B
A
x
y
5 mi
9 mi
υ
B
Solution 5.138
44
480 704 ft sec
30
44
360 528 ft sec
30
B
v
v
vv rv
ω
==
==
=+×+
rel 373 686 ft secij
Solution 5.139
Refer to solution for Prob. 5.138.
rel 19.20 ft seca=
y
x
A
R
r
60°
Solution 5.140
()
cos30 sin30
rR i j
=°−°
=°−°
rel
2
R
With 88 ft sec
υ
= and R = 200 ft:
rel 22 38.1 ft secij
υ
=−
O
r
A
60°y
‘
Solution 5.141
()
cos30 sin30
2
rR i j
=− °− °
′′
C x′y′ is attached to car C, so 2k
R
υ
ω
′
=−
rel 22 38.1 ft secij
5.14: 2
rel rel
2
AC
aa r r V a
ωωω
=+×− + × +
rel 27.6 10.19 ft secaij=− +
(aP)t
2ω × υrel
υA
C
0.12 m
N = 4 rad/s
45°
Solution 5.142
Let P = point on ODE coincident with A.
υυυ
=+
APAP
()
=
rel
2
7.68 m s
Pt
a
α
α
== = 2
7.68 0.12 64.0 rad s CCW
ODE
Alternatively, with
2
2
OP
y
xω = –4 k rad/s
ω
Solution 5.143
ω
==
0
0.2 m, 20 rad/s
r
rel rel 2
rel
y
AC
r
Solution 5.144
ω
==
0
0.2 , 20 rad s
rm
A
y
240 km
B
υ
A
Solution 5.145
For circular orbit: At equator g = 9.814 m/s2
() ()
υ
=+= =
9.814 1000
g 6378 3600 27 960 km/hr
ARRh
υ
2
22
2
6378
A
R
rel rel
AB
()
()
()
ω
3−
=− =− =−
2
22
24
6378 10 0.7292 10 0.03391 m s
B
aRj j j
=−
2
rel
8.02 m s
aj
υ
A
30°
60°60°
AB
ED
y
6″
ω
Solution 5.146
Let B = point on DO coincident with A
rel
12
tan30
ABAB
B
AB B
υ
υυυ
==
== °
()
2
0
2
2
rel
6
0, 6 3 ft sec
12
6
22 2ftsec
12
arkij
rkk i i
ω
ωω
=×=× =
××= × × =−
At
Solution 5.147
ω
α
==−
==− +
3 rad sec 6 rad s
0.1 0.15 m
p
kk
rr i j
()
()
()
υω
α
ωυ
=×= ×− + =− −
=×=− ×− + = +
=×= ×− − = −
2
2
3 0.1 0.15 0.45 0.3 m s
6 0.1 0.15 0.9 0.6 m s
3 0.45 0.3 0.9 1.35 m s
t
n
rk i j i j
ark i jij
akijij
Solution 5.148
()
;2 0.8 0.6
rkxiyj ij
ωυ
×= × + =− −
Solution 5.149
2
dKk
dt
ω
α
ω
==−−
y
Solution 5.150
BABAA BA
r
L
υυυ υω
=+ =+×
()
or speed to the left
i
υυ
=−
Solution 5.151
0.624
C (ICR)
B
A
υ
B
= 0.75 m/s
υ
A
60°
30°
x = 6′
Solution 5.152
sin
xx
θ
==
()
υ
ω
2
== ==
2
2
0
48 14.4 ft sec
nt
PO n
aPOPO
P
a
A
= 0.5 ft/sec
a
B
= 0.5 ft/sec
Solution 5.154
()()
2
po po po
nt
aa a a
=+ +
Solution 5.156
sin sin105
θ
°
υ
B
Solution 5.158
Let P be a point on BC coincident with A.
222
361