Solution 2.1
0t, s
–100
0
6
0t, s
–100
0
a, m/s
6
2
20 100 50
See plots
=−+
vt t
dv
Solution 2.2
(a) Particle displacement
10
8
4
2
–2
0
Velocity, v, ft/sec
(c) Particle acceleration
6
4
012345
2
0.81 1.3 2.35 ft/sec
1.62 1.3 ft/sec
vtt
dt
at
dt
21.080sec
0 0.81 1.3 2.35 2.69sec
vtt t
== − − →
=
Solution 2.3
32
Solution 2.4
2
210
==−
vt
dv
at
dt
Solution 2.6
At 00
0, and tss vv== =
+
0
222 2120
tv
v
2
1ct ds
22
1
11
t
st
ct ct
c
00
12 2
v
v
vv
ccvc
+
() ()
0 12 1 12 1201 120
22
11
ln ln
SS ccvc ccv ccvc ccv
cc
−= + − + − + − +
Solution 2.7
()
22
00
2
2( )
5280
060 2120
vv ass
a
−= −
−=
()
080 232.3
s
−=−
Solution 2.8
12
111
222
1
0
0.2 mm s , 3 mm/s
−
==
ts
k
v
ds ds
12
2
12
00 0 0
So,
()
=+
2
10.2 30 mm
4
st
12
1
ds k
15 0.2 5620 mm
ss
Solution 2.9
2
1
2.51, 7.43 sec
t
=
So impact must be when ball 1 is descending.
0
2
0
2
1
: g @ 300 ft:
2
y
yy vt t
=+ −
Solution 2.10
y
υ, ft/sec
Solution 2.11
()
22
2
0.985s
−= −
=
BA B A
BC
vv aSS
t
Solution 2.12
()
2 16.1
t
==
=4.24 sect represents the second time at which =50 ft.
y
Solution 2.13
0
100 3.6 27.8 m/s
==
v
14
Solution 2.14
()
22
2
3.6
393m
=
s
Solution 2.15
2
12 4 1.00 m/s
8
Δ−
== =
Δ
AV
v
at
Solution 2.16
22
02vv=+
as, where g
a
=
Solution 2.17
()
22
00
2
=+ −
vv ass
Solution 2.18
=
24.25 sect
vis constant between points and
16.28 ft/sec
a
Solution 2.19
The area under the a–s curve is
2
2
1
1
1
So 950, 43.6 m/s
2
40.0918s
43.6
−
==
== =
v
vv
dv a
ds v
Solution 2.20
()
22 2 2
2
1
So 0.01847 60 66.5sec
Car : 66.5 4 62.5sec
t
t
==
=−=
1
3600 62.5 mi
at 50 1.168 99.8
5280 3600 hr
vv=+ = + =
Solution 2.21
1g
Solution 2.22
()
()
acc
130 36.1 m/s constant
3.6
36.1
So 3 7.41 164.6 m
160
T
==
==
A
AA
p
svt t
S
()
hen 36.1 19.75 713 m==s
Solution 2.23
y
B
4.5 m/s
Solution 2.24
At 5 ft, 4 ft/secxv=− =
12
20249.81
1
2
−
=−
t
Solving, 125.4 mh=
23 23 23
total 1 1 2 2 3
total
125.4 0.85 147.5 s
8 2.45 147.5
157.9s
−−−
−−
=→= →=
=+ + =+ +
∴
=
f
hvt t t
tttt
t
3
a
0
1000
0246
t, sec
810
–200
Solution 2.26
2
ds
Solution 2.27
2
400
3.6
111.1
2
111.1
v
s
2
υ
Solution 2.28
22
0.0003
dv
avkvv
ds
=− =− =
Solution 2.29
()
2
1
a 200 100 mi/hr:
→=−=
dv
akvv
ds
()()
0 35. 2 5
−=−
s
Solution 2.30
0max
mid max
3
3.25 m/s 475 mm
2.85 m/s At 2
==
==
=− − =
vx
vxx
dv
akxkxv
()()
()
12
max
At , 0.475 0.475 0 3.25
242
−=−
x
2
Solution 2.31
2
() ()
() () ()( )
22 2 2
2
2
0
2
0
(
b) g moon radius
g
2 5.32 750 5280
2g 2
2160 750 5280
m
m
mm
Rv
mm
mh
Rdv
avR
rdr
dr
Rvdv
Rh
vR
+
=− = =
−=
==
+
Solution 2.32
For
()
H
y
t, assume a solution of form
()
t
y
te
λ
=
2
ttt t
λλλ λ
y
12
H
For
()
p
y
t, assume a solution of the form
()
3p
y
tc=
This yields,
() ( )
123
00 gkkc++ =
So… 3
g
ck
=
12
g 00and 0
tt
λλ
12
2
k
()
01122
0
y
vc c
λλ
′== +
Solving…
20 2
gkv
ck
λ
+
=− AND
20 1
gkv
ck
λ
+
=−
Hp
()
()
2
112
2
112
14
222
120 1 2
2
221 2
14
222
21 2
g2 g 4
g…
24
kkkt
kkkt
kkv k ke
y
tkkk k
−
+−
−
−−
−+ − −
=+
−
With numbers,
()
2.54 9.46
16.1 18.38 2.28 in.
tt
yt e e
−−
=− +
2
112
14
21 2
kkkt
y
()
2.54 t 9.46t
t
−−
()
2
lim
tyt k
→∞ = so,
2
24
ff
yy
k
== →=
0.0 0.5 1.0 1.5
Time, t, sec
2.0 2.5 3.0
00.0 0.5 1.0 1.5
Time, t, sec
2.0 2.5 3.0
1.234 sec
f
t
=
Solution 2.33
21
0
2
2
0.7 m/s , 0.2 , 4 m/s
−
===
=−
yv
mv
uv
vdv vdv
σ
η
ση
IF
2
0
2
2 m/s, ln 8.05 m
== = →=
vy y
Solution 2.35
Up: 2
g
u
dv
a
kv v dy
=− − =
()
0
0
2
0
2
0
20
g
g
11
ln g ln
2 0.006 9.81
h
v
vdv
dy kv
kv
hkv
=− +
+
=− + =
Down: 2
g
d
dv
akvv
dt
=− + =
0
f
v
y
Solution 2.36
Up: 2
g
u
dv
a
kv dt
=− − =
()
0
0
0
2
0
11
0
0
1
g
g
11
tan tan
gg
gg
10.006
tan 30 2.63 s
9.81
9.81 0.006
−−
−
=− +
==
==
u
t
v
v
u
u
dv
dt kv
vk k
tv
kk
t
Down: 2
g
d
dv
a
kv dt
=− + =
−−
−
=−+
==
=
2
00
11
0
1
g
g
11
tan tan
gg
gg
1 0.006
tan 24.1 9.81
Refer to solution
f
d
f
v
t
v
df
dv
dt kv
vk k
th hv
kk
h
g
Solution 2.37
()
2
10 10
1
10 32.2 , 0.788sec
′′
==
()
100 100
10 100 90
100 32.2 , 2.49sec
2
tt
′′
′′
==
()
1000
1000 32.2
2
7.88 7.84 0.0395sec
t
tt t
′
=
=−=−=