12–2.
When a train is traveling along a straight track at 2 m/s, it
begins to accelerate at a = 160 v-42 m>s2, where v is in m/s.
Determine its velocity v and the position 3 s after the
acceleration.
s
SOLUTION
a =
dv
dt
dt =
dv
a
v
3
dt =
L0
3 =
dv
-4
L2 60v
1
(v5 - 32)
300
v = 3.925 m>s = 3.93 m>s
Ans.
ads = vdv
ds =
s
L0
ds =
s =
1 5
vdv
=
v dv
a
60
1
60 L2
3.925
v5 dv
1 v6 3.925
a b`
60 6 2
= 9.98 m
Ans.
v
12–13.
The acceleration of a rocket traveling upward is given by ap = b + c sp. Determine the rocket’s
velocity when sp = sp1 and the time needed to reach this altitude. Initially, vp = 0 and sp = 0 when
t = 0.
b
Given:
6
m
c
2
s
0.02
1
sp1
2
2000 m
s
Solution:
ap
´
µ
¶
b c sp
vp
vp dvp
0
´
µ
¶
dvp
dsp
sp
b c sp dsp
0
vp
2
b sp 2
dsp
vp
t
vp
dt
µ́
µ
µ
¶
sp
0
c 2
sp
2
2b sp c sp
1
2
2b sp c sp
2
dsp
2
vp1
t1
2b sp1 c sp1
µ́
µ
µ
¶
sp1
0
vp1
1
2b sp c sp
2
dsp
t1
322.49
m
s
19.27 s
Ans.
Ans.
12–14.
The acceleration of a rocket traveling upward is given by ap = b + c sp. Determine the time
needed for the rocket to reach an altitute sp1. Initially, vp = 0 and sp = 0 when t = 0.
b
Given:
6
m
c
2
s
0.02
1
sp1
2
100 m
s
Solution:
ap
´
µ
¶
b c sp
vp
´
µ
¶
vp dvp
0
dvp
dsp
sp
b c sp dsp
0
vp
2
b sp 2
dsp
vp
t
vp
dt
µ́
µ
µ
¶
sp
0
c 2
sp
2
2b sp c sp
1
2
2b sp c sp
2
dsp
2
vp1
t1
2b sp1 c sp1
µ́
µ
µ
¶
sp1
0
vp1
1
2b sp c sp
2
dsp
t1
37.42
m
s
5.62 s
Ans.
12–26.
When two cars A and B are next to one another, they are
traveling in the same direction with speeds vA and vB ,
respectively. If B maintains its constant speed, while A
begins to decelerate at aA , determine the distance d
between the cars at the instant A stops.
A
d
SOLUTION
Motion of car A:
v = v0 + act
0 = vA - aAt
t =
vA
aA
v2 = v20 + 2ac(s - s0)
0 = v2A + 2( - aA)(sA - 0)
sA =
v2A
2aA
Motion of car B:
sB = vBt = vB a
vA
vAvB
b =
aA
aA
The distance between cars A and B is
sBA = |sB - sA| = `
v2A
vAvB
2vAvB - v2A
` = `
`
aA
2aA
2aA
B
Ans.
12–28.
A particle travels to the right along a straight line with a
velocity v = [5>14 + s2] m>s, where s is in meters.
Determine its deceleration when s = 2 m.
SOLUTION
v =
5
4+s
v dv = a ds
dv =
-5 ds
(4 + s)2
5
- 5 ds
b = a ds
a
(4 + s) (4 + s)2
a =
- 25
(4 + s)3
When s = 2 m
a = - 0.116 m>s2
Ans.
12–34.
A boy throws a ball straight up from the top of a 12-m high
tower. If the ball falls past him 0.75 s later, determine the
velocity at which it was thrown, the velocity of the ball when
it strikes the ground, and the time of flight.
SOLUTION
Kinematics: When the ball passes the boy, the displacement of the ball in equal to zero.
Thus, s = 0. Also, s0 = 0, v0 = v1, t = 0.75 s, and ac = - 9.81 m>s2.
A+cB
s = s0 + v0t +
1 2
at
2 c
0 = 0 + v110.752 +
1
1 - 9.81210.7522
2
v1 = 3.679 m>s = 3.68 m>s
Ans.
When the ball strikes the ground, its displacement from the roof top is s = - 12 m.
Also, v0 = v1 = 3.679 m>s, t = t2, v = v2, and ac = - 9.81 m>s2.
A+cB
s = s0 + v0t +
1 2
at
2 c
-12 = 0 + 3.679t2 +
1
1 - 9.812t22
2
4.905t22 - 3.679t2 - 12 = 0
t2 =
3.679 ; 21- 3.67922 - 414.90521 -122
214.9052
Choosing the positive root, we have
t2 = 1.983 s = 1.98 s
Ans.
Using this result,
A+cB
v = v0 + act
v2 = 3.679 + 1 -9.81211.9832
= - 15.8 m>s = 15.8 m>s T
Ans.
12–85.
A particle travels along the curve from A to B in 1 s. If it
takes 3 s for it to go from A to C, determine its average
velocity when it goes from B to C.
y
B
20 m
SOLUTION
A
Time from B to C is 3 - 1 = 2 s
vang =
(rAC - rAB)
40i - (20i + 20j)
¢r
=
=
= {10i - 10j} m>s
¢t
¢t
2
Ans.
C
x
12–87.
Pegs A and B are restricted to move in the elliptical slots
due to the motion of the slotted link. If the link moves with
a constant speed of 10 m/s, determine the magnitude of the
velocity and acceleration of peg A when x = 1 m.
y
A
D
C
SOLUTION
v
Velocity: The x and y components of the peg’s velocity can be related by taking the
first time derivative of the path’s equation.
x2
+ y2 = 1
4
1
#
#
(2xx) + 2yy = 0
4
1 #
#
xx + 2yy = 0
2
or
1
xv + 2yvy = 0
2 x
(1)
At x = 1 m,
(1)2
+ y2 = 1
4
y =
23
m
2
Here, vx = 10 m>s and x = 1. Substituting these values into Eq. (1),
23
1
(1)(10) + 2 ¢
≤ vy = 0
2
2
vy = -2.887 m>s = 2.887 m>s T
Thus, the magnitude of the peg’s velocity is
v = 2vx 2 + vy 2 = 2102 + 2.8872 = 10.4 m>s
Ans.
Acceleration: The x and y components of the peg’s acceleration can be related by
taking the second time derivative of the path’s equation.
1 # #
##
##
# #
(xx + xx) + 2(yy + yy) = 0
2
1 #2
##
##
#
A x + xx B + 2 A y 2 + yy B = 0
2
or
1
A v 2 + xax B + 2 A vy 2 + yay B = 0
2 x
Since vx is constant, ax = 0. When x = 1 m, y =
(2)
23
m, vx = 10 m>s, and
2
vy = -2.887 m>s. Substituting these values into Eq. (2),
23
1
a d = 0
A 102 + 0 B + 2 c ( - 2.887)2 +
2
2 y
ay = - 38.49 m>s2 = 38.49 m>s2 T
Thus, the magnitude of the peg’s acceleration is
a = 2ax 2 + ay 2 = 202 + ( -38.49)2 = 38.5 m>s2
Ans.
B
x2
4
y2
1
x
10 m/s
12–9 1.
The pitching machine is adjusted so that the
baseball is launched with a speed of W" 30 ms. If the ball
strikes the ground at B, determine the two possible angles u"
at which it was launched.
vA 30 m/s
A
uA
1.2 m
B
30 m
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with point A.
x-Motion: Here, (v")Y 30 cos u", Y" 0 and Y# 30 m. Thus,
Y# Y"
30 0
U (v")YU
30 cos u"U
1
cos u"
(1)
y-Motion: Here, (v")Z 30 sin u", BZ H 9.81 ms2, and Z# 1.2 m. Thus,
D
Z# Z"
1.2 0
1
B U2
2 Z
1
( 9.81)U2
30 sin u"U
2
(v")ZU
4.905U2 30 sin u" U 1.2 0
(2)
Substituting Eq. (1) into Eq. (2) yields
4.905 A
2
1
1
I 30 sin u" A
I 1.2 0
cos u"
cos u"
1.2 cos2 u"
30 sin u" cos u" 4.905 0
Solving by trial and error,
u" 7.19° and 80.5°
Ans.
12–93.
The balloon A is ascending at rate vA and is being carried horizontally by the wind at vw. If a
ballast bag is dropped from the balloon when the balloon is at height h, determine the time
needed for it to strike the ground. Assume that the bag was released from the balloon with the
same velocity as the balloon. Also, with what speed does the bag strike the ground?
Given:
vA
12
km
hr
vw
20
km
hr
h
50 m
g
9.81
m
2
s
Solution:
Thus
ax
0
ay
g
vx
vw
vy
g t vA
sx
vw t
sy
1 2
g t vA t h
2
0
vx
1 2
g t vA t h
2
vw
vy
g t vA
t
2
vA vA 2g h
g
v
2
2
vx vy
t
3.55 s
v
32.0
m
s
Ans.
Ans.
12–95.
A projectile is given a velocity v0 at an angle f above the
horizontal. Determine the distance d to where it strikes the
sloped ground. The acceleration due to gravity is g.
y
v0
φ
SOLUTION
+ B
A:
d cos u = 0 + v0 (cos f) t
s = s0 + v0 t +
1 2
a t
2 c
d sin u = 0 + v0 (sin f)t +
1
( - g) t2
2
Thus,
d sin u = v0 sin f a
d cos u
1
d cos u 2
b - ga
b
v0 cos f
2 v0 cos f
sin u = cos u tan f -
gd cos2 u
2v20 cos2 f
d = (cos u tan f - sin u)
d =
x
d
s = s0 + v0 t
A+cB
θ
2v20 cos2 f
g cos2 u
v 20
sin 2f - 2 tan u cos2 f
g cos u
Ans.
A projectile is fired from the platform at B. The
12–98.
shooter fires his gun from point A at an angle of 30°.
Determine the muzzle speed of the bullet if it hits the
projectile at C.
B
C
vA
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with point A.
Y$ Y"
20 0
(v")YU
v" cos 30° U
(1)
y-Motion: Here, Z" 1.8, (v")Z v" sin 30°, and BZ H 9.81 ms2. Thus,
D
1
BZU2
2
Z$ Z"
(v")ZU
10 1.8
v" sin 30°(U)
1
( 9.81)(U)2
2
Thus,
10 1.8 A
20 sin 30°
I (U) 4.905(U)2
cos 30°(U)
U 0.8261 s
So that
v" 20
28.0 ms
cos 30°(0.8261)
10 m
30
1.8 m
20 m
x-Motion: Here, Y" 0 and Y$ 20 m. Thus,
A
Ans.
12–101.
It is observed that the skier leaves the ramp A at an angle
uA = 25° with the horizontal. If he strikes the ground at B,
determine his initial speed vA and the speed at which he
strikes the ground.
vA
uA
A
4m
3
5
4
100 m
SOLUTION
Coordinate System: x - y coordinate system will be set with its origin to coincide
with point A as shown in Fig. a.
4
x-motion: Here, xA = 0, xB = 100 a b = 80 m and (vA)x = vA cos 25°.
5
+ B
A:
xB = xA + (vA)xt
80 = 0 + (vA cos 25°)t
80
t =
vA cos 25°
(1)
3
y-motion: Here, yA = 0, yB = - [4 + 100 a b ] = - 64 m and (vA)y = vA sin 25°
5
and ay = - g = - 9.81 m>s2.
A+ c B
yB = yA + (vA)y t +
1
a t2
2 y
- 64 = 0 + vA sin 25° t +
1
( - 9.81)t2
2
4.905t2 - vA sin 25° t = 64
(2)
Substitute Eq. (1) into (2) yieldS
4.905 ¢
¢
2
80
80
≤ = vA sin 25° ¢
≤ = 64
vA cos 25°
yA cos 25°
2
80
≤ = 20.65
vA cos 25°
80
= 4.545
vA cos 25°
vA = 19.42 m>s = 19.4 m>s
Substitute this result into Eq. (1),
t =
80
= 4.54465
19.42 cos 25°
Ans.
B
12–101. continued
Using this result,
A+ c B
(vB)y = (vA)y + ay t
= 19.42 sin 25° + ( -9.81)(4.5446)
= - 36.37 m>s = 36.37 m>s T
And
+ B
A:
(vB)x = (vA)x = vA cos 25° = 19.42 cos 25° = 17.60 m>s :
Thus,
vB = 2(vB)2x + (vB)2y
= 236.372 + 17.602
= 40.4 m>s
Ans.
12–122.
If the roller coaster starts from rest at A and its speed
increases at at = (6 – 0.06s) m>s2, determine the magnitude
of its acceleration when it reaches B where sB = 40 m.
y
y
1 x2
100
A
s
SOLUTION
B
Velocity: Using the initial condition v = 0 at s = 0,
x
v dv = at ds
v
L0
s
vdv =
L0
A 6 - 0.06s B ds
30 m
v = a 212s - 0.06s2 b m>s
Thus,
(1)
vB = 412 A 40 B - 0.06 A 40 B 2 = 19.60 m>s
Radius of Curvature:
1 2
x
100
dy
1
=
x
dx
50
y =
d2y
dx2
=
1
50
B1 + a
r =
2
dy 2 3>2
b R
dx
d2y
dx2
B1 + a
=
2
2 3>2
1
xb R
50
2 1 2
50
5
= 79.30 m
x = 30 m
Acceleration:
#
a t = v = 6 - 0.06(40) = 3.600 m>s2
an =
19.602
v2
= 4.842 m>s2
=
r
79.30
The magnitude of the roller coaster’s acceleration at B is
a = 2at 2 + an 2 = 23.6002 + 4.8422 = 6.03 m>s2
Ans.
12–126.
y
If the car passes point A with a speed of 20 m>s and begins
to increase its speed at a constant rate of at = 0.5 m>s2,
determine the magnitude of the car’s acceleration when
s = 100 m.
y
B
16 m
SOLUTION
Velocity: The speed of the car at C is
vC 2 = vA 2 + 2a t (sC - sA)
vC 2 = 202 + 2(0.5)(100 - 0)
vC = 22.361 m>s
Radius of Curvature:
y = 16 -
1 2
x
625
dy
= -3.2 A 10-3 B x
dx
d2y
dx2
= -3.2 A 10-3 B
B1 + a
r =
2
dy 2 3>2
b R
dx
d2y
dx2
c1 + a -3.2 A 10-3 B x b d
2 3>2
=
2
- 3.2 A 10-3 B
4
= 312.5 m
x=0
Acceleration:
#
a t = v = 0.5 m>s
an =
vC 2
22.3612
=
= 1.60 m>s2
r
312.5
The magnitude of the car’s acceleration at C is
a = 2a2t + a2n = 20.52 + 1.602 = 1.68 m>s2
16
Ans.
1 2
x
625
s
A x
12–133.
v
Car B turns such that its speed is increased by
(at)B = (0.5et) m>s2, where t is in seconds. If the car starts
from rest when u = 0°, determine the magnitudes of its
velocity and acceleration when t = 2 s. Neglect the size of
the car.
B
SOLUTION
5m
dv
Velocity: The speed v in terms of time t can be obtained by applying a =
.
dt
A
dv = adt
v
L0
When t = 2 s,
t
dv =
t
L0
0.5e dt
v = 0.5 A et - 1 B
v = 0.5 A e2 - 1 B = 3.195 m>s = 3.19 m>s
Ans.
Acceleration: The tangential acceleration of the car at t = 2 s is
at = 0.5e2 = 3.695 m>s2. To determine the normal acceleration, apply Eq. 12–20.
an =
v2
3.1952
= 2.041 m>s2
=
r
5
The magnitude of the acceleration is
a = 2a2t + a2n = 23.6952 + 2.0412 = 4.22 m>s2
Ans.
u
12–139.
The motorcycle is traveling at a constant speed of 60 km> h.
Determine the magnitude of its acceleration when it is at
point A.
y
y2 ⫽ 2x
A
x
25 m
SOLUTION
Radius of Curvature:
y = 22x1>2
dy
1
= 22x - 1>2
dx
2
d2y
dx2
= -
1
22x - 3>2
4
B1 + a
r =
`
dy 2 3>2
b R
dx
d2y
`
dx2
1
2
2 3>2
B 1 + ¢ 22x - 1>2 ≤ R
=
1
` - 22x - 3>2 `
4
4
= 364.21 m
x = 25 m
Acceleration: The speed of the motorcycle at a is
v = ¢ 60
an =
1h
km 1000 m
≤¢
≤¢
≤ = 16.67 m>s
h
1 km
3600 s
v2
16.672
=
= 0.7627 m>s2
r
364.21
Since the motorcycle travels with a constant speed, at = 0. Thus, the magnitude of
the motorcycle’s acceleration at A is
a = 2at 2 + an2 = 202 + 0.76272 = 0.763 m>s2
Ans.
12–145.
The particle travels with a constant speed v along the curve. Determine the particle’s
acceleration when it is located at point x = x1.
Given:
mm
s
v
300
k
20 u 10 mm
3
x1
2
200 mm
Solution:
y ( x)
k
x
y' ( x)
d
y ( x)
dx
y'' ( x)
d
y' ( x)
dx
1 y' ( x)
y'' ( x)
U ( x)
2
§ sin T 1 ·
¨
¸
U x1 © cos T 1 ¹
3
T ( x)
atan ( y' ( x) )
2
a
v
a
§ 144 · mm
¨
¸
© 288 ¹ s2
T1
T x1
a
T1
322
mm
2
s
26.6 deg
Ans.
12–150.
The two particles A and B start at the origin O and travel in
opposite directions along the circular path at constant
speeds vA = 0.7 m>s and vB = 1.5 m>s, respectively.
Determine the time when they collide and the magnitude of
the acceleration of B just before this happens.
y
SOLUTION
st = 2p(5) = 31.4159 m
5m
sA = 0.7 t
B
sB = 1.5 t
A
vB
Require
0.7 t + 1.5 t = 31.4159
aB =
(1.5)2
v2B
= 0.45 m>s2
=
r
5
x
O
vA
sA + sB = 31.4159
t = 14.28 s = 14.3 s
1.5 m/s
Ans.
Ans.
0.7 m/s
12–163.
At the instant shown, the watersprinkler is rotating with an angular speed T' and an angular
acceleration T''. If the nozzle lies in the vertical plane and water is flowing through it at a
constant rate r', determine the magnitudes of the velocity and acceleration of a water particle as it
exits the open end, r.
Given:
T'
2
T''
3
rad
s
rad
2
s
m
s
r'
3
r
0.2 m
Solution:
v
a
2
r' rT'
r T'
2
2
2
rT'' 2r' T'
2
m
s
v
3.03
a
12.63
Ans.
m
2
s
Ans.
12–167. If arm OA rotates counterclockwise with a
constant angular velocity of u 2 rads, determine the
magnitudes of the velocity and acceleration of peg P at
u 30°. The peg moves in the fixed groove defined by the
lemniscate, and along the slot in the arm.
r2 (4 sin 2 u)m2
r
O
Time Derivatives:
S2 4 sin 2u
2SS 8 cos 2uu
4 cos 2uu
S ! ms
S
2 SS
S2 8 2 sin 2uu2
u 2 rads
cos 2u u 4 cos 2uu 2 sin 2uu2 S2
S " ms2
S
u 0
At u 30°,
S u 30° 24 sin 60° 1.861 m
(4 cos 60°)(2)
S u 30° 2.149 ms
1.861
4 0 2 sin 60° 22 (2.149)2
S u 30° 17.37 ms2
1.861
Velocity:
vu Su 1.861(2) 3.722 ms
vS S 2.149 ms
Thus, the magnitude of the peg’s velocity is
v 2
2vS 2
2
vu 2 22.149
2
3.7222 4.30 ms
Ans.
Acceleration:
B S S Su2 17.37 1.861 22 24.82 ms2
B u S u 2S u 0 2(2.149)(2) 8.597 ms2
Thus, the magnitude of the peg’s acceleration is
B 2B
2 S2
Bu 2 2(
2 24.82)2
8.5972 26.3 ms2
Ans.
u
P
The peg moves in the curved slot defined by the
12–168.
lemniscate, and through
the slot in the arm. At u 30°, the
angular
velocity is u 2 rads, and the angular acceleration
is u 1.5 rads2. Determine the magnitudes of the velocity
and acceleration of peg P at this instant.
r2 (4 sin 2 u)m2
r
O
Time Derivatives:
2SS 8 cos 2uu
4 cos 2uu
S A
I ms
S
2 3SS
S2 4 8 3 2 sin 2uu
u 2 rads
cos 2uu2 4
4 cos 2uu 2 sin 2uu2 S2
S " ms2
S
u 1.5 rads2
At u 30°,
S u 30° 24 sin 60° 1.861 m
(4 cos 60°)(2)
S u 30° 2.149 ms
1.861
4 cos 60°(1.5) 2 sin 60° 22 (2.149)2
S u 30° 15.76 ms2
1.861
Velocity:
vu Su 1.861(2) 3.722 ms
vS S 2.149 ms
Thus, the magnitude of the peg’s velocity is
v 2
2BS 2
2
Bu 2 22.149
2
3.7222 4.30 ms
Ans.
Acceleration:
B S S S u2 15.76 1.861 22 23.20 ms2
Bu Su 2S u 1.861(1.5) 2(2.149)(2) 11.39 ms2
Thus, the magnitude of the peg’s acceleration is
B 2B
2 S2
Bu 2 2(
2 23.20)2
11.392 25.8 ms2
Ans.
u
P
12– 170.
The pin follows the path described by the equation r = a + bcos T. At the instant T = T1. the
angular velocity and angular acceleration are T' and T''. Determine the magnitudes of the pin’s
velocity and acceleration at this instant. Neglect the size of the pin.
Given:
a
0.2 m
b
0.15 m
T1
30 deg
T'
0.7
T''
0.5
rad
s
rad
2
s
T
Solution:
T1
r
a b cos T
v
r' rT'
a
r'' rT'
2
2
r'
b sin T T'
2
2
rT'' 2r' T'
2
r''
2
b cos T T' b sin T T''
v
0.237
a
0.278
m
s
m
2
s
Ans.
Ans.
12–171.
The slotted link is pinned
# at O, and as a result of the
constant angular velocity u = 3 rad>s it drives the peg P for
a short distance along the spiral guide r = 10.4u2 m, where
u is in radians. Determine the radial and transverse
components of the velocity and acceleration of P at the
instant u = p>3 rad.
0.5 m
P
r
·
u
r
3 rad/s
SOLUTION
#
u = 3 rad>s r = 0.4 u
#
#
r = 0.4 u
$
$
r = 0.4 u
At u =
p
,
3
u
O
r = 0.4189
#
r = 0.4(3) = 1.20
$
r = 0.4(0) = 0
#
v = r = 1.20 m>s
#
vu = r u = 0.4189(3) = 1.26 m>s
#
$
ar = r - ru2 = 0 - 0.4189(3)2 = - 3.77 m>s2
$
##
au = r u + 2ru = 0 + 2(1.20)(3) = 7.20 m>s2
Ans.
Ans.
Ans.
Ans.
0.4 u
12–172.
Solve Prob. # 12–171
if the slotted
link has . an angular
#
#
acceleration u = 8 rad>s 2 when u = 3 rad>s at u = p>3 rad.
SOLUTION
#
u = 3 rad/s
u =
0.5 m
r = 0.4 u
#
#
r = 0.4 u
$
$
r = 0.4 u
r
·
u
r
3 rad/s
u
p
3
O
#
u = 3
$
u = 8
r = 0.4189
#
r = 1.20
$
r = 0.4(8) = 3.20
#
vr = r = 1.20 m>s
#
vu = r u = 0.4189(3) = 1.26 m>s
#
$
ar = r - ru2 = 3.20 - 0.4189(3)2 = - 0.570 m>s2
$
##
au = r u + 2 ru = 0.4189(8) + 2(1.20)(3) = 10.6 m>s2
P
Ans.
Ans.
Ans.
Ans.
0.4 u
12–173.
The slotted link is pinned
# at O, and as a result of the
constant angular velocity u = 3 rad>s it drives the peg P for
a short distance along the spiral guide r = 10.4u2 m, where
u is in radians. Determine the velocity and acceleration of
the particle at the instant it leaves the slot in the link, i.e.,
when r = 0.5 m.
0.5 m
P
r
·
u
SOLUTION
u
r = 0.4 u
#
#
r = 0.4 u
$
$
r = 0.4 u
#
u = 3
$
u = 0
O
At r = 0.5 m,
u =
0.5
= 1.25 rad
0.4
#
r = 1.20
$
r = 0
#
vr = r = 1.20 m>s
#
vu = r u = 0.5(3) = 1.50 m>s
#
$
ar = r - r(u)2 = 0 - 0.5(3)2 = - 4.50 m>s2
$
##
au = ru + 2ru = 0 + 2(1.20)(3) = 7.20 m>s2
r
3 rad/s
Ans.
Ans.
Ans.
Ans.
0.4u
12–179.
A block moves outward along the slot in the platform with
#
a speed of r = 14t2 m>s, where t is in seconds. The platform
rotates at a constant rate of 6 rad/s. If the block starts from
rest at the center, determine the magnitudes of its velocity
and acceleration when t = 1 s.
·
θ = 6 rad/s
r
θ
SOLUTION
#
r = 4t|t = 1 = 4
#
$
u = 6
u = 0
1
L0
dr =
$
r = 4
1
L0
4t dt
r = 2t2 D 10 = 2 m
#
#
v = 3 A r B 2 + A ru B 2 = 2 (4)2 + [2(6)]2 = 12.6 m>s
#
$
$
##
a =
r - ru2 2 + ru + 2 ru 2 =
[4 - 2(6)2 ]2 + [0 + 2(4)(6)]2
= 83.2 m s
2
Ans.
Ans.
12–180.
The rod OA rotates counterclockwise with a constant angular velocity of T'. Two
pin-connected slider blocks, located at B, move freely on OA and the curved rod whose shape
is a limaçon described by the equation r = b(c cos(T)). Determine the speed of the slider
blocks at the instant T T1.
Given:
rad
s
T'
5
b
100 mm
c
2
T1
120 deg
Solution:
T
T1
r
b c cos T
r'
b sin T T'
v
2
r' rT'
2
v
1.323
m
s
Ans.
12–181.
The rod OA rotates counterclockwise with a constant angular velocity of T'. Two
pin-connected slider blocks, located at B, move freely on OA and the curved rod whose shape
is a limaçon described by the equation r = b(c cos(T)). Determine the acceleration of the
slider blocks at the instant T T1.
Given:
rad
s
T'
5
b
100 mm
c
2
T1
120 deg
Solution:
T
T1
r
b c cos T
r'
b sin T T'
r''
b cos T T'
a
r'' rT'
2
2
2
2r' T'
2
a
8.66
m
2
s
Ans.
12 –183. The slotted arm AB drives pin C through the spiral
groove described by the
equation R A .. If the angular
velocity is constant at ., determine the radial and transverse
components of velocity and acceleration of the pin.
B
C
Time Derivatives: Since . is constant, then . 0.
R A.
R A.
R A. 0
r
.
Velocity: Applying Eq. 12–25, we have
2R R A.
2. R. A..
Acceleration: Applying Eq. 12–29, we have
AR R R.2 0 A..2 A..2
A. R. 2R . 0 2(A.)(.) 2A.2
Ans.
Ans.
Ans.
Ans.
A
12–184. The slotted arm AB drives pin C through the
spiral groove described by the equation R (1.5 .) m, where
. is in radians. If the arm starts from
rest when . 60° and
is driven at an angular velocity of . (4T) rads, where t is
in seconds, determine the radial and transverse components
of velocity and acceleration of the pin C when T 1 s.
B
C
r
Time Derivatives: Here, . 4 T and . 4 rads2.
R 1.5 .
R 1.5 . 1.5(4T) 6T
R 1.5 . 1.5 (4) 6 ms2
.
Velocity: Integrate the angular rate,
'
)
3
T
D. '0
4TDT, we have . 1
1
Then, R ( (6T2 )) 8 m. At T 1 s, R 6(12)
2
2
and . 4(1) 4 rads. Applying Eq. 12–25, we have
1
(6T2
3
.
)) rad.
) 4.571 m, R 6(1) 6.00 ms
2R R 6.00 ms
2. R. 4.571 (4) 18.3 ms
Ans.
Ans.
Acceleration: Applying Eq. 12–29, we have
R.2 6 4.571 42 67.1 ms2
A. R . 2R. 4.571(4) 2(6) (4) 66.3 ms2
AR R
Ans.
Ans.
A
12 –194 . At the instant u 30°, the cam rotates with a
clockwise angular
velocity of u 5 rads and and angular
acceleration of u 6 rads2. Determine the magnitudes of
the velocity and acceleration of the follower rod AB at this
instant. The surface of the cam has a shape of a limaçon
defined by S (200 100 cos u) mm.
r (200 100 cos u) mm
u
Time Derivatives:
S (200 100 cos u) mm
S (100 sin uu) mms
S 100 sin uu cos uu2 mms2
When u 30°,
S u 30° 200
100 cos 30° 286.60 mm
S u 30° 100 sin 30°(5) 250 mms
S u 30° 100 sin 30°(6)
cos 30° 52 2465.06 mms2
Velocity: The radial component gives the rod’s velocity.
vS S 250
250 mms
Acceleration: The radial component gives the rod’s acceleration.
BS S Su2 2465.06 286.60 52 9630
9630 mms2
Ans.
Ans.
A
B
Determine the displacement of the log if the
12 –203.
truck at C pulls the cable 1.2 m to the right.
B
S#) L
2S"
(S"
3S"
S# L
3bS"
bS# 0
Since bS# 3bS" bS" 1.2, then
1.2
0.4 m 0.4 m Ans.
C
12–209.
If the point A on the cable is moving upwards at vA, determine the speed of block B.
Given:
vA
m
s
14
Solution:
L1
0
sD sA sD sE
2vD vA vE
L2
0
sD sE sC sE
vD vC 2vE
L3
0
sC sD sC sE
2vC vD vE
m
s
Guesses
vC
Given
0
2vD vA vE
0
vD vC 2vE
0
2vC vD vE
§ vC ·
¨ ¸
¨ vD ¸
¨v ¸
© E¹
vB
1
vD
vB
2
m
s
§ vC ·
¨ ¸
¨ vD ¸
¨v ¸
© E¹
Find vC vD vE
vC
1
m
s
vE
1
m
s
§ 2 ·
¨ 10 ¸ m
¨
¸ s
© 6 ¹
Positive means down,
Negative means up
Ans.
12–210.
The motor draws in the cable at C with a constant velocity vC. The motor draws in the cable at
D with a constant acceleration of aD. If vD = 0 when t = 0, determine (a) the time needed for
block A to rise a distance h, and (b) the relative velocity of block A with respect to block B
when this occurs.
Given:
vC
4
aD
8
m
s
m
2
s
h
3m
Solution:
L1
sD 2sA
0
vD 2vA
0
aD 2aA
L2
sB sB sC
0
2vB vC
0
2aB aC
aA
aD
2
vA
aA t
sA
h
vA
aA t
§ t2 ·
¸
©2¹
aA¨
t
vB
2h
aA
1
vC
2
t
vAB
1.225 s
vA vB
Ans.
vAB
2.90
m
s
Ans.
12–217.
The crate C is being lifted by moving the roller at A
downward with a constant speed of vA = 2 m>s along the
guide. Determine the velocity and acceleration of the crate
at the instant s = 1 m. When the roller is at B, the crate
rests on the ground. Neglect the size of the pulley in the
calculation. Hint: Relate the coordinates xC and xA using
the problem geometry, then take the first and second time
derivatives.
4m
B
xA
xC
4m
SOLUTION
s
xC + 2x2A + (4)2 = l
#
#
1
xC + (x2A + 16) - 1/2(2xA)(xA) = 0
2
1
# 2
$
# 2
$
xC - (x2A + 16) - 3/2 (2x2A)(xA ) + (x2A + 16) - 1/2 (xA) + (x2A + 16) - 1/2 (xA)(xA) = 0
2
l = 8 m , and when s = 1 m ,
xC = 3 m
xA = 3 m
#
vA = xA = 2 m>s
$
aA = x A = 0
Thus,
vC + [(3)2 + 16] - 1/2 (3)(2) = 0
vC = -1.2 m>s = 1.2 m>s c
Ans.
aC - [(3)2 + 16] - 3/2 (3)2(2)2 + [(3)2 + 16] - 1/2 (2)2 + 0 = 0
aC = -0.512 m>s2 = 0.512 m>s2 c
A
C
Ans.
12–221.
At a given instant, two particles A and B are moving with a speed of v0 along the paths
shown. If B is decelerating at v'B and the speed of A is increasing at v'A, determine the
acceleration of A with respect to B at this instant.
Given:
v0
a
8
m
s
1m
v'A
5
m
2
s
v'B
6
m
2
s
Solution:
3
2
y ( x)
§x·
a¨ ¸
© a¹
U
1 y' ( a)
y'' ( a)
d
y ( x)
dx
y' ( x)
2
d
y' ( x)
dx
y'' ( x)
3
T
U
atan ( y' ( a) )
2
aA
aAB
§ cos T · v0 § sin T ·
v'A ¨
¸
¨
¸
© sin T ¹ U © cos T ¹
aA aB
aAB
§ 0.2 · m
¨
¸
© 4.46 ¹ s2
aB
aAB
7.812 m
v'B § 1
·
¨ ¸
2 © 1 ¹
4.47
m
2
s
Ans.
12–222.
Two planes, A and B, are flying at the same altitude. If their
velocities are vA = 600 km>h and vB = 500 km>h such that
the angle between their straight-line courses is u = 75°,
determine the velocity of plane B with respect to plane A.
A
vA
B
u
SOLUTION
vB
vB = vA + vB/A
75°
[500 ; ] = [600 c u ] + vB/A
+ )
(;
500 = - 600 cos 75° + (vB/A)x
(vB/A)x = 655.29 ;
(+ c)
0 = -600 sin 75° + (vB/A)y
(vB/A)y = 579.56 c
1vB>A2 = 31655.2922 +1579.5622
vB/A = 875 km/h
u = tan-1 a
579.56
b = 41.5° b
655.29
Ans.
Ans.
12 –231. At the instant shown, cars A and B travel at
*12–224.
At the instant shown, cars A and B travel at speeds
speeds of 70 kmh and 50 kmh, respectively. If B is
of 70 mih and 50 mih, respectively. If2 B is increasing its speed
increasing its speed by 1100 kmh , while A maintains a
by 1100 mih2, while A maintains a constant speed, determine
constant speed, determine the velocity and acceleration of
the velocity and acceleration of B with respect to A. Car B
B with respect to A. Car B moves along a curve having a
moves along a curve having a radius of curvature of 0.7 mi.
radius of curvature of 0.7 km.
A
VA 70 kmh
Relative Velocity:
v" v!
v"!
30°
50 cos 30°j 70j
50 sin 30°i
v"! {25.0i
v"!
26.70j} kmh
Thus, the magnitude of the relative velocity vB/A is
2"! 25.02
( 26.70)2 36.6 kmh
Ans.
The direction of the relative velocity is the same as the direction of that for relative
acceleration. Thus
1
. tan
26.70
46.9° 25.0
Ans.
Relative Acceleration: Since car B is traveling along a curve, its normal
22"
502
acceleration is (A")N 3571.43 kmh2. Applying Eq. 12–35 gives
+
0.7
a" a!
(1100 sin 30°
3571.43 cos 30°)i
a"!
(1100 cos 30°
a"! {3642.95i
3571.43 sin 30°)j 0
a"!
833.09j} kmh2
Thus, the magnitude of the relative velocity aB/A is
A"! 3642.952
( 833.09)2 3737 kmh2
Ans.
And its direction is
tan
1
833.09
12.9° 3642.95
VB 50 kmh
Ans.
B
•12–225.
12 –232. At the instant shown, cars A and B travel at
speeds of 70 kmh
mih and 50 kmh,
mih, respectively.
respectively. If B is
decreasing its speed
speed by
at 1400 kmh
mih22,while
while A
A is
is increasing its
speed at 800 mih
kmh22,, determine
determine the
the acceleration
acceleration of B with
respect to A. Car B moves along a curve having a radius of
curvature of 0.7 mi.
km.
A
VA 70 kmh
Relative Acceleration: Since car B is traveling along a curve, its normal acceleration
22"
502
is (A")N 3571.43 kmh2. Applying Eq. 12–35 gives
+
0.7
a" a!
(3571.43 cos 30°
1400 sin 30°)i
a"!
( 1400 cos 30°
a"! {2392.95i
3571.43 sin 30°)j 800j
a"!
3798.15j} kmh2
Thus, the magnitude of the relative acc. aB/A is
A"! 2392.952
( 3798.15)2 4489 kmh2
Ans.
And its direction is
tan
1
3798.15
57.8° 2392.95
Ans.
VB 50 kmh
30°
B
12–239.
Both boats A and B leave the shore at O at the same time. If
A travels at vA and B travels at vB, write a general
expression to determine the velocity of A with respect to B.
A
B
SOLUTION
O
Relative Velocity:
vA = vB + vA>B
vA j = vB sin ui + vB cos uj + vA>B
vA>B = - vB sin ui + (vA - vB cos u)j
Thus, the magnitude of the relative velocity vA>B is
vA>B = 2( -vB sin u)2 + (vA - vB cos u)2
= 2vA2 + vB2 - 2vA vB cos u
Ans.
And its direction is
u = tan-1
vA - vB cos u
vB sin u
b
Ans.