6 1
5
SOLUTION
12–58.
A two-stage rocket is fired vertically from rest with the
acceleration shown. After 15 s the first stage A burns out and
the second stage B ignites. Plot the v–t and s–t graphs which
describe the motion of the second stage for 0 … t … 40 s. A
B
t (s)
a (m/s2)
15
20
12–59.
T
h
e spee
d
of a tra
i
n
d
ur
i
ng t
h
e f
i
rst m
i
nute
h
as
b
een
recorded as follows:
Plot the graph, approximating the curve as straight-line
segments between the given points. Determine the total
distance traveled.
v–t
SOLUTION
t(s) 0 20 40 60
() m>sv0 16 21 24
*12–60.
SOLUTION
For elevator:
An
s.
s=110 ft
s=100 +4(2.620)
(+c)s2=s0+vt
t=2.620 s
–80.35 =4+(–32.2)t
(+c)v=v0+act
v=80.35 ft>sT
v2=(4)2+2(–32.2)( 0 –100)
(+c)v2=v0
A man riding upward in a freight elevator accidentally
drops a package off the elevator when it is 100 ft from the
ground. If the elevator maintains a constant upward speed
of determine how high the elevator is from the
ground the instant the package hits the ground. Draw the
v–tcurve for the package during the time it is in motion.
Assume that the package was released with the same
upward speed as the elevator.
4ft>s,
6 5
12–61.
Two cars start from rest side by side and travel along a
straight road. Car Aaccelerates at for 10 s and then
maintains a constant speed. Car Baccelerates at
until reaching a constant speed of 25 m/s and then
maintains this speed. Construct the a–t,v–t, and s–tgraphs
for each car until What is the distance between the
two cars when t=15 s?
t=15 s.
5m>s2
4m>s2
SOLUTION
At
Car B:
sB=0+0+1
2(5)t2=2.5t2
s=s0+v0t+1
2act2
When
When t = 10 s, vA = (vA)max = 40 m/s and sA = 200 m.
When t = 5 s, sB = 62.5 m.
When t = 15 s, sA = 400 m and sB = 312.5 m.
vB=25 m/s,
t=25
5=5s
vB=0+5t
v=v0+act
t=15 s,
sA=400 m
sA=40t–200
LsA
200
ds =Lt
10
40 dt
t710 s ,
ds =vdt
At t=10 s,
sA=200 m
sA=0+0+1
2(4)t2=2t2
s=s0+v0t+1
2act2
At t=10 s,
vA=40 m>s
vA=0+4t
When
Distance between the cars is
Ans.
Car Ais ahead of car B.
¢s=sA–sB=400 –312.5 =87.5 m
t=15 s,
sB=312.5
sB=25t–62.5
sB–62.5 =25t–125
LsB
62.5
ds =Lt
5
25 dt
t75s,ds =vdt
12–62.
If the position of a particle is defined as
where tis in seconds, construct the s–t,v–t, and a–tgraphs
for 0…t…10 s.
s
=15
t
–3
t
22ft,
SOLUTION
6 8
12–63.
From experimental data, the motion of a jet plane while
traveling along a runway is defined by the graph.
Construct the and graphs for the motion. When
,.s=0t=0
a–ts –t
n–t
SOLUTION
t(s)
60
v(m/s)
20
6 9
300 600
3
s (m)
a (m/s2)
*12–64.
The motion of a train is described by the a–s graph shown.
Draw the v–s graph if v = 0 at s = 0.
10
12–66.
The boat travels along a straight line with the speed
described by the graph. Construct the and graphs.
Also, determine the time required for the boat to travel a
distance if .s=0 when t=0s=400 m
a–ss–t
Lt
0
dt =Ls
0
ds
2s1>2
A
:
+
B
dt =ds
v
v (m/s)
20
80
v2 4s
7 2
12–67.
The graph of a cyclist traveling along a straight road is
shown. Construct the graph.a–s
vs
Thus at and
For
Thus at
The graph is shown in Fig. a.
Thus at and
At , achanges from to .amin =-0.6 ft>s2
amax =1.5 ft>s2
s=100 ft
aƒs=100 ft =0.01
A
100
B
+0.5 =1.5 ft>s2
aƒs=0=0.01
A
0
B
+0.5 =0.5 ft>s2
100 fts=0
a–s
aƒs=350 ft =0.0016
A
350
B
–0.76 =-0.2 ft>s2
aƒs=100 ft =0.0016
A
100
B
–0.76 =-0.6 ft>s2
s=100 ft and 350 ft
a=v dv
ds =
A
–0.04s+19
BA
–0.04
B
=
A
0.0016s–0.76
B
ft>s2
A
+
:
B
100 ft 6s…350 ft,
aƒs=100 ft =0.01
A
100
B
+0.5 =1.5 ft>s2
aƒs=0=0.01
A
0
B
+0.5 =0.5 ft>s2
100 fts=0
a=v dv
ds =
A
0.1s+5
BA
0.1
B
=
A
0.01s+0.5
B
ft>s2
A
+
:
B
v (ft/s)
15
5
v 0.1s 5
v 0.04 s 19
–
*12–68.
T
h
e v–sgrap
h
for a test ve
hi
c
l
e
i
s s
h
own. Determ
i
ne
i
ts
acceleration when and when s=175 m.s=100 m
v(m/s)
50
7 4
SOLUTION
2
1>2
12–69.
If the velocity of a particle is defined as v(t) = {0.8t2i +
12t1
>
2j + 5k} m
>
s, determine the magnitude and coordinate
direction angles
a
, b,
g
of the particle’s acceleration when
t = 2 s.
Ans:
a
=
5.31 m
>
s
2
a=53.0°
b
=37.0°
g
=90.0°
12–70.
The velocity of a particle is ,where t
is in seconds.If ,determine the
displacement of the particle during the time interval
.t=1sto t=3s
r=0when t=0
v=
5
3i+
(
6–2t
)
j
6
m
>
s
SOLUTION
7 6
12–71.
SOLUTION
Position:The position expressed in Cartesian vector form can be obtained by
applying Eq. 12–7.
When .
The coordinates of the particle are
Ans.(4 ft, 2 ft, 6 ft)
t=1s,r =(13+3)i+2j+(14+5)k={4i+2j+6k}ft
r={(t3+3) i+2j+(t4+5)k}ft
r–(3i+2j+5k)=t3i+t4k
Lr
r1
dr =Lt
013t2i+4t3k2dt
dr =vdt
v={3t2i+4t3k} ft/s
Lv
0
dv=Lt
016ti+12t2k2dt
Aparticle,originally at rest and located at point
(
3 ft, 2 ft, 5 ft
)
,
is subjected to an acceleration of
Determine the particle’s position (x, y, z)at t=1s.
a=56ti+12t2k6ft>s2.
Ans:
(4 ft, 2 ft,
6 ft)
*12–72.
The velocity of a particle is given by
, where tis in seconds.If
the particle is at the origin when , determine the
magnitude of the particle’s acceleration when . Also,
what is the x, y, zcoordinate position of the particle at this
instant?
t=2s
t=0
v=516t2i+4t3j+(5t+2)k6m>s
SOLUTION
SOLUTION
vx=15 cos 60°=7.5 ft>s
v
y
=15 sin 60°=12.99 ft>s
1
+
S
2
s=v0 t
x=7.5t
1
+
c
2
s=so+vot+
1
2
ac t2
y=0+12.99t+
1
2
(–32.2)t2
y
=
1.732x
–
0.286x
2
Since y
=
0.05x
2
,
0.05x2=1.732x–0.286x2
x(0.336x–1.732) =0
x=5.15 ft
Ans.
y
=
0.05(5.15)
2=
1.33 ft Ans.
Also,
1
+
S
2
s=v0t
x=15 cos 60°t
1
+
c
2
s=s0+v0t+
1
2
ac t2
y=0+15 sin 60°t+
1
2
(–32.2)t2
Since y
=
0.05x
2
12.99t–16.1t2=2.8125t2
t=0.6869 s
So that,
x=15 cos 60° (0.6868) =5.15 ft
Ans.
y
=
0.05(5.15)
2=
1.33 ft Ans.
12–73.
The water sprinkler, positioned at the base of a hill, releases
a stream of water with a velocity of 15 ft
>
s as shown.
Determine the point B(x, y) where the water strikes the
ground on the hill. Assume that the hill is defined by the
equation y = (0.05x2) ft and neglect the size of the sprinkler.
y
x
60
15 ft/s
B
y (0.05x2) ft
SOLUTION
a=6ti+12t2
k
Lv
0
dv=
Lt
0
(6ti+12t2
k) dt
v=3t2i+4t3
k
L
r
r
0
dr=L
t
0
(3t2
i+4t3
k) dt
r
–
(3i
+
2j
+
5k)
=
t
3
i
+
t
4
k
When t=2 s
r={11i+2j+21k} ft
Ans.
12–74.
A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft),
is subjected to an acceleration a = {6t i + 12t2 k} ft
>
s2.
Determine the particle’s position (x, y, z) when t = 2 s.
12–75.
A particle travels along the curve from A to B in
2 s. It takes 4 s for it to go from B to C and then 3 s to go
from C to D. Determine its average speed when it goes from
A to D.
SOLUTION
1