12–133.
A particle travels around a circular path having a radius of
50 m. If it is initially traveling with a speed of 10 m s and its
speed then increases at a rate of
determine the magnitude of the particle’s acceleraton four
seconds later.
v
#=10.05 v2 m>s2,
>
SOLUTION
12–134.
The motorcycle is traveling at a constant speed of 60 km h.
Determine the magnitude of its acceleration when it is at
point .A
>
SOLUTION
Acceleration: The speed of the motorcycle at ais
Since the motorcycle travels with a constant speed, Thus, the magnitude of
the motorcycle’s acceleration at is
Ans.a
=2
a
t
2
+
a
n
2
=2
02
+
0.76272
=
0.763 m
>
s2
A
at=0.
an=v2
r=16.672
364.21 =0.7627 m>s2
v=
¢
60 km
h
≤¢
1000 m
1 km
≤¢
1 h
3600 s
≤
=16.67 m>s
r=
B
1+ady
dx b2
R
3>2
`
d2y
dx2
`
=
B
1+
¢
1
222x–1>2
≤
2
R
3>2
`
– 1
422x–3>2
`
4x=25 m
=364.21 m
d2y
dx2=- 1
4 22x–3>2
dy
dx =1
222x–1>2
y
y2 2x
x
25 m
A
12–135.
When t = 0, the train has a speed of 8 m
>
s, which is increasing
at 0.5 m
>
s2. Determine the magnitude of the acceleration of
the engine when it reaches point A, at t = 20 s. Here the radius
of curvature of the tracks is
r
A = 400 m.
SOLUTION
vt
8 m
/
s
A
*12–136.
At a given instant the jet plane has a speed of
550 m
>
s and an acceleration of 50 m
>
s2 acting in the
direction shown. Determine the rate of increase in the
plane’s speed, and also the radius of curvature
r
of the path.
SOLUTION
550 m/s
70
a fi 50 m/s2
r
12–137.
SOLUTION
Ans.
When ,
Ans.
Ans.
Ans.at=9.81 sin 17.04° =2.88 m>s2
an=9.81 cos 17.04° =9.38 m>s2
ax=0ay=9.81 m>s2
u=tan–1a2.4525
8b=17.04°
v=31822+12.452522=8.37 m>s
vy=-2.4525 m>s
t=0.25 s
vy=0–9.81t
v=v0+act
y=-0.0766x2(Parabola)
y=-4.905ax
8b2
y=-4.905t2
y=0+0+1
2(–9.81)t2
A
+c
B
s=s0+v0t+1
2act2
x=8t
(:
+)s=v0t
vx=8m>s
The ball is ejected horizontally from the tube with a speed
of Find the equation of the path, and then
find the ball’s velocity and the normal and tangential
components of acceleration when t=0.25 s.
y=f1x2,8m>s.
v
A
8m/s
y
x
A
12–138.
The motorcycle is traveling at 40 m
>
s when it is at A. If the
speed is then decreased at
v
#
= – (0.05 s) m
>
s2, where s is in
meters measured from A, determine its speed and
acceleration when it reaches B.
SOLUTION
A
B
150 m
150 m
60
12–139.
Cars move around the “traffic circle” which is in the shape of
an ellipse. If the speed limit is posted at 60 km
>
h, determine
the minimum acceleration experienced by the passengers.
SOLUTION
2
2
x
60 m
40 m
y
(40)2 1y2
(60)2
x2
(60
)
2
x
40
y
0)
2
1
2
12–139. Continued
149
*12–140.
Cars move around the “traffic circle” which is in the shape of
an ellipse. If the speed limit is posted at 60 km
>
h, determine
the maximum acceleration experienced by the passengers.
SOLUTION
2
2
(40)2=10.4 m
x
60 m
40 m
y
(40)2 1y2
(60)2
x2
(60
)
2
x
40
y
0)
2
1
2
Ans:
a
max =
10.4 m
>
s
2
150
12–141.
A package is dropped from the plane which is flying with a
constant horizontal velocity of Determine
the normal and tangential components of acceleration and
the radius of curvature of the path of motion (a) at the
moment the package is released at A, where it has a
horizontal velocity of and (b) just before it
strikes the ground at B.
vA=150 ft>s,
vA=150 ft>s.
v
A
A
1500 ft
12–142.
The race car has an initial speed at A. If it
increases its speed along the circular track at the rate
where sis in meters, determine the time
needed for the car to travel 20 m. Take r=150 m.
at=10.4s2m>s2,
vA=15 m
>
s
SOLUTION
At ,
An
s.
t=1.21 s
s=20 m
1n (s+2s2+562.5)–3.166 196 =0.632 456t
1n (s+2s2+562.5)
`
s
0
=0.632 456t
Ls
0
ds
2s2+562.5
=0.632 456t
Ls
0
ds
20.4s2+225
=Lt
0
dt
n=
ds
dt
=20.4s2+225
n2=0.4s2+225
0.4s2
2=n2
2–225
2
0.4s2
2
`
s
0
=n2
2
`
15
n
Ls
0
0.4sds=Ln
15
ndn
ads=ndn
at=0.4s=ndn
ds
A
s
r
152
12–143.
The motorcycle travels along the elliptical track at a
constant speed v. Determine its greatest acceleration if
a7b
.
SOLUTION
2
2
a
b
a
y
x
fi 1
x2
a2
y2
b2
b
153
*12–144.
The motorcycle travels along the elliptical track at a
constant speed v. Determine its smallest acceleration if
a7b
.
SOLUTION
2
2
b
b
a
y
x
fi 1
x2
a2
y2
b2
a
12–145.
SOLUTION
When , Btravels a distance of
The distance traveled by particle Ais determined as follows:
(1)
Thus the distance between the two cyclists after is
Ans.
Acceleration:
For A, when ,
The magnitude of the A’s acceleration is
Ans.
For B, when ,
B
A
at
B
B=v
#
A=0
t=1s
aA=23.41762+18.642=19.0 m>s2
(an)A=v2
A
r=9.6552
5=18.64 m>s2
vA=0.632528.5442+160 =9.655 m>s
A
at
B
A=v
#
A=0.4
A
8.544
B
=3.4176 m>s2
t=1s
d=10.47 +8.544 –8=11.0 m
t=1s
s=8.544 m
1=1
0.6325
£
InC2s2+160 +s
2160 S≥
Lt
0
dt =Ls
0
ds
0.63252s2+160
dt =ds
v
v=0.63252s2+160
Lv
8m>s
vdv=Ls
0
0.4 sds
vdv=ads
dB=8(1) =8m
t=1s
Particles Aand Bare traveling counter-clockwise around a
circular track at a constant speed of 8 . If at
the instant shown the speed of Abegins to increase by
where sAis in meters, determine the
distance measured counterclockwise along the track from B
to Awhen .What is the magnitude of the
acceleration of each particle at this instant?
t=1s
(at)A=(0.4sA)m>s2,
m>s
r5m
120fi
sB
sA
A
B
u
12–146.
Particles Aand Bare traveling around a circular track at a
speed of 8 at the instant shown. If the speed of Bis
increasing by ,and at the same instant Ahas an
increase in speed of ,determine how long it
takes for a collision to occur.What is the magnitude of the
acceleration of each particle just before the collision occurs?
(at)A=0.8tm>s2
(at)B=4m>s2
m>s
SOLUTION
Distance Traveled: Initially the distance between the two particles is
. Since particle Btravels with a constant acceleration,
distance can be obtained by applying equation
[1]
The distance traveled by particle Acan be obtained as follows.
[2]
In order for the collision to occur
Solving by trial and error Ans.
Note: If particle Astrikes Bthen, .This equation will result in
.
Acceleration: The tangential acceleration for particle Aand Bwhen are
and ,respectively.When
,from Eq. [1], and
.Todetermine the normal acceleration, apply Eq. 12–20.
(an)B=y2
B
r=18.032
5=65.01 m>s2
(an)A=y2
A
r=10.512
5=22.11 m>s2
=8+4(2.5074) =18.03 m>s
yB=(y0)B+actyA=0.4
A
2.50742
B
+8=10.51 m>st=2.5074 s
(at)B=4m>s2
(at)A=0.8 t=0.8 (2.5074) =2.006 m>s2
t=2.5074
t=14.6 s 72.51 s
sA=5a240°
180° pb+sB
t=2.5074 s =2.51 s
0.1333t3+8t+10.47 =8t+2t2
sA+d0=sB
sA=0.1333t3+8t
LsA
0
dsA=Lt
0
A
0.4 t2+8
B
dt
dsA=yAdt
yA=
A
0.4 t2+8
B
m>s
LyA
8m>s
dyA=Lt
0
0.8 tdt
dyA=aAdt
sB=0+8t+1
2(4) t2=
A
8t+2t2
B
m
sB=(s0)B+(y0)Bt+1
2act2
=5a120°
180° pb=10.47 m
d0=ru
r5m
120fi
sB
sA
A
B
u
156
12–147.
The jet plane is traveling with a speed of 120 m
>
s which is
decreasing at 40 m
>
s2 when it reaches point A. Determine
the magnitude of its acceleration when it is at this point.
Also, specify the direction of flight, measured from the
x axis.
y fi 15 lnQ
R
80 m
y
x
A
x
80
Ans:
u=10.6°
y fi 15 lnQ R
80 m
y
x
A
x
80
*12–148.
The jet plane is traveling with a constant speed of 110 m
>
s
along the curved path. Determine the magnitude of the
acceleration of the plane at the instant it reaches
point A(y = 0).
SOLUTION
x
12–149.
SOLUTION
Acceleration:
The magnitude of the train’s acceleration at Bis
Ans.a=2at
2+an
2=2
A
–0.5
B
2+0.10502=0.511 m
>
s2
an=v2
r=202
3808.96 =0.1050 m>s2
at=v
#=-0.5 m>s2
r=c1+ady
dx b2d3>2
2d2y
dx22=C1+
¢
0.2 e
x
1000
≤
2S3>2
`
0.2
A
10–3
B
e
x
1000
`
6x=400 m
=3808.96 m
d2y
dx2=0.2a1
1000 be
x
1000 =0.2
A
10–3
B
e
x
1000.
dy
dx =200a1
1000 be
x
1000 =0.2e
x
1000
The train passes point Bwith a speed of which is
decreasing at . Determine the magnitude of
acceleration of the train at this point.
=– 0.5 m>s2
at
20 m>sy
y200 e
x
1000
B
A
12–150.
The train passes point Awith a speed of and begins
to
decrease its speed at a constant rate of .
Determine
the magnitude of the acceleration of the train
when it reaches point
B, where .sAB =412 m
at=– 0.25 m>s2
30 m>s
SOLUTION
Radius of Curv
ature:
Acceleration:
T
he magnitude of the train’s acceleration at Bis
Ans.a=2at
2+an
2=2
A
–0.5
B
2+0.18222=0.309 m
>
s2
an=v2
r=26.342
3808.96 =0.1822 m>s2
at=v
#=-0.25 m>s2
r=
B
1+ady
dx b2
R
3>2
2d2y
dx22=C1+
£
0.2ex
1000 ≥2S3>2
20.2
A
10–3
B
ex
1000 26x=400 m
=3808.96 m
d2y
dx2=0.2
A
10–3
B
ex
1000
dy
dx =0.2ex
1000
y=200ex
1000
vB=26.34 m>s
vB2=302+2(–0.25)(412 –0)
vB2=vA2+2at(sB–sA)
y
y200 e
x
1000
B
A
160
12–151.
SOLUTION
Ans.
Since ,
u=tan–1(–0.5) =26.6° g
dy
dx =-0.5
=2(0)2+(322)2=322 mm>s2
a=2at
2+an
2
an=n2
r=(300)2
279.5 =322 mm>s2
r=
C
1+
A
dy
dx
B
2
D
3
2
`
d2y
dx2
`
=[1 +(–0.5)2]3
2
`
5(10–3)
`
=279.5 mm
d2y
dx2
`
x=200
=40(103)
x3=5(10–3)
dy
dx
`
x=200
=-
20(103)
x2=-0.5
y=20(103)
x
at=dn
dt =0
n=300 mm>s
The particle travels with a constant speed of
along the curve. Determine the particle’s acceleration when
it is located at point (200 mm, 100 mm) and sketch this
vector on the curve.
300 mm
>
s
y
(mm)
x(mm)
P
v
y20(10
3
)
x
Ans:
a
=
322 mm
>
s
2
u=26.6°
g
Ans.