Two-Dimensional Motion and Vectors
Chapter
3
I
Section Review, p. 87
Givens
2. ∆x1 = 85 m
d2 = 45 m
q2 = 30.0°
Solutions
Students should use graphical techniques. Their answers can be checked using the
techniques presented in Section 3-2. Answers may vary.
∆x2 = d2(cos q2) = (45 m)(cos 30.0°) = 39 m
∆y2 = d2(sin q2) = (45 m)(sin 30.0°) = 22 m
∆xtot = ∆x1 + ∆x2 = 85 m + 39 m = 124 m
∆ytot = ∆y2 = 22 m
d = (∆
m)2
+(22
m)2
xtot)2+(∆
ytot)2 = (1
24




d = 15
m2+
m2 = 15
m2 = 126 m
400

480

900



!
!
∆ytot
22 m
q = tan−1 
= tan−1  = (1.0 × 101)° above the horizontal
∆xtot
124 m
3. vy, 1 = 2.50 × 102 km/h
Students should use graphical techniques.
v2 = 75 km/h
vx,2 = v2(cos q2) = (75 km/h)[cos (−45°)] = 53 km/h
q2 = −45°
vy,2 = v2(sin q2) = (75 km/h)[sin (−45°)] = −53 km/h
vy,tot = vy,1 + vy,2 = 2.50 × 102 km/h − 53 km/h = 197 km/h
vx,tot = vs,2 = 53 km/h
2
2
v = (v
vy,to
)2
+(1
)2
x,to
(

3km
/h

97
km
/h

t)+
t) = (5
Copyright © by Holt, Rinehart and Winston. All rights reserved.


v = 28
00
km
2/h2+3880
0km
2/h2 = 41
600
km
2/h2 = 204 km/h


!
!
v tot
204 km/h
q = tan−1 y,
= tan−1  = 75° north of east
vx,tot
53 km/h
2.50 × 102 km/h
4. vy,1 = 
2
= 125 km/h
vx,2 = 53 km/h
vy,2 = −53 km/h
Students should use graphical techniques.
vy,dr = vy,1 + vy,2 = 125 km/h − 53 km/h = 72 km/h
vx,dr = vx,2 = 53 km/h
2
v = (v
)2
=vy,d
)2
+
(72km
)2
x,dr


3km
/h

/h

r) = (5


v = 28
03km
00
km
2/h2+520
0km
2/h2 = 8.
0
1
2/h2 = 89 km/h


!
!
vy dr
72 km/h
q = tan−1 ,
= tan−1  = 54° north of east
vx,dr
53 km/h
Section One—Pupil’s Edition Solutions
I Ch. 3–1
Practice 3A, p. 91
Givens
Solutions
1. ∆x1 = 8 km east
a. d = ∆x1 + ∆x2 + ∆x3 = 8 km + 3 km + 12 km = 23 km
∆x1 = 8 km
I
∆x2 = 3 km west = −3km, east
b. ∆xtot = ∆x1 + ∆x2 + ∆x3 = 8 km + (−3 km) + 12 km = 17 km east
∆x2 = 3 km
∆x3 = 12 km east
∆x3 = 12 km
∆y = 0 km
2. ∆x = 7.5 m
∆y = 45.0 m
d= ∆
x2
+
∆
y 2 = (7
m
)2
+(45
m
)2

.5

.0



d = 56
m2+
m2 = 45.6 m

202
0m
2 = 20
80



Measuring direction with respect to y (north),
!
!
∆x
7.5 m
q = tan−1  = tan−1  = 9.5° east of due north
∆y
45.0 m
3. ∆x = 6.0 m
∆y = 14.5 m
d= ∆
∆y2 = (6
m
)2
+(14
m
)2
x2+
.0

.5



d = 36
m2+
02
m2 = 24

2.1
0×1
6m
2 = 15.7 m


Measuring direction with respect to the length of the field (down the field),
!
!
∆x
6.0 m
q = tan−1  = tan−1  = 22° to the side of downfield
∆y
14.5 m
∆y = –1.4 m
d= ∆
∆y2 = (1
m
)2
+–
(1.
)2
x2+
.2

4m



d = 1.
m2 = 3.
4m
2+2.0

4m
2 = 1.8 m


!
!
∆y
–1.4 m
q = tan−1  = tan−1  = –49° = 49° below the horizontal
∆x
1.2 m
Practice 3B, p. 94
1. v = 105 km/h
vx = v(cos q ) = (105 km/h)(cos 25°) = 95 km/h
q = 25°
2. v = 105 km/h
vy = v(sin q) = (105 km/h)(sin 25°) = 44 km/h
q = 25°
3. v = 22 m/s
vx = v(cos q ) = (22 m/s)(cos 15°) = 21 m/s
q = 15°
vy = v(sin q) = (22 m/s)(sin 15°) = 5.7 m/s
4. d = 5 m
q = 90°
I Ch. 3–2
∆x = d(cos q) = (5 m)(cos 90°) = 0 m
∆y = d(sin q) = (5 m)(sin 90°) =
Holt Physics Solution Manual
5m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. ∆x = 1.2 m
Practice 3B, p. 94 cont.
Givens
5. d = 125 m
q = −25°
6. d = 23.0 m
q = −30.5°
7. a = 2.5 m/s2
q = −18°
Solutions
∆x = d(cos q ) = (125 m)[cos (−25°)] = 1.1 × 102 m
∆y = d(sin q) = (125 m)[sin (−25°)] = −53 m
I
∆x = d(cos q) = (23.0 m)[cos (−30.5°)] = 19.8 m
∆y = d(sin q) = (23.0 m)[sin (−30.5°)] = −11.7 m
ax = a(cos q) = (2.5 m/s2)[cos (−18°)] = 2.4 m/s2
ay = a(sin q) = (2.5 m/s2)[sin (−18°)] = −0.77 m/s2
Practice 3C, p. 97
1. d1 = 35 m
∆x1 = d1(cos q1) = (35 m)(cos 0.0°) = 35 m
q1 = 0.0°
∆y1 = d1(sin q1) = (35 m)(sin 0.0°) = 0.0 m
d2 = 15 m
∆x2 = d2(cos q2 ) = (15 m)(cos 25°) = 14 m
q2 = 25°
∆y2 = d2(sin q2 ) = (15 m)(sin 25°) = 6.3 m
∆xtot = ∆x1 + ∆x2 = 35 m + 14 m = 49 m
∆ytot = ∆y1 + ∆y2 = 0.0m + 6.3m = 6.3 m
dtot = (∆
xtot
)2
+(∆
ytot
)2 = (4
)2
+(6.
)2




9m

3m



dtot = 24
m2+
m2 = 49 m
00

40m
2 = 24
00



!
!
Copyright © by Holt, Rinehart and Winston. All rights reserved.
6.3 m
∆ytot
qtot = tan−1 
= tan−1  = 7.3° to the right of downfield
49 m
∆xtot
2. d1 = 2.5 km
∆x1 = d1(cos q1) = (2.5 km)(cos 35°) = 2.0 km
q1 = 35°
∆y1 = d1(sin q1) = (2.5 km)(sin 35°) = 1.4 km
d2 = 5.2 km
∆x2 = d2(cos q2) = (5.2 km)(cos 22°) = 4.8 km
q2 = 22°
∆y2 = d2(sin q2) = (5.2 km)(sin 22°) = 1.9 km
∆xtot = ∆x1 + ∆x2 = 2.0 km + 4.8 km = 6.8 km
∆ytot = ∆y1 + ∆y2 = 1.4 km + 1.9 km = 3.3 km
xt
)2
+(∆
yt
)2 = (6
)2
+(3.
)2
dtot = (∆


.8
km

3km

ot
ot


dtot = 46
km
2=11
km
2 = 57
km
2 = 7.5 km


!
!
∆ytot
3.3 km
qtot = tan−1 
= tan−1  = 26° above the horizontal
∆xtot
6.8 km
Section One—Pupil’s Edition Solutions
I Ch. 3–3
Givens
Solutions
3. d1 = 8.0 m
I
Measuring direction with respect to y = (north),
q1 = 0.0°
∆x1 = d1(sin q1) = (8.0 m)(sin 0.0°) = 0.0 m
d2 = 3.5 m
∆y1 = d1(cos q1) = (8.0 m)(cos 0.0°) = 8.0 m
q2 = 35°
∆x2 = d2(sin q2) = (3.5 m)(sin 35°) = 2.0 m
d3 = 5.0 m
∆y2 = d2(cos q2) = (3.5 m)(cos 35°) = 2.9 m
q3 = 90.0°
∆x3 = d3(sin q3) = (5.0 m)(sin 90.0°) = 5.0 m
∆y3 = d3(cos q3) = (5.0 m)(cos 90.0°) = 0.0 m
∆xtot = ∆x1 + ∆x2 + ∆x3 = 0.0. m + 2.0 m + 5.0 m = 7.0 m
∆ytot = ∆y1 + ∆y2 + ∆y3 = 8.0. m + 2.9 m + 0.0 m = 10.9 m
dtot = (∆
xt
)2(∆
yt
)2 = (7
m
)2
+(10
m
)2


.0

.9

ot
ot


dtot = 49
m2+
m2 = 16

119

8m
2 = 13.0 m


!
!
∆xtot
7.0 m
qtot = tan−1 
= tan−1  = 33° east of north
∆ytot
10.9 m
4. d1 = 75 km
Measuring direction with respect to y (north),
q1 = −30.0°
∆x1 = d1(sin q1) = (75 km)(sin − 30.0°) = −38 km
d2 = 155 km
∆y1 = d1(cos q1) = (75 km)(cos − 30.0°) = 65 km
q2 = 60.0°
∆x2 = d2(sin q2) = (155 km)(sin 60.0°) = 134 km
∆y2 = d2(cos q2) = (155 km)(cos 60.0°) = 78 km
∆xtot = ∆x1 + ∆x2 = −38 km + 134 km = 96 km
∆ytot = ∆y1 + ∆y2 = 65 km + 78 km = 143 km
2
2
dtot = 
(∆
xt
)2
+(
∆
yt
)2 = 
(9
6
km
)2
+(
14
3
km
)2 = 
9.
2
×1
03k
m
.0
×1
04 km
+2

ot
ot



= 2.
92
×104km
2 = 171 km
!
!
∆xtot
96 km
q = tan−1 
= tan−1  = 34° east of north
∆ytot
143 km
Section Review, p. 97
2. vx = 3.0 m/s
vy = 5.0 m/s
2
a. v = vx2+
m/s
)2
=(5.
)2
v
.0


0m
/s

y = (3


v = 9.
s2 
+25m
s2 = 34
m2/
s2 = 5.8 m/s
0m
2/
2/



!
!
vy
5.0 m/s
q = tan−1  = tan−1  = 59° downriver from its intended path
vx
3.0 m/s
vx = 1.0 m/s
vy = 6.0 m/s
2
b. v = vx2+
m/s
)2
+(6.
)2
v
.0


0m
/s

y = (1


v = 1.
s2
+36m
s2 = 37
m2/
s2 = 6.1 m/s
0m
2/
2/



!
!
v
1.0 m/s
q = tan−1 x = tan−1  = 9.5° from the direction the wave is traveling
vy
6.0 m/s
I Ch. 3–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.

Givens
Solutions
3. d = 10.0 km
a. ∆x = d(cos q) = (10.0 km)(sin 45.0°) = 7.07 km
q = 45.0°
∆y = d(sin q) = (10.0 km)(cos 45.0°) = 7.07 km
a = 2.0 m/s2
b. ax = a(cos q) = (2.0 m/s2)(cos 35°) = 1.6 m/s2
q = 35°
I
ay = a(sin q) = (2.0 m/s2)(sin 35°) = 1.1 m/s2
4. d1 = 10.0 m
∆x1 = d1(cos q) = (10.0 m)(cos 55°) = 5.7 m
q1 = 55°
∆y1 = d1(sin q) = (10.0 m)(sin 55°) = 8.2 m
d2 = 5.0 m
∆x2 = d2(cos q 2) = (5.0 m)(cos 0.0°) = 5.0 m
q2 = 0.0°
∆y2 = d2(sin q 2) = (5.0 m)(sin 0.0°) = 0.0 m
∆xtot = ∆x1 + ∆x2 = 5.7 m + 5.0 m = 10.7 m
∆ytot = ∆y1 + ∆y2 = 8.2 m + 0.0 m = 8.2 m
dtot = (∆
xt
)2
+(∆
yt
)2 = (1
)2
+(8.
)2


0.
7m

2m

ot 
ot


dtot = 11
4m
2+67m
2 = 18
1m
2 = 13.5 m


!
!
∆ytot
8.2 m
qtot = tan−1 
= tan−1  = 37° north of west
∆xtot
10.7 m
Practice 3D, p. 102
1. ∆y = −0.70 m
2∆y ∆x
x
∆x = 0.25 m
2
g = 9.81 m/s
2. ∆y = −1.0 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
"#−#g = v 
−g
v = "
#2∆#y ∆x = "
#(2−#)9(.#−8#01.
#7m#0/#ms#) (0.25 m) =
∆t =
∆x = 2.2 m
2
x
"#−#g = v 
−g
 ∆x =  (2.2 m) =
v = −"
"#(−29#)(.8#−11.#m#0#/ms#)
2#∆#y
∆t =
2∆y ∆x
x
2
g = 9.81 m/s
3. ∆y = −5.4 m
∆x = 8.0 m
"#−#g = v 
−g
m/s
v = "
#2∆#y ∆x = "
#(−29#)(.8#−1#5.
#4#m#) (8.0 m) =
g = 9.81 m/s
4. vx = 7.6 m/s
∆y = −2.7 m
2
g = 9.81 m/s
2
x
"#−#g = v 
2∆y
(2)(−2.7 m)
∆x = "
#−#g v = "
#−#9.#81#
#m#/s# (7.6 m/s) =
∆t =
4.9 m/s
2∆y ∆x
x
2
2
x
∆t =
0.66 m/s
7.6 m/s
2∆y ∆x
x
x
2
5.6 m
Section One—Pupil’s Edition Solutions
I Ch. 3–5
Practice 3E, p. 104
Givens
Solutions
1. ∆x = 4.0 m
∆x = vi (cos q)∆t
q = 15°
I
vi = 5.0 m/s
∆x
4.0 m
∆t =  =  = 0.83 s
vi (cos q ) (5.0 m/s)(cos 15°)
∆ymax = −2.5 m
∆y = vi (sin q)∆t − 2 g ∆t 2 = (5.0 m/s)(sin 15°)(0.83 s) − 2 (9.81 m/s2)(0.83 s)2
g = 9.81 m/s2
∆y = 1.1 m − 3.4 m = −2.3 m
2. ∆x = 301.5 m
q = 25.0°
2
g = 9.81 m/s
1
1
yes
At maximum height, vy,f = 0 m/s.
vy,f = vi (sin q ) − g∆t = 0
vi (sin q)

∆t = 
g
v2
∆x = vi (cos q)∆t = (cos q)(sin q)i
g
x
(9.81 m/s )(301.5 m)
=  = 87.9 m/s
"
#(c#o#s #qg#)∆
(#
n#
q #) "##
si#
(cos 25.0°)(sin 25.0°)
2
vi =
vy,f 2 = vi 2(sin q )2 − 2g∆ymax = 0
2
2
vi 2(sin q)2 (87.9 m/s) (sin 25.0°)
 = 
= 70.3 m
∆y max = 
2
(2)(9.81 m/s )
2g
q = 25°
∆x
∆x
42.0 m
∆t =  =  =  = 2.0 s
vx vi (cos q) (23.0 m/s)(cos 25°)
vi = 23.0 m/s
At maximum height, vy,f = 0 m/s.
3. ∆x = 42.0 m
g = 9.81 m/s2
vy,f 2 = vy,i2 − 2g∆ymax = 0
4. ∆x = 2.00 m
∆y = 0.55 m
q = 32.0°
g = 9.81 m/s2
I Ch. 3–6
∆x = vi(cos q)∆t
∆x
∆t = 
vi(cos q)
$
% $
%
∆x
∆x
∆y = vi(sin q)∆t − 12g∆t 2 = vi(sin q)  − 12 g 
vi(cos q)
vi(cos q)
g∆x2

∆y = ∆x(tan q) − 
2vi2(cos q)2
g∆x2
∆x(tan q) − ∆y = 
2 2
2vi (cos q)
g∆x2
2vi2(cos q)2 =  
∆x(tan q) − ∆y
2
vi =
"##
vi =
"###
vi =
"## "##
g∆x2

2
2(cos q) [∆x(tan q) − ∆y]
(9.81 m/s2)(2.00 m)2

(2)(cos 32.0°)2[(2.00 m)(tan 32.0°) − 0.55 m]
(9.81 m/s2)(2.00 m)2

=
(2)(cos 32.0°)2(1.25 − 0.55 m)
Holt Physics Solution Manual
(9.81 m/s2)(2.00 m)2

= 6.2 m/s
(2)(cos 32.0°)2(0.70 m)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vy,i2 vi2(sin q)2 (23.0 m/s)2(sin 25°)2
 = 
∆ymax =  = 
= 4.8 m
(2)(9.81 m/s2)
2g
2g
Givens
5. ∆x = 31.5 m
q = 40.0°
g = 9.81 m/s2
Solutions
∆x = vi(cos q)∆t
∆x
∆t = 
vi(cos q)
∆y = vi(sin q)∆t− 12 g∆t2 = 0
I
∆y = vi(sin q) − 12 g∆t = 0
g∆x
∆y = vi(sin q) −  = 0
2vi(cos q)
(9.81 m/s2)(31.5 m)
g∆x
vi =  =  = 17.7 m/s
(2)(cos 40.0°)(sin 40.0°)
2(cos q)(sin q)
"######### "##
At maximum height, vy,f = 0 m/s.
vy,f2 = vi2(sin q)2− 2g∆ymax = 0
2
2
vi2(sin q)2 (17.7 m/s) (sin 40.0°)
 = 
∆ymax = 
= 6.60 m
2
(2)(9.81 m/s )
2g
Section Review, p. 105
3. vx = 100.0 m/s
∆y = −50.0 m
g = 9.81 m/s2
4. vx = 100.0 m/s
∆x = 319 m
∆y = −50.0 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
1
∆y = − 2g∆t2
∆t =
"#−#g
2∆y
"##
2∆y
∆x = vx∆t = vx  = (100.0 m/s)
−g
"(2
#−)#9(#.−85#10
#m.0#/m#s )# =
2
319 m
2
vy = vy
=2g∆
m/s
)2
−(2)
)2(−
m) = ±31.3 m/s = −31.3 m/s
y = (0


(9
.8
1m
/s

50
.0

,i 


2
vtot = vy2+
m/s
)2
+(−
m/s
)2 = 1.
04
m2s2+
m2/
s2
00
.0


31
.3


00
0×1
9.8
0×102
v
y = (1



vtot = 10
m2/
s2 = 104.8 m/s
980


!
!
v
−31.3 m/s
q = tan−1 y = tan−1  = −17.4°
vx
100.0 m/s
q = 17.4° below the horizontal
5. ∆y = −125 m
vx = 90.0 m/s
g = 9.81 m/s2
1
∆y = − 2g∆t 2
2∆y
∆t =  =
−g
"
#−#9#.8#1
#m#/s# =
(2)(−125 m)
2
5.05 s
∆x = vx ∆t = (90.0 m/s)(5.05 s) = 454 m
6. ∆t = 0.50 s
∆x = 1.5 m
q = 33°
7. ∆t = 0.35 s
q = 67°
vi = 5.0 m/s
g = 9.81 m/s2
∆x = vi (cos q)∆t
∆x
1.5 m
vi =  =  = 3.6 m/s
(cos q )∆t (cos 33°)(0.50 s)
1
∆y = vi (sin q )∆t − 2 g∆t 2
1
∆y = (5.0 m/s)(sin 67°)(0.35 s) − 2 (9.81 m/s2)(0.35 s)2
∆y = 1.6 m − 0.60 m = 1.0 m
Section One—Pupil’s Edition Solutions
I Ch. 3–7
Practice 3F, p. 109
Givens
Solutions
1. vte = +15 m/s
vbe = vbt + vte = −15 m/s + 15 m/s = 0 m/s
vbt = −15 m/s
I
2. vaw = +18.0 m/s
vsa = −3.5 m/s
3. vfw = 2.5 m/s north
vwe = 3.0 m/s east
vsw = vsa + vaw = − 3.5 m/s
18.0 m/s
vsw = 14.5 m/s in the direction that the aircraft carrier is moving
vfe = vfw + vwe
2 v 2 = (2.5 m/s)2 + (3.0 m/s)2
vtot = vf
w

w+
e


vtot = 6.
s2
+9.0
m2/
s2 = 15
.2
m2/
s2 = 3.90 m/s
2m
2/




!
!
vfw
2.5 m/s
q = tan−1  = tan  = (4.0 × 101)° north of east
vwe
3.0 m/s
4. vtr = 25.0 m/s north
vdt = 1.75 m/s at 35.0° east
of north
vdr = vdt + vtr
vx,tot = vx,dt = (1.75 m/s)(sin 35.0°) = 1.00 m/s
vy,dt = (1.75 m/s)(cos 35.0°) = 1.43 m/s
vy,tot = vtr + vy,dt = 25.0 m/s + 1.43 m/s = 26.4 m/s
2 (v
2
vtot = (v
)2
+(26
m/s
)2
x,
to
y 
.0
0m
/s

.4


t)+
,to
t) = (1


vtot = 1.
m2/
s2
+697
m2/
s2 = 69
s2 = 26.4 m/s
00


8m
2/


!
!
vx, tot
1.00 m/s
 = tan−1  = 2.17° east of north
q = tan−1 
vy, tot
26.4 m/s
2. vwg = −9 m/s
vbg = 1 m/s
3. vbw = 0.15 m/s north
vwe = 1.50 m/s east
vbw = vbg vgw = vbg − vwg = (1 m/s) – (–9 m/s) = 1 m/s + 9 m/s
vbw =
10 m/s away in the oppposite direction
vbe = vbw + vwe
vtot = vbw
)2
+(1.
m/s
)2
2+vwe2 = (0
.1
5m
/s

50




vtot = 0.
m2/s2 = 1.51 m/s
02
2m
2/s2+2.2
5m
2/s2 = 2.
27



!
!
vw
0.15 m/s
q = tan−1 b
= tan−1  = 5.7° north of east
vwe
1.50 m/s
I Ch. 3–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Section Review, p. 109
Chapter Review and Assess, pp. 113–119
Givens
6. A = 3.00 units (u)
B = −4.00 units (u)
Solutions
Students should use graphical techniques.
A2+
B 2 = (3
)2
+(−
)2
a. A + B = 

.0
0u
4.
00
u


A + B = 9.
00
u2+16.
0u2 = 25
.0
u2 = 5.00 units


!
I
!
B
− 4.00 u
q = tan−1  = tan−1  = 53.1° below the positive x-axis
A
3.00 u
A2+
)2 = (3
)2
+(4.
)2
b. A − B = 
(−B
.0
0u
00
u


A − B = 9.
00
u2+16.
0u2 = 25
.0
u2 = 5.00 units


!
!
−B
4.00 u
q = tan−1  = tan−1  = 53.1° above the positive x-axis
A
3.00 u
A2+
)2 = (3
)2
+(−
)2
c. A + 2B = 
(2B

.0
0u
8.
00
u


A + 2B = 9.
00
u2+64.
0u2 = 73
.0
u2 = 8.54 units


!
!
2B
−8.00 u
q = tan−1  = tan−1  = 69.4° below the positive x-axis
A
3.00 u
B 2+
)2 = (−
)2
+(−
)2 = 5.00 units
d. B − A = 
(−A

40
0u
3.
00
u


!
!
B
− 4.00 u
q = tan−1  = tan−1  = 53.1° below the negative x-axis
−A
−3.00 u
or 127° clockwise from the positive x-axis
Copyright © by Holt, Rinehart and Winston. All rights reserved.
7. A = 3.00 m
Students should use graphical techniques.
B = 3.00 m
Ax = A(cos q) = (3.00 m)(cos 30.0°) = 2.60 m
q = 30.0°
Ay = A(sin q) = (3.00 m)(sin 30.0°) = 1.50 m
2 + (A + B)2 = (2.60 m)2 + (4.50 m)2
a. A + B = A

y 
x 


A + B = 6.
m2+
m2 = 5.20 m
76

20.
2m
2 = 27
.0



!
!
Ay + B
4.50 m
q = tan−1  = tan−1  = 60.0° above the positive x-axis
Ax
2.60 m
2 + (A − B)2 = (2.60 m)2 + (−1.50 m)2
b. A − B = A

y 
x 


A − B = 6.
m2+
m2 = 3.00 m
76

2.2
5m
2 = 9.
01



!
!
Ay − B
−1.50 m
q = tan−1  = tan−1  = 30.0° below the positive x-axis
Ax
2.60 m
Section One—Pupil’s Edition Solutions
I Ch. 3–9
Chapter Review and Assess, pp. 113–119 continued
Givens
Solutions
2 (−A )2 = (1.50 m)2 + (−2.60 m)2
c. B − A = (B
A
−


y)+
x 


B − A = 3.00 m
I
!
!
B − Ay
1.50 m
q = tan−1  = tan−1  = 30.0° above the negative x-axis
−Ax
−2.60 m
or 150° counterclockwise from the positive x-axis
2 + (A − 2B)2 = (2.60)2 + (−4.50)2 = 5.20 m
d. A − 2B = A

y  
x 


!
!
Ay − 2B
− 4.50 m
q = tan−1  = tan−1  = 60.0° below the positive x-axis
Ax
2.60 m
8. ∆y1 = −3.50 m
Students should use graphical techniques.
d2 = 8.20 m
∆x2 = d2 (cos q2 ) = (8.20 m)(cos 30.0°) = 7.10 m
q2 = 30.0°
∆y2 = d2 (sin q2 ) = (8.20 m)(sin 30.0°) = 4.10 m
∆x3 = 15.0 m
∆xtot = ∆x2 + ∆x3 = 7.10 m − 15.0 m = −7.9 m
∆ytot = ∆y1 + ∆y2 = −3.50 m + 4.10 m = 0.60 m
d = (∆
xt
)2
+(∆
yt
)2 = (−
)2
+(0.
m
)2


7.
9m

60

ot 
ot 


d = 62
m2+

0.3
6m
2 = 62
m
2= 7.9 m


!
!
∆ytot
0.60 m
q = tan−1 
= tan−1  = 4.3° north of west
∆xtot
−7.9 m
∆y = 13.0 m
Students should use graphical techniques.
d= ∆
x2
+
∆
y 2 = (−
m
)2
+(13
m
)2

8.
00

.0



d = 64
m2+
m2 = 23
.0

169

3m
2 = 15.3 m


!
!
∆y
13.0 m
q = tan−1  = tan−1  = 58.4° south of east
∆x
−8.00 m
22. ∆x1 = 3 blocks west
= −3 blocks east
∆y = 4 blocks north
a. ∆xtot = ∆x1 + ∆x2 = −3 blocks + 6 blocks = 3 blocks
∆ytot = ∆y = 4 blocks
d = (∆
xtot
)2
+(∆
ytot
)2 = (3




blo
ck
s)
2+(4blo
ck
s)
2


∆x2 = 6 blocks east
d = 9b
s2
+16b
s2 = 25
s2 = 5 blocks
lo
ck

lo
ck

blo
ck



!
!
∆ytot
4 blocks
q = tan−1 
= tan−1  = 53° north of east
∆xtot
3 blocks
b. distance traveled = 3 blocks + 4 blocks + 6 blocks = 13 blocks
I Ch. 3–10
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
9. ∆x = −8.00 m
Givens
Solutions
23. ∆x = 6.00 m
d= ∆
∆y2 = (6
)2
+ (−5.
m
)2
x2+
.0
0m

40

∆y = −5.40 m


d = 36
m2+
m2 = 8.07 m
.0

29.
2m
2 = 65
.2



!
!
∆y
−5.40 m
q = tan−1  = tan−1  = 42.0° south of east
∆x
6.00 m
24. ∆y1 = −10.0 yards
I
∆ytot = ∆y1 + ∆y2 = −10.0 yards + 50.0 yards = 40.0 yards
∆x = 15.0 yards
∆xtot = ∆x = 15.0 yards
∆y2 = 50.0 yards
d = (∆
xt
)2
+(∆
yt
)2 = (1


5.
0ya
rd
s)
2+(40
.0
yar
ds)
2
ot
ot


d = 22
s2
+1.6
03yar
s2 = 18
s2 = 42.7 yards
5ya
rd

0×1
d
20
yar
d

25. ∆y1 = −40.0 m

Case 1: ∆ytot = ∆y1 + ∆y2 = −40.0 m − 20.0 m = −60.0 m
∆x = ±15.0 m
∆xtot = ∆x = +15.0 m
∆y2 = ±20.0 m
d = (∆
m)2
+(15
m)2
ytot)2+(∆
xtot)2 = (−
60
.0

.0



d = 3.
03
m2+
m2 = 38
m2 = 61.8 m
60
×1
225

20



!
!
∆ytot
−60.0 m
q = tan−1 
= tan−1  = 76.0° south of east
∆xtot
15.0 m
Case 2: ∆ytot = ∆y1 + ∆y2 ! −40.0 m+ 20.0 m+ −20.0 m
∆xtot = ∆x = +15.0 m
d = (∆
yt
)2
+)∆
xt
)2 = (−
m
)2
+(15
m
)2


20
.0

.0

ot
ot


d = 4.
m2 = (6
m)2 = 25.0 m
00
×102+225

25



!
!
∆ytot
−20.0 m
∆ = tan−1 
= tan−1  = 53.1° south of east
∆tot
15.0 m
Case 3: ∆ytot = ∆y1 + ∆y2 = −40.0 m − 20.0 m = −60.0 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆xtot = ∆x = −15.0 m
d = (∆
yt
)2
+(∆
xt
)2 = (−
m
)2
+(−
m
)2


60
.0

15
.0

ot
ot


d = 61.8 m
!
!
∆ytot
−60.0 m
q = tan−1 
= tan−1  = 76.0° south of west
∆xtot
−15.0 m
Case 4: ∆ytot = ∆y1 + ∆y2 = −40.0 m + 20.0 m = −20.0 m
∆xtot = ∆x = −15.0 m
d = (∆
yt
)2
+(∆
xt
)2 = (−
m
)2
+(−
m
)2


20
.0

15
.0

ot
ot


d = 25.0 m
!
!
∆ytot
−20.0 m
q = tan−1 
= tan−1  = 53.1° south of west
∆xtot
−15.0 m
26. d = 110.0 m
∆ = −10.0°
∆x = d(cos q) = (110.0 m)[cos(−10.0°)] = 108 m
∆x = d(sin q) = (110.0 m)[sin(−10.0°)] = −19.1 m
Section One—Pupil’s Edition Solutions
I Ch. 3–11
Givens
Solutions
27. q = 25.0°
∆x = d(cos q) = (3.10 km)(cos 25.0°) = 2.81 km east
d = 3.10 km
I
28. d = 41.1 m
∆y = d(sin q) = (3.10 km)(sin 25.0°) = 1.31 km north
∆x = d(cos q) = (41.1 m)(cos 40.0°) = 31.5 m
q = 40.0°
∆y = d(sin q) = (41.1 m)(sin 40.0°) = 26.4 m
29. d1 = 100.0 m
∆x1 = d1(cos q1) = (100.0 m)(cos 0.00°) = 100.0 m
q1 = 0.00° east = 0.00°
∆y1 = d1(sin q1) = (100.0 m)(sin 0.00°) = 0.000 m
d2 = 300.0 m
∆x2 = d2(cos q2) = (300.0 m)[cos (−90.0°)] = 10.00 m
q2 = 90.0° south = −90.0°
∆y2 = d2(sin q2) = (300.0 m)[sin (−90.0°)] = −300.0 m
d3 = 150.0 m
∆x3 = d3(cos q3) = (150.0 m)[cos (−150°)] = −129.9 m
q3 = 30.0° south of west
∆y3 = d3(sin q3) = (150.0 m)[sin (−150°)] = −75.00 m
= 180.0° − 30.0° south of
east
= −150°
d4 = 200.0 m
q4 = 60.0 ° north of west
= 180° − 60.0° north of
east
= 120°
∆x4 = d4(cos q4) = (200.0 m)(cos (120°)] = −100.0 m
∆y4 = d4(sin q4) = (200.0 m)[sin (120°)] = 173.2 m
∆xtot = ∆x1 + ∆x2 + ∆x3 + ∆x4 = (100.0 m) + (0.00 m) + (−129.9 m) + (−100.0 m)
= −129.9 m
∆ytot = ∆y1 + ∆y2 + ∆y3 + ∆y4 = (0.000 m) + (−300.0 m) + (−75.00 m) + (173.2 m)
= −201.8 m
d = (∆
xtot)2+(∆
ytot)2 = (−
12
9.
9m
)2+(−20
1.
8m
)2


d = 16
m2+
m2 = 240.0 m
870

4072
0m
2 = 57
590



!
!
34. ∆y = −0.809 m
∆x = 18.3 m
2
g = 9.81 m/s
∆x
∆t =  
vx
!
−g
m/s
v = "
#2∆#y ∆x = "
#(2−#)#(9−.8#01#.8
#0#9 #m#) (18.3 m) =
1
1 ∆x
∆y = − 2g∆t 2 = − 2 g 
vx
2
2
x
35. vi = 1.70 × 103 m/s
q = 55.0°
g = 9.81 m/s2
1
45.1 m/s, or 162 km/h
1
a. ∆y = vi(sin q)∆t − 2g∆t2 = vi(sin q ) − 2g∆t = 0
2vi(sin q) (2)(1.70 × 103 m/s)(sin 55.0°)
 = 
∆t = 
= 284 s
9.81 m/s2
g
∆x = vi(cos q)∆t = (1.70 × 103 m/s)(cos 55.0°)(284 s) = 2.77 × 105 m
b. ∆t = 284 s (See a.)
I Ch. 3–12
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆ytot
−201.8 m
= tan−1  = 57.23° south of west
q = tan−1 
∆xtot
−129.9 m
Givens
Solutions
36. vx = 18 m/s
∆y = − 2 g∆t2
1
∆y = −52 m
g = 9.81 m/s2
∆t =
"#−#g = "
#−#9.#81#
#m#/#s =
2∆y
(2)(−52 m)
2
3.3 s
When the stone hits the water,
I
vy = −g∆t = (−9.81 m/s)(3.3 s) = −32 m/s
2
vtot = vx2+
)2+(−
m/s
)2
v
8m
/s

32


y = (1


vtot = 32
m2/s2 = 36 m/s
0m
2/s2+100
0m
2/s2 = 13
00


37. vx,s = 15 m/s
vx,o = 26 m/s
∆y = −5.0 m
g = 9.81 m/s2

1
∆y = − 2g∆t2
2∆y
∆t =  =
−g
"## "
#−#9.#81#
##/s# = 1.0 s
m
2(−5.0 m)
2
∆xs = vx,s ∆t = (15 m/s)(1.0 s) = 15 m
∆xo = vx,o ∆t = (26 m/s)(1.0 s) = 26 m
∆xo − ∆xs = 26 m − 15 m = 11 m
38. ∆x = 36.0 m
vi = 20.0 m/s
q = 53°
a. ∆x = vi(cos q)∆t
∆x
36.0 m
∆t =  =  = 3.0 s
vi(cos q) (20.0 m/s)(cos 53°)
1
1
∆ybar = 3.05 m
∆y = vi(sin q)∆t − 2g∆t2 = (20.0 m/s)(sin 53°)(3.0 s) − 2(9.81 m/s2)(3.0 s)2
g = 9.81 m/s2
∆y = 48 m − 44 m = 4 m
∆y = ∆ybar = 4 m − 3.05 m = 1 m
The ball clears the goal by 1 m.
b. vy,f = vi(sin q) − g∆t = (20.0 m/s)(sin 53°) − (9.81 m/s2)(3.0 s)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vx,f = 16 m/s − 29 m/s = −13 m/s
The velocity of the ball as it passes over the crossbar is negative; therefore, the ball
is falling.
39. vi = 25.0 m/s
q = 45.0°
∆x = 50.0 m
2
g = 9.81 m/s
∆x = vi (cos q)∆t
∆x
∆t = 
vi (cos q)
$
% $
%
∆x
∆x
1
1
∆y = vi (sin q)∆t − 2g∆t 2 = vi (sin q)  − 2g 
vi(cos q)
vi(cos q)
2
g∆x2
(9.81 m/s2)(50.0 m)2


∆y = ∆x(tan q) − 
=
(50.0
m)(tan
45.0°)
−
2vi2(cos q)2
(2)(25.0 m/s)(cos 45.0)2
∆y = 50.0 m − 39.2 m = 10.8 m
Section One—Pupil’s Edition Solutions
I Ch. 3–13
Givens
Solutions
40. ∆y = −1.00 m
Find the initial velocity of the water when shot at rest horizontally 1 m above the
ground.
∆x = 5.00 m
1
v = 2.00 m/s
∆y = − 2g∆t2
2∆ y
∆t =  
−g
∆t = 0.329 s
∆x = vx∆t
q = 45.0°
I
"##
g = 9.81 m/s2
∆x
5.00 m
∆x
vx =  =  =  = 11.1 m/s
y
2
∆
(2
)(−1.00 m)
∆t


−g
−9.81 m/s2
"## "########
Find how far the water will go if it is shot horizontally 1 m above the ground while
the child is sliding down the slide.
vx, tot = vx + v(cos q)
∆x = vx, tot∆t = [vx + v(cos q)]∆t = 11.1 m/s + (2.00 m/s)(cos 45.0°)](0.329 s)
∆x = [11.1 m/s + 1.41 m/s](0.329 s) = (12.5 m/s)(0.329 s) = 4.11 m
41. ∆x1 = 2.50 × 103 m
∆x2 = 6.10 × 102 m
∆ymountain = 1.80 × 103 m
2
vi = 2.50 × 10 m/s
q = 75.0°
g = 9.81 m/s2
For projectile’s full flight,
∆x
∆t = 
vi(cos q)
1
1
∆y = vi (sin q)∆t − 2g∆t 2 = vi(sin q) − 2g∆t = 0
$
%
∆x
1
vi (sin q) − 2g  = 0
vi(cos q)
2vi2(sin q)(cos q) (2)(2.50 × 102 m/s)2(sin 75.0°)(cos 75.0°)
 = 
∆x = 
= 3190 m
9.81 m/s2
g
Distance between projectile and ship = ∆x − ∆x1 − ∆x2
= 3190 m − 2.50 × 103 m − 6.10 × 102 m = 80 m
$
% $
%
∆x1
∆x1
1
1
∆y = vi(sin q)∆t′ − 2g∆t′2 = vi(sin q) 
− g 
vi(cos q) 2 vi(cos q)
2
g∆x12
∆y = ∆x1(tan q) − 
2 2
2vi (cos q)
∆y = (2.50 × 103 m)(tan 75.0°)
(9.81 m/s2)(2.50 × 103 m)2
− 
(2)(2.50 × 102 m/s)2 (cos 75.0°)2
∆y = 9330 m − 7320 = 2010 m
distance above peak = ∆y − ∆ymountain = 2010 m − 1.80 × 103 m =
47. vap = 165 km/h south
= −165 km/h north
vpe = 145 km/ north
I Ch. 3–14
vae = vap + vpe
vae = −165 km/h + 145 km/h = −20 km/h north = 20 km/h south
Holt Physics Solution Manual
210 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
For projectile’s flight to the mountain,
∆xi
∆t′ = 
vi (cos q)
Givens
Solutions
48. vre = 1.50 m/s east
a. vbe = vbr + vre
vbr = 10.0 m/s north
2
2
vbe ! vb
vr
)2
=(1.
m/s
)2
0.
0m
/s

50


r 
e = (1


vbe! 1.
02
m2/
s2
+2.2
s2 = 10
s2 = 10.1 m/s
00
×1
5m
2/
2m
2/


!
!
v
1.50 m/s
 = tan−1  = 8.53° east of north
q = tan−1 re
vbr
10.0 m/s
I
∆x 325 m
b. ∆t =  =  = 32.5 s
vbr 10.0 m/s
∆x = 325 m
49. vwe = 50.0 km/h south
vaw = 205 km/h
vae is directed due west
∆y = vre ∆t = (1.50 m/s)(32.5 s) = 48.8 m
a. vaw = vae + (−vwe)
!
!
v
we = sin q
vaw
v
50.0 km/h
q = sin−1 we = sin−1  = 14.1° north of west
vaw
205 km/h
b. vaw2 = vae2 + vwe2
2
vae = va
)2
−(50
)2
vwe2 = (2
05
km
/h

.0
km
/h

w−


vae = 4.
04km
03km
20
×1
2/h2−2.5
0×1
2/h2

vae = 3.
04km
95
×1
2/h2 = 1.99 km/h

50. ∆x = 1.5 km
vre = 5.0 km/h
vbr = 12 km/h
51. vre = 3.75 m/s downstream
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vsr = 9.50 m/s
vse is directed across the river
The boat’s velocity in the x direction is greatest when the boat moves directly across
the river with respect to the river.
1.5 km
∆x
∆tmin =  =  = 7.5 min
vbr (12 km/h)(1 h/60 min)
vre

= sin q
vsr
a. vsr = vse + (−vre)
!
3.75 m/s
q = sin−1  = 23.2° upstream from straight across
9.50 m/s
b. vsr2 = vse2 + vre2
2
vse = vs
−vre
)2
−(3.
m/s
)2
2 = (9
.5
0m
/s

75


r 


vse = 90
m2/
s2
−14.
s2 = 76
m2/
s2 = 8.72 m/s
.2

1m
2/
.1



vse = 8.72 m/s directly across the river
52. vbr = 12.0 m/s east
vre = 3.5 m/s south
a. vbe = vbr + vre
2
vbe = vb
+vre
2 = (1
2.
0m
/s
)2+(3.
5m
/s
)2
r 


vbe = 14
s2 
+12m
s2 = 15
s2 = 12.5 m/s
4m
2/
2/
6m
2/


!
!
v
3.5
 = tan−1  = 16° south of east
q = tan−1 re
vbr
12.0
∆x = 1360 m
∆x 1360 m
b. ∆t =  =  = 113 s
vbr 12.0 m/s
Section One—Pupil’s Edition Solutions
I Ch. 3–15
Givens
Solutions
53. ∆x = 130.0 m
a. ∆x = vi(cos q)∆t
q = 35.0°
∆y = 21.0 m − 1.0 m
= 20.0 m
I
g = 9.81 m/s2
∆x
∆t = 
vi(cos q )
$
% $
∆x
∆x
1
1
∆y = vi(sin q)∆t − 2 g∆t2 = vi(sin q)  − 2 g 
vi(cos q)
vi(cos q)
%
2
2
g∆x
∆y = ∆x(tan q) − 
2vi2(cos q)2
2vi2(cos q)2[∆x(tan q) − ∆y] = g∆x2
vi =
"##
g∆x2

2
2(cos q) [∆x(tan q) − ∆y]
"###
(9.81 m/s )(130.0 m)
 = 41.7 m/s
v = "##
(2)(cos 35.0°)(91.0 m − 20.0 m)
(9.81 m/s2)(130.0 m)2

(2)(cos 35.0)2[(130.0 m)(tan 35.0°) − 20.0 m]
vi =
2
2
i
130.0 m
∆x
b. ∆t =  =  = 3.81 s
vi(cos q) (41.7 m/s)(cos 35.0°)
c. vy,f = vi(sin q) − g∆t = (41.7 m/s)(sin 35.0°) − (9.81 m/s2)(3.81 s)
vy,f = 23.9 m/s − 37.4 m/s = −13.5 m/s
vx,f = vx = vi(cos q) = (41.7 m/s)(cos 35.0°) = 34.2 m/s
2
2
vf = (v
vy
)2
+(−
m/s
)2
x,
(
4.
2m
/s

13
.5


f )+
,f) = (3


vf = 11
m/s
m2/
s2 = 13
m2/
s2 = 36.7 m/s
70

2+182

50


54. ∆x = 12 m

q = 15°
2
g = 9.81 m/s
∆x
∆t = 
vi(cos q)
1
1
∆y = vi(sin q)∆t − 2 g∆t2 = vi(sin q) − 2 g∆t = 0
$
%
∆x
1
vi(sin q) − 2 g  = 0
vi(cos q)
2vi2(sin q)(cos q) = g∆x
vi =
"# "##
g∆x
 =
2(sin q)(cos q)
(9.81 m/s2)(12 m)
 = 15 m/s
(2)(sin 15°)(cos 15°)
12 m
∆x
b. ∆t =  =  = 0.83 s
(15 m/s)(cos 15°)
vi(cos q)
vy,f = vi(sin q) − g∆t = (15 m/s)(sin 15°) − (9.81 m/s2)(0.83 s)
vy,f = 3.9 m/s − 8.1 m/s = −4.2 m/s
I Ch. 3–16
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
a. ∆x = vi(cos q )∆t
Givens
Solutions
vx,f = vx = vi(cos q ) = (15 m/s)(cos 15°) = 14 m/s
2
vf = (v
x,f)2+(vy
4m
/s
)2+(−4.
2m
/s
)2
,f) = (1


vf = 2.
02
m2/
s2 
+18m
s2 = 22
s2 = 15 m/s
0×1
2/
0m
2/


See solution to Chapter 3 Review and Assess problem 53 for a derivation of the following equation.
55. ∆x = 10.0 m
q = 45.0°
∆y = 3.05 m − 2.00 m
= 1.05 m
vi =
"## "###
vi =
 =  = 10.5 m/s
"##
(2)(cos 45.0°) (10.0 m − 1.05 m) "##
(2)(cos 45.0°) (8.95 m)
g = 9.81 m/s2
g∆x2

=
2
2(cos q) [∆x(tan q) − ∆y]
I
(9.81 m/s2)(10.0 m)2

(2)(cos 45.0°)2[(10.0 m)(tan 45.0°) − 1.05 m]
(9.81 m/s2)(10.0 m)2
(9.81 m/s2)(10.0 m)2
2
2
1
∆y = vi(sin q)∆t − 2 g∆t2 = vi(sin q) − g∆t = 0
56. ∆t = 3.00 s
q = 30.0°
2
g = 9.81 m/s
57. ∆x = 20.0 m
∆t = 50.0 s
vpe = ±0.500 m/s
g∆t
(9.81 m/s2)(3.00 s)
vi =  =  = 29.4 m/s
2(sin q )
(2)(sin 30.0°)
∆x 20.0 m
veg =  =  = 0.400 m/s
∆t 50.0 s
vpg = vpe + veg
a. Going up:
vpg = vpe + veg = 0.500 m/s + 0.400 m/s = 0.900 m/s
20.0 m
∆x
∆tup =  =  = 22.2 s
vpg 0.900 m/s
b. Going down:
vpg = −vpe + veg = −0.500 m/s + 0.400 m/s = −0.100 m/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
−20.0 m
−∆x
∆tdown =  =  = 2.00 × 102 s
vpg −0.100 m/s
58. ∆y = −1.00 m
∆x = 1.20 m
g = 9.81 m/s2
∆x
∆t = 
vx
a. ∆x = vx ∆t
1
1 ∆x
∆y = − 2 g∆t2 = − 2 g 
vx
vx =
!
2
g∆x2
= 
2vx2
"# "##
−g∆x2
=
2y
−(9.81 m/s2)(1.20 m)2
 = 2.66 m/s
(2)(−1.00 m)
b. The ball’s velocity vector makes a 45° angle with the horizontal when vx = vy .
vx
vx = vy,f = −g∆t
∆t = 
g
vx 2
vx 2
1
1
2
∆y = − 2 g∆t = − 2 g  = − 
2g
g
!
2
(2.66 m/s)
∆y = − 
= − 0.361 m
(2)(9.81 m/s2)
h = 1.00 m − 0.361 m = 0.64 m
Section One—Pupil’s Edition Solutions
I Ch. 3–17
Givens
Solutions
59. v1 = 40.0 km/h
For lead car:
v2 = 60.0 km/h
∆xi = 125 m
∆xtot = v1∆t + ∆xi
For chasing car:
∆xtot = v2∆t
I
v2∆t = v1∆t + ∆xi
(125 m)(10−3 km/m)
∆xi
∆t =  = 
v2 − v1
(60.0 km/h − 40.0 km/h)(1 h/3600 s)
125 × 10−3 km
∆t =  = 22.5 s
(20.0 km/h)(1 h/3600 s)
60. q = 60.0°
d1 = v1∆t = (41.0 km/h)(3.00 h) = 123 km
v1 = 41.0 km/h
∆x1 = d1(cos q) = (123 km)(cos 60.0°) = 61.5 km
v2 = 25.0 km/h
∆y1 = d1(sin q) = (123 km)(sin 60.0°) = 107 km
∆t1 = 3.00 h
∆t2 = ∆t − ∆t1 = 4.50 h − 3.00 h = 1.50 h
∆t = 4.50 h
∆y2 = v2∆t2 = (25.0 km/h)(1.50 h) = 37.5 km
∆xtot = ∆x1 = 61.5 km
∆ytot = ∆y1 + ∆y2 = 107 km + 37.5 km = 144 km
d = (∆
xtot
)2
+(∆
ytot
)2 = (6
)2
+(14
)2




1.
5km

4km



d = 37
80
km
2+2070
0km
2 = 24
500
km
2 = 157 km

61. q = −24.0°
a = 4.00 m/s2
d = 50.0 m

1
a. d = 2 a∆t2
∆t1 =
 = 5.00 s
"#a = "#
4.00 m/s
2d
(2)(50.0 m)
2
∆y = −30.0 m
vi = a∆t1 = (4.00 m/s2)(5.00 s) = 20.0 m/s
g = 9.81 m/s2
2
vy,f = 
vi
(
si
n
q
)2
−2
g∆y = (2
]2
−(2)
m)
0.
0m
/s
2)[s
in
(−
24
.0
°)

(9
.8
1m
/s
2)(−
30
.0



vy,f = 66
m2/
s2
=589
m2/
s2 = 65
s2 = ±25.6 m/s = −25.6 m/s
.2


5m
2/

vy,f = vi(sin q) − g∆t2
vy,f − vi(sin q) −25.6 m/s − (20.0 m/s)(sin −24.0°)
∆t2 =  = 
−9.81 m/s2
−g
−25.6 m/s + 8.13 m/s −17.5 m/s
= 2 = 1.78 s
∆t2 = 
−9.81 m/s2
−9.81 m/s
∆x = vi(cos q)∆t2 = (20.0 m/s)[cos(−24.0°)](1.78 s) = 32.5 m
b. ∆t2 = 1.78 s (See a.)
62. vbw = ±7.5 m/s
vwe = 1.5 m/s
∆x d = 250 m
vbe = vbw + vwe
Going downstream:
vbe,d = 7.5 m/s + 1.5 m/s = 9.0 m/s
∆x u = −250 m
Going upstream:
vbe,u = −7.5 m/s + 1.5 m/s = −6.0 m/s
∆xd ∆xu
250 m
−250 m
∆t = 
+  =  +  = 28 s + 42 s = 7.0 × 101 s
vbe,d vbe,u 9.0 m/s − 6.0 m/s
I Ch. 3–18
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.

Givens
Solutions
63. q = 34°
a. ∆x = vi (cos q )∆t
∆x
∆t = 
vi (cos q )
∆x = 240 m
g = 9.81 m/s2
1
1
∆y = vi (sin q)∆t − 2 g∆t 2 = vi (sin q) − 2 g∆t = 0
$
%
∆x
g ∆x
1
vi (sin q) − 2 g  = vi2(sin q) −  = 0
vi(cos q)
2(cos q)
vi =
I
"######### "##
g ∆x
 =
2(cos q )(sin q )
(9.81 m/s2)(240 m)
 = 5.0 × 101 m/s
(2)(cos 34°)(sin 34°)
vy,f 2 − vy,i2
b. ∆ymax = 
−2g
Because vy,f = 0 m/s,
−vi2(sin q)2 (5.0 × 101 m/s)2(sin 34°)2
∆ymax =  = 
−2g
(2)(9.81 m/s2)
∆ymax = 4.0 × 101 m
$
% $
∆x
∆x
1
vi(sin q)  − g 
2vi(cos q) 2 2vi(cos q)
%
2
2
∆x
g∆x

∆ymax =  (tan q) − 
2
8 vi2(cos q)2
(240 m)(tan 34°)
(9.81 m/s2)(240 m)2
∆ymax =  − 
2
(8)(5.0 × 101 m/s)2(cos 34°)
∆ymax = 81 m − 41 m = 4.0 × 101 m
64. vwe = −0.500 m/s
∆x = 0.560 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆tupstream = 0.800 s
a. vse = vsw + vwe
Going upstream:
∆x
0.560 m
vse =  =  = 0.700 m/s
∆tupstream 0.800 s
0.700 m/s = vsw + (−0.500 m/s)
vsw = 0.700 m/s + 0.500 m/s = 1.20 m/s
Going downstream:
vse = −0.500 m/s (same as the water)
−0.500 m/s = vsw + (−0.500 m/s)
vsw = −0.500 m/s + 0.500 m/s = 0.00 m/s
b. d = vsw ∆t = (1.200 m/s)(0.800 s) = 0.960 m
−∆x
− 0.560 m
c. ∆tdownstream =  =  = 1.12 s
vse
− 0.500 m/s
∆ttotal = ∆tupstream + ∆tdownstream = 0.800 s + 1.12 s = 1.92 s
d
0.960 m
vsw, avg =  =  = 0.500 m/s
∆ttotal
1.92 s
Section One—Pupil’s Edition Solutions
I Ch. 3–19
Givens
Solutions
65. vce = 50.0 km/h east
a. vce = vrc (sin q )
v
50.0 km/h
 =  = 57.7 km/h
vrc = ce
(sin q) (sin 60.0°)
q = 60.0°
vrc = 57.7 km/h at 60.0° west of the vertical
I
b. vre = vrc (cos q) = (57.7 km/h)(cos 60.0°) = 28.8 km/h
vre = 28.8 km/h straight down
66. ∆twalk = 30.0 s
∆tstand = 20.0 s
L
L
vpe =  = 
∆twalk 30.0 s
L
L
veg =  = 
∆tstand 20.0 s
vpg = vpe + veg
vpg = vpe + veg
L
L
2L + 3L
5L
vpg =  +  =  = 
30.0 s 20.0 s
60.0 s
60.0 s
L
5L
  = 
∆t 60.0 s
60.0 s
∆t =  = 12.0 s
5
67. ∆x Earth = 3.0 m
2
g = 9.81 m/s
1
1
∆y = vi (sin q)∆t − 2 g∆t 2 = vi (sin q) − 2 g∆t = 0
2vi (sin q)

∆t = 
g
$
%
2vi (sin q)

∆xEarth = vi (cos q)∆t = vi (cos q) 
g
2vi2(cos q)(sin q)
Because vi and q are the same for all locations,
k
∆xEarth = , where k = 2vi2(cos q )(sin q)
g
!
g
k = g∆xEarth =  ∆xmoon = (0.38g)∆xMars
6
∆xmoon = 6∆xEarth = (6)(3.0 m) = 18 m
∆xEarth 3.0 m
∆xMars =  =  = 7.9 m
0.38
0.38
68. vx = 10.0 m/s
q = 60.0°
g = 9.81 m/s2
The observer on the ground sees the ball rise vertically, which indicates that the
x-component of the ball’s velocity is equal and opposite the velocity of the train.
vx = vi(cos q )
10.0 m/s
vx
vi =  =  = 20.0 m/s
(cos q) (cos 60.0°)
At maximum height, vy = 0, so
vy,f 2 − vy,i2 vi2(sin q)2
∆ymax =  = 
−2g
2g
(20.0 m/s)2(sin 60.0°)2
∆ymax = 
= 15.3 m
(2)(9.81 m/s2)
I Ch. 3–20
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆xEarth = 
g
Givens
Solutions
69. vi = 18.0 m/s
∆y = vi (sin q)∆t − 2 g∆t 2 = vi (sin q) − 2 g∆t = 0
q = 35.0°
1
1
∆xi = 18.0 m
2vi (sin q) 2(18.0 m/s)(sin 35.0°)
 = 
∆t = 
= 2.10 s
g
9.81 m/s2
g = 9.81 m/s2
∆x = vi (cos q)∆t = (18.0 m/s)(cos 35.0°)(2.10 s) = 31.0 m
I
∆xrun = ∆x − ∆xi = 31.0 m − 18.0 m = 13.0 m
∆x un 13.0 m
vrun = r
=  = 6.19 m/s downfield
∆t
2.10 s
ay = a(sin q) = (25 m/s2)(sin 53°) = 2.0 × 101 m/s2
70. q = 53°
1
1
vi = 75 m/s
∆y = vi (sin q)∆t + 2 ay ∆t 2 = (75 m/s)(sin 53°)(25 s) + 2 (2.0 × 101 m/s2)(25 s)2
∆t = 25 s
∆y = 1500 m + 6200 m = 7700 m
2
a = 25 m/s
vf = vi + a∆t = 75 m/s + (25 m/s2)(25 s) = 75 m/s + 620 m/s = 7.0 × 102 m/s
For the motion of the rocket after the boosters quit:
vi = vf = 7.0 × 102 m/s
q = 53°
g = 9.81 m/s2
vy,f = vi (sin q ) − g∆t = 0
2
vi (sin q) (7.0 × 10 m/s)(sin 53°)
 = 
∆t = 
= 57 s
9.81 m/s2
g
1
1
∆y = vi (sin q)∆t − 2 g∆t 2 = (7.0 × 102 m/s)(sin 53°)(57 s) − 2 (9.81 m/s2)(57 s)2
∆y = 32 000 m − 16 000 m = 16 000 m
a. ∆ytotal = 7700 m + 16 000 m = 2.4 × 104 m
1
b. ∆y = − 2 g∆t 2
∆t =
 = 7.0 × 10 s
"#−g# = "##
−9.81 m/s
2∆y
(2)(−24 000 m)
2
1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
∆ttotal = 25 s + 57 s + 7.0 × 101 s = 152 s
c. ax = a(cos q) = (25 m/s2)(cos 53°) = 15 m/s2
1
1
∆x = vi (cos q)∆t + 2 a∆t 2 = (75 m/s)(cos 53°)(25 s) + 2 (75 m/s2)(25 s)2
∆x = 1.1 × 103 m + 2.3 × 104 m = 2.4 × 104 m
vi = 7.0 × 102 m/s
After the rockets quit:
q = 53°
∆t = 57 s + 7.0 × 101 s = 127 s
∆x = vi (cos q)∆t = (7.0 × 101 m/s)(cos 53°)(127 s) = 5.4 × 104 m
∆xtot = 2.4 × 104 m + 5.4 × 104 m = 7.8 × 104 m
Section One—Pupil’s Edition Solutions
I Ch. 3–21