h) �36
1
0
h
0 s��11
0
k
3
m
m�(3.07 s)
∆x = 12
�96 × 103 m
h�(8.53 + × 10−4 h) = 41 m
7. ∆x = 290.0 m
∆t = 10.0 s
vf = 0 km/h = 0 m/s
vi = 2
∆∆t
x − vf =
(2)(
1
2
0
9
.
0
0
.0
s
m) − 0 m/s =
(Speed was in excess of 209 km/h.)
58.0 m/s = 209 km/h
8. ∆x = 5.7 × 103 km
∆t = 86 h
vf = vi + (0.10) vi
vf + vi = 2
∆∆t
x
vi (1.00 + 0.10) + vi = 2
∆∆t
x
vi = (2)
(
(
2
5
.
.
1
7
0
×)(
1
8
0
6
3
h
k
)
m) = 63 km/h
9. vi = 2.60 m/s
vf = 2.20 m/s
∆t = 9.00 min
∆x = 12
(vi + vf)∆t = 12
(2.60 m/s + 2.20 m/s)(9.00 min)�m
60
in
s� = 1
2(4.80 m/s)(5.40 × 102 s)
∆x = 1.30 × 103 m = 1.30 km
2. vi = −15.0 m/s
vf = 0 m/s
a = +2.5 m/s2
vi = 0 m/s
vf = +15.0 m/s
a = +2.5 m/s
For stopping:
∆t1 = vf
a
− vi = 0 m/s
2
−.5
(−m
1
/
5
s
.2
0 m/s) =
1
2
5
.5
.0
m
m
/s
/2s
= 6.0 s
For moving forward:
∆t2 = vf
a
− vi = = 1
2
5
.5
.0
m
m
/s
/2s
= 6.0 s
∆t tot = ∆t1 + ∆t2 = 6.0 s + 6.0 s = 12.0 s
15.0 m/s − 0.0 m/s
2.5 m/s2
Holt Physics Solution ManualII Ch. 2–6
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olt,
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t and
Win
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serv
ed.
3. vi = 24.0 km/h
vf = 8.0 km/h
a = −0.20 m/s2
∆t = vf −
a
vi
∆t =
∆t = = 22 s
�−16.0 k
h
m��36
1
0
h
0 s��11
0
k
3
m
m�
−0.20 m/s2
(8.0 km/h − 24.0 km/h) �36
1
0
h
0 s� �11
0
k
3
m
m�
−0.20 m/s2
4. v1 = 65.0 km/h
vi,2 = 0 km/h
a2 = 4.00 × 10−2 m/s2
∆x = 2072 m
For cage 1:
∆x = v1∆t1
∆t1 = ∆v1
x = =
For cage 2:
∆x = vi,2∆t2 + 12
a2∆t22
Because vi,2 = 0 km/h,
∆t2 = �2
a
∆�2
x� = �
4.�0
(2
0�)
×(�2
1
0�0
7−�2
2� m
m�)
/s�2� =
Cage 1 reaches the bottom of the shaft in nearly a third of the time required for cage 2.
322 s
115 s2072 m
(65.0 km/h)�36
1
0
h
0 s� �110k
3
m
m�
5. ∆x = 2.00 × 102 m
v = 105.4 km/h
vi,car = 0 m/s
a. ∆t = = =
b. ∆x = vi,car ∆t + 12
acar∆t2
acar = 2
∆∆t2
x =
(2)(2
(
.
6
0
.
0
83
×s
1
)
02
2 m) = 8.57 m/s2
6.83 s2.00 × 102 m
�105.4 k
h
m� �36
1
0
h
0 s� �110k
3
m
m�
∆xv
Givens Solutions
7. vi = 3.17 × 102 km/h
vf = 2.00 × 102 km/h
∆t = 8.0 s
a = vf
∆−t
vi =
a = =
∆x = vi∆t + 12
a∆t2 = (3.17 × 102 km/h)�36
1
0
h
0 s��11
0
k
3
m
m�(8.0 s) + 1
2(−4.1 m/s2)(8.0 s)2
∆x = (7.0 × 102 m) + (−130 m) = +570 m
−4.1 m/s2
(−117 km/h)�36
1
0
h
0 s��11
0
k
3
m
m�
8.0 s
(2.00 × 102 km/h − 3.17 × 102 km/h)�36
1
0
h
0 s��11
0
k
3
m
m�
8.0 s
6. vi = 6.0 m/s
a = 1.4 m/s2
∆t = 3.0 s
∆x = vi∆t + 12
a∆t2 = (6.0 m/s)(3.0 s) + 12
(1.4 m/s2)(3.0 s)2 = 18 m + 6.3 m = 24 m
Section Two — Problem Workbook Solutions II Ch. 2–7
II
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yrig
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olt,
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t and
Win
ston
.All
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s re
serv
ed.
10. vi = 24.0 m/s
a = −0.850 m/s2
∆t = 28.0 s
vf = vi + a∆t = 24.0 m/s + (− 0.850 m/s2)(28.0 s) = 24.0 m/s − 23.8 m/s = +0.2 m/s
Givens Solutions
11. a = +2.67 m/s2
∆t = 15.0 s
∆x = +6.00 × 102m
vi ∆t = ∆x − 12
a∆t2
vi = ∆∆
x
t − 1
2a∆t =
6.00
15
×.0
10
s
2 m − 1
2(2.67 m/s2)(15.0 s) = 40.0 m/s − 20.0 m/s = +20.0 m/s
12. a = 7.20 m/s2
∆t = 25.0 s
vf = 3.00 × 102 ms
vi = vf − a∆t
vi = (3.00 × 102 m/s) − (7.20 m/s2)(25.0 s) = (3.00 × 102 m/s) − (1.80 × 102 m/s)
vi = 1.20 × 102 m/s
13. vi = 0 m/s
∆x = 1.00 × 102 m
∆t = 12.11 s
∆x = vi∆t + 12
a∆t2
Because vi = 0 m/s,
a = 2
∆∆t2
x =
(2)(
(
1
1
.0
2
0
.1
×1
1
s)
02
2 m) = 1.36 m/s2
8. vi = 0 m/s
vf = 3.06 m/s
a = 0.800 m/s2
∆t2 = 5.00 s
∆t1 = vf
a
− vi = 3.0
0
6
.
m
80
/
0
s
m
−/
0
s2m/s
= 3.82
∆x1 = vi∆t1 + 12
a∆t12 = (0 m/s) (3.82 s) + 1
2(0.800 m/s2) (3.82 s)2 = 5.84 m
∆x2 = vf∆t2 = (3.06 m/s)(5.00 s) = 15.3 m
∆xtot = ∆x1 + ∆x2 = 5.84 m + 15.3 m = 21.1 m
9. vf = 3.50 × 102 km/h
vi = 0 km/h = 0 m/s
a = 4.00 m/s2
∆t = (vf
a
− vi) = =
∆x = vi∆t + 12
a∆t2 = (0 m/s)(24.3 s) + 12
(4.00 m/s2)(24.3 s)2
∆x = 1.18 × 103 m = 1.18 km
24.3 s
(3.50 × 102 km/h − 0 km/h) �36
1
0
h
0 s� �11
0
k
3
m
m�
(4.00 m/s2)
14. vi = 3.00 m/s
∆x = 1.00 × 102 m
∆t = 12.11 s
a = 2(∆x
∆−t 2
vi∆t) =
a =
a = (
(
2
1
)
2
(
.
6
1
4
1
m
s)2)
= 0.87 m/s2
(2)(1.00 × 102 m − 36.3 m)
(12.11 s)2
(2)[1.00 × 102 m − (3.00 m/s)(12.11 s)]
(12.11 s)2
15. vf = 30.0 m/s
vi = 18.0 m/s
∆t = 8.0 s
a = vf
∆−t
vi = = 12
8
.0
.0
m
s
/s = 1.5 m/s230.0 m/s − 18.0 m/s
8.0 s
Holt Physics Solution ManualII Ch. 2–8
II
Cop
yrig
ht ©
by H
olt,
Rin
ehar
t and
Win
ston
.All
right
s re
serv
ed.
1. vi = 0 km/h
vf = 965 km/h
a = 4.0 m/s2∆x =
vf2
2
−a
vi2
=
∆x = 7.19
8
×.0
1
m
04
/
m
s2
2/s2
= 9.0 × 103 m = 9.0 km
�(965 km/h) 2 − (0 km/h)2��36
1
0
h
0 s�
2
�110k
3
m
m�
2
(2)(4.0 m/s2)
2. vi = (0.20) vmax
vmax = 2.30 × 103 km/h
vf = 0 km/h
a = −5.80 m/s2
∆x = vf
2
2
−a
vi2
=
∆x = −1.6
−3
11
×.6
1
m
04
/
m
s2
2/s2
= 1.41 × 103 m = 1.41 km
�(0 km/h)2 − (0.20)2 (2.30 × 103 km/h)2� �36
1
0
h
0 s�
2
�110k
3
m
m�
2
(2)(−5.80 m/s2)
3. vf = 9.70 × 102 km/h
vi = (0.500)vf
a = 4.8 m/s2
∆x = vf
2
2
−a
vi2
=
∆x =
∆x =
∆x = = 5.7 × 103 m = 5.7 km5.45 × 104m2/s2
9.6 m/s2
(7.06 × 105 km2/h2)�36
1
0
h
0 s�
2
�110k
3
m
m�
2
(2)(4.8 m/s2)
(9.41 × 105 km2/h)2 −2.35 × 105 km2/h2)�36
1
0
h
0 s�
2
�110k
3
m
m�
2
(2)(4.8 m/s2)
�(9.70 × 102 km/h)2 −(0.50)2 (9.70 × 102 km/h)2��36
1
0
h
0 s�
2
�110k
3
m
m�
2
(2)(4.8 m/s2)
Additional Practice E
Givens Solutions
4. vi = 8.0 m/s
∆x = 40.0 m
a = 2.00 m/s2
vf =√
2a�∆�x� +� v�i2� =
√(2�)(�2.�0�m�/s�2)�(4�0.�m�)�+� (�8.�0�m�/s�)2� =
√1.�60� ×� 1�02�m�2/�s2� +� 6�4�m�2/�s2�
vf =√
22�4�m�2/�s2� = ± 15 m/s = 15 m/s
5. ∆x = +9.60 km
a = −2.0 m/s2
vf = 0 m/s
vi = vf�2−� 2�a∆�x� = (0� m�/s�)2� −� (�2)�(−�2.�0�m�/s�2)�(9�.6�0�×� 1�03� m�)�
vi =√
3.�84� ×� 1�04�m�2/�s2� = ±196 m/s = +196 m/s
7. ∆x = 44.8 km
∆t = 60.0 min
a = −2.0 m/s2
∆x = 20.0 m
vi = 12.4 m/s
a. vavg = ∆∆
x
t = =
b. vf =√
2a�∆�x�+� v�i2� =
√(2�)(�−�2.�0�m�/s�2)�(2�0.�0�m�)�+� (�12�.4� m�/s�)2�
vf =√
(−�80�.0� m�2/�s2�)�+� 1�54� m�2/�s2� =√
74� m�2/�s2� = ±8.6 m/s = 8.6 m/s
12.4 m/s44.8 × 103 m
(60.0 min)(60 s/min)
6. a = +0.35 m/s2
vi = 0 m/s
∆x = 64 m
vf =√
2a�∆�x� +� v�i2� =
√(2�)(�0.�35� m�/s�2)�(6�4�m�)�+� (�0�m�/s�)2�
vf =√
45� m�2s�2� = ±6.7 m/s = +6.7 m/s
Section Two — Problem Workbook Solutions II Ch. 2–9
II
Cop
yrig
ht ©
by H
olt,
Rin
ehar
t and
Win
ston
.All
right
s re
serv
ed.
3. ∆y = −443 m + 221 m = −222 m
a = −9.81 m/s2
vi = 0 m/s
vf =√
2a�∆�y� −� v�i2� =
√(2�)(�−�9.�81� m�/s�2)�(−�22�2�m�)�−� (�0�m�/s�)2� =
√43�60� m�2/�s2�
vf = ±66.0 m/s = −66.0 m/s
Givens Solutions
4. ∆y = +64 m
a = −9.81 m/s2
∆t = 3.0 s
∆y = vi∆t + 12
a∆t2
vi = = = 64 m
3.
+0
4
s
4 m
vi = 1
3
0
.
8
0
m
s = 36 m/s initial speed of arrow = 36 m/s
3.0 s
∆t
5. ∆y = −111 m
∆t = 3.80 s
a = −9.81 m/s2
∆y = vi ∆t + 12
a∆t2
vi = = =
vi = −
3
4
.
0
8
.
0
2
s
m = −10.6 m/s
−111 m + 70.8 m
3.80 s
3.80 s
∆t
9. ∆x = 4.0 × 102 m
∆t = 11.55
vi = 0 km/h
vf = 2.50 × 102 km/h
1. ∆y = −343 m
a = −9.81 m/s2
vi = 0 m/s
vf =√
2a�∆�y� +� v�i2� =
√(2�)(�−�9.�81� m�/s�2)�(−�34�3�m�)�+� (�0�m�/s�)2� =
√67�30� m�2/�s2�
vf = ±82.0 m/s = −82.0 m/s
Additional Practice F
2. ∆y = +4.88 m
vi = +9.98 m/s
a = −9.81 m/s2
vf =√
2a�∆�y� +� v�i2� =
√(2�)(�−�9.�81� m�/s�2)�(4�.8�8�m�)�+� (�9.�98� m�/s�)2� =
√−�95�.7� m�2/�s2� +� 9�9.�6�m�2/�s2�
vf =√
3.�90� m�2/�s2� = ±1.97 m/s = ±1.97 m/s
a = vf
2
2∆−
x
vi2
=
a = 4.8
8
2
.0
××1
1
0
0
3
2m
m
2/s2
= 6.0 m/s2
�(2.50 × 102 km/h)2 − (0 km/h)2��36
1
0
h
0 s�
2
�110
k
3
m
m�
2
(2)(4.0 × 102 m)
8. ∆x = 2.00 × 102 m
a = 1.20 m/s2
vf = 25.0 m/s
vi =√
vf�2�−� 2�a∆�x� =√
(2�5.�0�m�/s�)2� −� (�2)�(1�.2�0�m�/s�2)�(2�.0�0�×� 1�02� m�)�
vi =√
62�5�m�2/�s2� −� 4�.8�0�×� 1�02�m�2/�s2�
vi =√
14�5�m�2/�s2� = ±12.0 m/s = 12.0 m/s
10. vi = 25.0 km/h
vf = 0 km/h
∆x = 16.0 m
a = vf
2
2∆−
x
vi2
=
a = = −1.51 m/s2−4.82 m2/s2
32.0 m
�(0 km/h)2 − (25.0 km/h)2��36
1
0
h
0 s�
2
�110
k
3
m
m�
2
(2)(16.0 m)
64 m − 12
(−9.81 m/s2)(3.0 s)2
∆y − 12
a∆t2 −111 m − 12
(−9.81 m/s2)(3.80 s)2
∆y − 12
a∆t2
Holt Physics Solution ManualII Ch. 2–10
II
Cop
yrig
ht ©
by H
olt,
Rin
ehar
t and
Win
ston
.All
right
s re
serv
ed.
9. ∆ymax = +21 cm
a = −9.81 m/s2
vf = 0 m/s
∆y = +7.0 cm
vi =√
vf�2�−� 2�a�∆�ym�ax� =√
(0� m�/s�)2� −� (2�)(�−9�.8�1�m�/s�2)�(2�.1� ×� 1�0−�1�m�)� =√
4.�1�m�2/�s2�
vi = +2.0 m/s
For the flea to jump +7.0 cm = +7.0 × 10−2 m = ∆y ,
∆y = vi∆t + 12
a∆t2 or 12
a∆t2 + vi∆t − ∆y = 0
Solving for ∆t by means of the quadratic equation,
∆t =
∆t =
∆t = =
∆t = = 0.37 s or 0.04 s
To choose the correct value for ∆t, insert ∆t, a, and vi into the equation for vf .
vf = a∆t + vi = (−9.81 m/s2)(0.37 s) + 2.0 m/s
vf = (−3.6 m/s) + 2.0 m/s = −1.6 m/s
vf = a∆t + vi = (−9.81 m/s2)(0.04 s) + 2.0 m/s
vf = (−0.4 m/s) + 2.0 m/s = +1.6 m/s
Because vf is still directed upward, the shorter time interval is correct. Therefore,
∆t = 0.04 s
2.0 m/s ± 1.6 m/s
9.81 m/s2
2.0 m/s ±√
2.�6�m�2/�s2�9.81 m/s2
−2.0 m/s ±√
4.�0�m�2/�s2� −� 1�.4� m�2/�s2�−9.81 m/s2
−2.0 m/s ±√
(2�.0� m�/s�)2� −� (�2)�(−�9.�81� m�/s�2)�(−�7.�0�×� 1�0−�2�m�)�−9.81 m/s2
−vi ± �(v�i)�2�−� 4��2
a���(−�∆�y�)�
2�2
a�
6. ∆y = −228 m
a = −9.81 m/s2
vi = 0 m/s
When vi = 0 m/s,
∆t = �2
a
∆�y� = �� =
In the presence of air resistance, the sandwich would require more time to fall be-cause the downward acceleration would be reduced.
6.82 s(2)(−228 m)−9.81 m/s2
Givens Solutions
7. vi = 12.0 m/s, upward =+12.0 m/s
vf = 3.0 m/s, upward =+3.0 m/s
a = −9.81 m/s2
yi = 1.50 m
∆y = vf
2
2
−a
vi2
= =
∆y = = 6.88 m
height of nest from ground = h
∆y = h − yi h = ∆y + yi = 6.88 m + 1.50 m = 8.38 m
−135 m2/s2
−19.6 m/s2
9.0 m3/s2 − 144 m2/s2
(2)(−9.81 m/s2)
(3.0 m/s)2 − (12.0 m/s)2
(2)(−9.81 m/s2)
8. ∆y = +43 m
a = −9.81 m/s2
vf = 0 m/s
Because it takes as long for the ice cream to fall from the top of the flagpole to theground as it does for the ice cream to travel up to the top of the flagpole, the free-fallcase will be calculated.
Thus, vi = 0 m/s, ∆y = −43 m, and ∆y = 12
a∆t2.
∆t = �2
a
∆�y� = �� = 3.0 s
(2)(−43 m)−9.81 m/s2
Section Five—Problem Bank V Ch. 2–1
V
Cop
yrig
ht ©
Hol
t, R
ineh
art a
nd W
inst
on.A
ll rig
hts
rese
rved
.
Motion In One Dimension
Problem Bank Answers
1. ∆x = 3.33 km forward∆t = 30.0 s vavg =
∆∆
x
t =
3.33
3
×0.
1
0
0
s
3 m =
vavg = (111 m/s)(3600 s/h)(10−3 km/m) = 4.00 × 102 km/h forward
111 m/s forward
Additional Practice A
Givens Solutions
2. ∆x = 15.0 km west∆t = 15.3 s
vavg = ∆∆
x
t = =
4.2
1
5
5
×.0
1
k
0
m−3 h
= 3.53 × 103 km/h west15.0 km
(15.3 s)�36
1
0
h
0 s�
3. ∆x = 4.0 m∆t = 5.0 min vavg =
∆∆
x
t = = 48 m/h
4.0 m
(5.0 min)�60
1
m
h
in�
4. ∆x = 3.20 × 104 km south∆t = 122 days vavg =
∆∆
x
t =
3.2
1
0
2
×2
1
d
0
a
4
ys
km = 262 km/day south
5. ∆x1 = 1.70 × 104 km south= +1.70 × 104 km
∆x2 = 6.0 × 102 km north= −6.0 × 102 km
∆x3 = 1.44 × 104 km south= +1.44 × 104 km
∆t = 122 days
d = total distance traveled = magnitude ∆x1 + magnitude ∆x2 + magnitude ∆x3
d = 1.70 × 104 km + 6.0 × 102 km + 1.44 × 104 km
d = (1.70 + 0.060 + 1.44) × 104 km
d = 3.20 × 104 km
average speed = ∆d
t =
3.2
1
0
2
×2
1
d
0
a
4
ys
km =
vavg = ∆∆xt
tot =
∆x1 + ∆∆x
t2 + ∆x3
vavg =
vavg =
vavg = 3.0
1
8
2
×2
1
d
0
a
4
ys
km = +252 km/day = 252 km/day south
(1.70 − 0.060 + 1.44) × 104 km
122 days
(1.70 × 104 km) + (−6.0 × 102 km) + (1.44 × 104 km)
122 days
262 km/day
6. ∆x1 = 20.0 km east = + 20.0 km
∆x2 = 20.0 km west = − 20.0 km
∆x3 = 0 km
∆x4 = 40.0 km east = +40.0 km
∆t = 60.0 min
a. vavg = =
vavg =
vavg = = + 40.0 km/h = 40.0 km/h east40.0 km
(60.0 min) �60
1
m
h
in�
(20.0 km) + (−20.0 km) + (0 km) + (40.0 km)
60.0 min
∆x1 + ∆x2 + ∆x3 + ∆x4∆t
∆xtot∆t
Holt Physics Solution ManualV Ch. 2–2
V
Cop
yrig
ht ©
Hol
t, R
ineh
art a
nd W
inst
on.A
ll rig
hts
rese
rved
.
b. d = total distance traveled
d = magnitude ∆x1 + magnitude ∆x2 + magnitude ∆x3 + magnitude ∆x4
d = 20.0 km + 20.0 km + 0 km + 40.0 km = 80.0 km
average speed = = = 80.0 km/h80.0 km
(60.0 min)�60
1
m
h
in�
d∆t
Givens Solutions
7. v = 89.5 km/h north
vavg = 77.8 km/h north
∆trest = 22.0 min
∆x = vavg ∆t = v(∆t − ∆trest)
∆t(vavg − v) = −v∆trest
∆t = = =
∆t = 2.80 h = 2 h, 48 min
(89.5 km/h)(22.0 min)�60
1
m
h
in�
11.7 km/h
(89.5 km/h)(22.0 min)�60
1
m
h
in�
89.5 km/h − 77.8 km/h
v∆trestv − vavg
8. v = 6.50 m/s downward = −6.50 m/s
∆t = 34.0 s
∆x = v∆t = (−6.50 m/s)(34.0 s) = −221 m = 221 m downward
9. vt = 10.0 cm/s
vh = 20 vt = 2.00 × 102 cm/s
∆trace = ∆tt
∆th = ∆tt − 2.00 min
∆xt = ∆xh + 20.0 cm = ∆xrace
∆xt = vt∆tt
∆xh = vh∆th = vh (∆tt − 2.00 min)
∆xt = ∆xrace = ∆xh + 20.0 cm
vt ∆tt = vh (∆tt − 2.00 min) + 20.0 cm
∆tt (vt − vh) = −vh (2.00 min) + 20.0 cm
∆tt =
∆trace = ∆tt =
∆trace = =
∆trace = 126 s
−2.40 × 104 cm−1.90 × 102 cm/s
20.0 cm − 2.40 × 104 cm
−1.90 × 102 cm/s
20.0 cm − (2.00 × 102 cm/s)(2.00 min)(60 s/min)
10.0 cm/s − 2.00 × 102 cm/s
20.0 cm − vh (2.00 min)
vt − vh
10. ∆xrace = ∆xt
vt = 10.0 cm/s
∆tt = 126 s
∆xrace = ∆xt = vt∆tt = (10.0 cm/s)(126 s) = 1.26 × 103 cm = 12.6 m
Additional Practice B
1. ∆t = 6.92 s
vf = 17.34 m/s
vi = 0 m/s
aavg = ∆∆
v
t =
vf
∆−t
vi = = 2.51 m/s217.34 m/s − 0 m/s
6.92 s
2. vi = 0 m/s
vf = 7.50 × 102 m/s
∆t = 2.00 min
aavg = = = = 6.25 m/s27.50 × 102 m/s − 0 m/s
(2.00 min)�16
m
0
i
s
n�
vf − vi∆t
∆v∆t
Section Five—Problem Bank V Ch. 2–3
V
Cop
yrig
ht ©
Hol
t, R
ineh
art a
nd W
inst
on.A
ll rig
hts
rese
rved
.
Givens Solutions
3. vi = 0 m/s
vf = 0.85 m/s forward
∆t = 3.7 s
aavg = = = = 0.23 m/s2 forward0.85 m/s − 0 m/s
3.7 s
vf − vi∆t
∆v∆t
4. vi = 13.7 m/s forward = +13.7 m/s
vf = 11.5 m/s backward = −11.5 m/s
∆t = 0.021 s
aavg = = = =
aavg = −1200 m/s2, or 1200 m/s2 backward
−25.2 m/s
0.021 s
(− 11.5 m/s) − (13.7 m/s)
0.021 s
vf − vi∆t
∆v∆t
5. vi = +320 km/h
vf = 0 km/h
∆t = 0.18 saavg = = =
aavg = = −490 m/s2−89 m/s
0.18 s
(0 km/h − 320 km/h)�36
1
0
h
0 s��11
0
k
3
m
m�
0.18 s
vf − vi∆t
∆v∆t
6. aavg = 16.5 m/s2
vi = 0 km/h
vf = 386.0 km/h∆t = = =
∆t = = 6.50 s107.2 m/s16.5 m/s2
(386.0 km/h − 0 km/h)�36
1
0
h
0 s��11
0
k
3
m
m�
16.5 m/s2
vf − viaavg
∆vaavg
7. vi = −4.0 m/s
aavg = −0.27 m/s2
∆t = 17 s
vf = aavg ∆t + vi
vf = (−0.27 m/s2)(17 s) + (−4.0 m/s) = −4.6 m/s − 4.0 m/s = −8.6 m/s
8. vi = 4.5 m/s
vf = 10.8 m/s
aavg = 0.85 m/s2
∆t = a
∆
a
v
vg = =
10.8
0
m
.8
/
5
s −m
4
/s
.52
m/s =
0
6
.8
.3
5
m
m
/
/
s
s2 = 7.4 svf − viaavg
9. vf = 296 km/h
vi = 0 km/h
aavg = 1.60 m/s2∆t =
a
∆
a
v
vg = = =
1
8
.
2
6
.
0
2
m
m
/
/
s
s2 = 51.4 s
(296 km/h − 0 km/h)�36
1
0
h
0 s��11
0
k
3
m
m�
1.60 m/s2
vf − viaavg
10. aavg = − 0.87 m/s2
∆t = 3.85
∆v = aavg ∆t = (−0.87 m/s2)(3.85 s) = –3.4 m/s
1. ∆t = 0.910 s
∆x = 7.19 km
vi = 0 km/s
vf = 2
∆∆t
x − vi =
(2)
0
(7
.9
.1
1
9
0
k
s
m) − 0 km/s = 15.8 km/s
Additional Practice C
Givens Solutions
Holt Physics Solution ManualV Ch. 2–4
V
Cop
yrig
ht ©
Hol
t, R
ineh
art a
nd W
inst
on.A
ll rig
hts
rese
rved
.
2. vi = 4.0 m/s
∆t = 18 s
∆x = 135 m
vf = 2
∆∆t
x − vi =
(2)(
1
1
8
35
s
m) − 4.0 m/s = 15 m/s − 4.0 m/s =
vf = (11 m/s)�36
1
0
h
0 s� �
1
1
0
k3m
m�
vf = 4.0 × 101 km/h
11 m/s
3. ∆x = 55.0 m
∆t = 1.25 s
vf = 43.2 m/s
vi = 2
∆∆t
x − vf =
(2)
1
(5
.2
5
5
.0
s
m) − 43.2 m/s = 88.0 m/s − 43.2 m/s = 44.8 m/s
4. ∆x = 38.5 m
∆t = 5.5 s
vi = 0 m/s
vf = 2
∆∆t
x − vi =
(2)(
5
3
.
8
5
.5
s
m) − 0 m/s = 14 m/s
5. ∆x = 478 km
∆vi = 72 km/h
∆t = 5 h, 39 min
vf = 2
∆∆t
x − vi = − 72 km/h =
(
5
2
h
)(4
+7
0
8
.6
k
5
m
h
) − 72 km/h
vf = 9
5
5
.
6
65
km
h − 72 km/h = 169 km/h − 72 km/h = 97 km/h
(2)(478 km)
5 h + 39 min�60
1
m
h
in�
6. ∆x = 4.2 m
∆t = 3.0 s
vf = 1.3 m/s
vi = 2
∆∆t
x − vf =
(2)
3
(4
.0
.2
s
m) − 1.3 m/s = 2.8 m/s − 1.3 m/s = 1.5 m/s
7. vi = 25 m/s west
vf = 35 m/s west
∆x = 250 m west
∆t = v
2
i
∆+
x
vf =
25
(
m
2)
/
(
s
2
+50
35
m
m
)
/s =
6
5
.0
.0
××1
1
0
01
2
m
m
/s = 8.3 s
8. vi = 755.0 km/h
vf = 777.0 km/h
∆t = 63.21 s
∆x = 1
2 (vi + vf)∆t =
1
2 (755.0 km/h + 777.0 km/h)(63.21 s)�36
1
0
h
0 s�
∆x = 1
2 (1532.0 km/h)(1.756 × 10−2 h) = 13.45 km
9. vi = 0 m/s
vf = 30.8 m/s
∆x = 493 m
∆t = v
2
i
∆+
x
vf =
0 m
(2
/s
)(
+49
3
3
0.
m
8 m
)
/s =
3
9
0
8
.8
6
m
m
/s = 32.0 s
10. ∆x = 1220 km
vi = 11.1 km/s
vf = 11.7 km/s
∆t = v
2
i
∆+
x
vf = =
2
2
2
4
.
4
8
0
k
k
m
m
/s = 107 s
(2)(1220 km)11.1 km/s + 11.7 km/s
Section Five—Problem Bank V Ch. 2–5
V
Cop
yrig
ht ©
Hol
t, R
ineh
art a
nd W
inst
on.A
ll rig
hts
rese
rved
.Givens Solutions
2. ∆t = 1.5 s
vi = 2.8 km/h
vf = 32.0 km/ha = =
a = = 5.4 m/s2
(29.2 km/h)�36
1
0
h
0 s��11
0
k
3
m
m�
1.5 s
(32.0 km/h − 2.8 km/h)�36
1
0
h
0 s��10
k
3
m
m�
1.5 s
vf − vi∆t
3. ∆x = 18.3 m
∆t = 2.74 s
vi = 0 m/s
Because vi = 0 m/s, a = �2
∆∆t
x2� =
(2
(
)
2
(
.
1
7
8
4
.3
s)
m2
) = 4.88 m/s2
4. vi = 2.3 m/s
vf = 46.7 m/s
∆t = 7.0 s
a = = 46.7 m/
7
s
.0
−s
2.3 m/s =
44
7
.4
.0
m
s
/s = 6.3 m/s2vf − vi
∆t
5. vi = 6.23 m/s
∆x = 255 m
∆t = 82 s
a = 2(∆x
∆−t2vi ∆t) =
a = = �(
6
2
.
)
7
(−×2
1
5
0
53m
s2)
� = −7.6 × 10−2 m/s2(2)(255 m − 510 m)
6.7 × 103 s2
(2)[255 m − (6.23 m/s)(82 s)]
(82 s)2
6. vi = 11 km/h
vf = 55 km/h
∆ = 4.1 sa = =
a = = 3.0 m/s2
(44 km/h)�36
1
0
h
0 s��11
0
k
3
m
m�
4.1 s
(55 km/h − 11 km/h)�36
1
0
h
0 s��11
0
k
3
m
m�
4.1 s
vf − vi∆t
7. vi = 42.0 m/s southeast
∆t = 0.0090 s
∆x = 0.020 m/s southeast
a = 2(∆x
∆−t2vi ∆t) =
a = =
a = −8.9 × 103 m/s2, or 8.9 × 103 m/s2 northwest
(2)(−0.36 m)8.1 × 10−5 s2
(2)(0.020 m/s − 0.38 m)
8.1 × 10−5 s2
(2)[0.020 m − (42.0 m/s)(0.0090 s)]
(0.0090 s)2
8. ∆t = 28 s
a = 0.035 m/s2
vi = 0.76 m/s
vf = a∆t + vi = (0.035 m/s2)(28.0 s) + 0.76 m/s = 0.98 m/s + 0.76 m/s = 1.74 m/s
Additional Practice D
1. ∆x = 12.4 m upward
∆t = 2.0 s
vi = 0 m/s
Because vi = 0 m/s, a = 2
∆∆t2x
= (2)
(
(
2
1
.0
2.
s
4
)2m)
= 6.2 m/s2 upward
Givens Solutions
Holt Physics Solution ManualV Ch. 2–6
V
Cop
yrig
ht ©
Hol
t, R
ineh
art a
nd W
inst
on.A
ll rig
hts
rese
rved
.
4. ∆x = 2.00 × 102 m
vi = 9.78 m/s
vf = 10.22 m/s
a = vf
2
2∆−
x
vi2
= =
a = 4.
8
0
.
0
8
×m
1
2
0
/2s2
m = 2.2 × 10−2 m/s2
104.4 m2/s2 − 95.6 m2/s2
4.00 × 102 m
(10.22 m/s)2 − (9.78 m/s)2
(2)(2.00 × 102 m)
10. vi = +4.42 m/s
vf = 0 m/s
a = −0.75 m/s2
∆t = 5.9 s
a. ∆t = vf −
a
vi = 0 m
−/
0
s
.7
−5
4
m
.42
/s2m/s
= −−0
4
.
.
7
4
5
2
m
m
/
/
s
s2 =
b. ∆x = vi∆t + 1
2a∆t2 = (4.42 m/s)(5.9 s) +
1
2(−0.75 m/s2)(5.9 s)2
∆x = 26 m − 13 m = 13 m
5.9 s
1. vi = 1.8 km/h
vf = 24.0 km/h
∆x = 4.0 × 102 ma =
vf2
2∆−
x
vi2
=
a =
a = = 5.5 × 10−2 m/s2
(573 km2/h2)�36
1
0
h
0 s�
2
�1
1
0
k
3
m
m�
2
8.0 × 102 m
(576 km2/h2 − 3.2 km2/h2)�36
1
0
h
0 s�
2
�1
1
0
k
3
m
m�
2
8.0 × 102 m
[(24.0 km/h)2 − (1.8 km/h)2]�36
1
0
h
0 s�
2
�1
1
0
k
3
m
m�
2
(2)(4.0 × 102 m)
2. vf = 0 m/s
vf = 8.57 m/s
∆x = 19.53 m
a = vf
2
2
−∆x
vi2
= = 7
3
3
9
.4
.0
m
6
2
m
/s2
= 1.88 m/s2(8.57 m/s)2 − (0 m/s)2
(2)(19.53 m)
9. vi = 0 m/s
vf = 72.0 m/s north
a = 1.60 m/s2 north
∆t = 45.0 s
a. ∆t = vf −
a
vi = 72.0
1.
m
60
/s
m
−/
0
s2m/s
=
b. ∆x = vi∆t + 1
2a∆t2 = (0 m/s)(45.0 s) +
1
2 (1.60 m/s2)(45.0 s)2 = 0 m + 1620 m
∆x = 1.62 km
45.0 s
Additional Practice E
3. vi = 7.0 km/h
vf = 34.5 km/h
∆x = 95 ma =
vf2
2
−∆x
vi2
=
a =
a = = 0.46 m/s2
(1140 km2/h2)�36
1
0
h
0 s�
2
�1
1
0
k
3
m
m�
2
190 m
(1190 km2/h2 − 49 km2/h2)�36
1
0
h
0 s�
2
�1
1
0
k
3
m
m�
2
190 m
[(34.5 km/h)2 − (7.0 km/h)2]�36
1
0
h
0 s�
2
�1
1
0
k
3
m
m�
2
(2)(95 m)
Section Five—Problem Bank V Ch. 2–7
V
Cop
yrig
ht ©
Hol
t, R
ineh
art a
nd W
inst
on.A
ll rig
hts
rese
rved
.
6. vi = 50.0 km/h forward= +50.0 km/h
vf = 0 km/h
a = 9.20 m/s2 backward= −9.20 m/s2
Givens Solutions
∆x = vf
2
2
−a
vi2
=
∆x =
∆x = 10.5 m = 10.5 m forward
−(2.50 × 103 km2/h2)�36
1
0
h
0 s�
2
�110
k
3
m
m�
2
−18.4 m/s2
[(0 km/h)2 − (50.0 km/h)2]�36
1
0
h
0 s�
2
�110
k
3
m
m�
2
(2)(−9.20 m/s2)
7. a = 7.56 m/s2
∆x = 19.0 m
vi = 0 m/s
vf =√
vi�2�+� 2�a∆�x� =√
(0� m�/s�)2� +� (�2)�(7�.5�6�m�/s�2)�(1�9.�0�m�)�
vf =√
28�7�m�2/�s2� = ±16.9 m/s = 16.9 m/s
8. vi = 1.8 m/s
vf = 9.4 m/s
a = 6.1 m/s2
∆x = vf
2
2
−a
vi2
= =
∆x = (2)
8
(
5
6.
m
1
2
m
/s
/
2
s2) = 7.0 m
88 m2/s2 − 3.2 m2/s2
(2)(6.1 m/s2)
(9.4 m/s)2 − (1.8 m/s)2
(2)(6.1 m/s2)
9. vi = 1.50 m/s to the right = +1.50 m/s
vf = 0.30 m/s to the right = +0.30 m/s
a = 0.35 m/s2 to the left = −0.35 m/s2
∆x = vf
2
2
−a
vi2
=
∆x =
∆x = −−2
0
.
.
1
7
6
0
m
m
2
/
/
s
s2
2
= +3.1 m = 3.1 m to the right
9.0 × 10−2 m2/s2 − 2.25 m2/s2
−0.70 m/s2
(0.30 m/s)2 − (1.50 m/s)2
(2)(−0.35 m/s2)
10. a = 0.678 m/s2
vf = 8.33 m/s
∆x = 46.3 m
vi =√
vf�2�−� 2�a∆�x� =√
(8�.3�3�m�/s�)2� −� (�2)�(0�.6�78� m�/s�2)�(4�6.�3�m�)�
vi =√
69�.4� m�2/�s2� −� 6�2.�8�m�2/�s2� =√
6.�6�m�2/�s2� = ±2.6 m/s = 2.6 m/s
1. vi = 0 m/s
vf = 49.5 m/s downward= 49.5 m/s
a = −9.81 m/s2
∆tot = −448 m
∆xi = vf
2
2
−a
vi2
= = (2)
2
(
4
−5
9
0
.8
m
1
2
m
/s
/
2
s2) = −125 m
∆x2 = ∆xtot − ∆x1 = (−448 m) − (−125 m) = −323 m
distance from net to ground = magnitude ∆x2 = 323 m
(−49.5 m/s)2 − (0 m/s)2
(2)(−9.81 m/s2)
Additional Practice F
5. ∆x = +42.0 m
vi = +153.0 km/h
vf = 0 km/ha =
vf2
2∆−
x
vi2
=
a = = −21.5 m/s2
−(2.34 × 104 km2/h2)�36
1
0
h
0 s�
2
�1
1
0
k
3
m
m�
2
(84.0 m)
[(0 km/h)2 − (153.0 km/h)2]�36
1
0
h
0 s�
2
�1
1
0
k
3
m
m�
2
(2) (42.0 m)
Givens Solutions
Holt Physics Solution ManualV Ch. 2–8
V
Cop
yrig
ht ©
Hol
t, R
ineh
art a
nd W
inst
on.A
ll rig
hts
rese
rved
.
2. vi = 0 m/s
a = −9.81 m/s2
∆t1 = 1.00 s
∆t2 = 2.00 s
∆t3 = 3.00 s
Because vi = 0 m/s, ∆x1 = 1
2a∆t1
2 = 1
2(−9.81 m/s2)(1.00 s)2 =
∆x2 = 1
2a∆t2
2 = 1
2(−9.81 m/s2)(2.00 s)2 =
∆x3 = 1
2a∆t3
2 = 1
2(−9.81 m/s2)(3.00 s)2 = −44.1 m
−19.6 m
−4.90 m
3. vi = 0 m/s
∆t = 2.0 s
a = −9.81 m/s2
Because vi = 0 m/s, ∆x = 1
2a∆t2 =
1
2(−9.81 m/s2)(2.0 s)2 = −2.0 × 101 m
distance of bag below balloon = 2.0 × 101 m
4. vi = +17.5 m/s
vf = 0.0 m/s
a = −9.81 m/s2
∆ttot = 3.60 s
∆xtot = 0 m
∆x = 1
2(vi + vf)∆t
∆ttop = ∆t
2tot =
3.6
2
0 s = 1.80 s
∆x = 1
2(17.5 m/s + 0.0 m/s)(1.80 s) = 15.8 m
5. vi = 0 m/s
vf = 11.4 m/s downward= −11.4 m/s
a = 3.70 m/s2 downward= −3.70 m/s2
∆x = vf
2
2
−a
vi2
= = = −17.6 m
∆x = 17.6 m downward
1.30 × 102 m2/s2
−7.40 m/s2
(−11.4 m/s)2 − (0 m/s)2
(2)(−3.70 m/s2)
6. ∆ttot = 5.10 s
∆tdown = 1
2 ∆ttop
vi = 0 m/s
a = −9.81 m/s2
To find the ball’s maximum height, calculate the displacement from that height to itsoriginal position. The time interval for this free-fall is 1
2 ∆ttot, and vi = 0 m/s.
∆xdown = 1
2 a∆tdown
2 = 1
2a�
1
2∆ttot�
2=
1
2(−9.81 m/s2)�
5.1
2
0 s�
2= −31.9 m
ball’s maximum height = 31.9 m
7. ∆ttot = 5.10 s
a = −9.81 m/s2
∆xtot = 0 m
∆ttot = vi ∆ ttot + 1
2a∆ttot
2
Because ∆xtot = 0,
vi = − 1
2a∆ttot = −
1
2(−9.81 m/s2)(5.10 s) = +25.0 m/s = 25.0 m/s upward
8. vi = 85.1 m/s upward= + 85.1 m/s
a = −9.81 m/s2
∆x = 0 m
Because ∆x = 0 m, vi ∆t + 1
2a∆t 2 = 0
∆t = − −2
a
vi = − −(
(
2
−)
9
(
.
8
8
5
1
.1
m
m
/s
/2s
)
) = 17.3 s
Section Five—Problem Bank V Ch. 2–9
V
Cop
yrig
ht ©
Hol
t, R
ineh
art a
nd W
inst
on.A
ll rig
hts
rese
rved
.
Givens Solutions
10. aup = 3.125 m/s2, upward= +3.125 m/s2
∆tup = 4.00 s
vi = 0 m/s
vf = aup ∆tup + vi = (3.125 m/s2) (4.00 s) + 0 m/s = 2.5 m/s
When the cable brakes, the upward-moving elevator undergoes free-fall acceleration.
vi = +12.5 m/s
a = −9.81 m/s2
∆t1 = 0.00 s
∆t2 = 1.00 s
∆t3 = 2.00 s
∆t4 = 3.00 s
vf, 1 = a∆t1 + vi = (−9.81 m/s2)(0.00 s) + 12.5 m/s =
vf, 2 = a∆t2 + vi = (−9.81 m/s2)(1.00 s) + 12.5 m/s = −9.81 m/s + 12.5 m/s =
vf, 3 = a∆t3 + vi = (−9.81 m/s2)(2.00 s) + 12.5 m/s = −19.6 m/s + 12.5 m/s =
vf, 4 = a∆t4 + vi = (−9.81 m/s2)(3.00 s) + 12.5 m/s = −29.4 m/s + 12.5 m/s = − 16.9 m/s
−7.1 m/s
+2.7 m/s
+12.5 m/s
9. ∆td = 4.00 s
∆tp = ∆td + 3.00 s= 7.00 s
a = −9.81 m/s2
∆xd = ∆xp
vi, p = 0 m/s
∆xd = vi, d ∆td + 1
2a∆td
2
∆xp = vi, p ∆tp + 1
2a∆tp
2
vi, d ∆td + 1
2a∆td
2 = vi, p∆tp + 1
2a∆tp
2
vi, d = + vi,
∆p
t
∆
d
tp
Because vi, p = 0 m/s,
vi, d = =
vi, d = − = − =
vf, d = a ∆td + vi, d = (−9.81 m/s2)(4.00 s) + (−40.5 m/s) = (−39.2 m/s) + (−40.5 m/s)
vf, d = −79.7 m/s
−40.5 m/s(9.81 m/s2)(33.0 s2)
8.00 s
(9.81 m/s2)(49.0 s2 − 16.0 s2)
8.00 s
1
2(−9.81 m/s2)[(7.00 s)2 − (4.00 s)2]
4.00 s
1
2a(∆tp
2 − ∆td2)
∆td
1
2a(∆tp
2 − ∆td2)
∆td
Original content Copyright © by Holt, Rinehart and Winston. Additions and changes to the original content are the responsibility of the instructor.
Holt Physics: Answer Key 4 Chapter 2
Holt Physics
Answer Key
CHAPTER 2
ESSAY Answers should include the following: In a graph that shows the distance an object traveled as a function of time, the slope of the line will tell you the velocity of the object. To find the slope of a line, you divide the change in y-coordinate factor (distance) by the change in the x-coordinate factor (time). In this case, the object had a velocity of about 10 m/s during the first second of the observation period. The object’s velocity slowly decreased over time, indicating a negative acceleration. In the last two seconds of the observation period, the object had a velocity of about 2.5 m/s. [SAI 2, PS 4a]
MATH Item Answer Correlation
1. C UCP 3, PS 4a 2. E UCP 3, PS 4a 3. A SAI 2, PS 4a 4. B SAI 2, PS 4a 5. B UCP 3, PS 4a 6. A UCP 3, PS 4a 7. D UCP 3, PS 4a 8. B UCP 3, PS 4b 9. C UCP 3, PS 4b 10. C UCP 3, PS 4a 11. B UCP 3, PS 4a 12. D UCP 3, PS 4b
SENTENCE COMPLETION Item Answer Correlation 13. C SAI 2, PS 4a 14. C SAI 2, PS 4a 15. E UCP 2, PS 4b 16. A UCP 2, PS 4b
READING PASSAGE Item Answer Correlation 17. B SAI 1, PS 4a 18. D SAI 1, PS 4a 19. D SAI 2, PS 4a 20. E SAI 2, PS 4a
IMPROVING SENTENCES Item Answer Correlation 21. B UCP 3, PS 4b 22. C UCP 2, PS 4b 23. E SAI 2, PS 4a 24. A UCP 3, PS 4a