*12–76.
A particle travels along the curve from A to B in 5 s. It
takes 8 s for it to go from B to C and then 10 s to go from
C to A. Determine its average speed when it goes around
the closed path.
B
x
y
20 m
12–77.
The position of a crate sliding down a ramp is given by
where t
is in seconds. Determine the magnitude of the crate’s
velocity and acceleration when .t=2 s
y=(1.5t2) m, z=(6 –0.75t5>2) m,x=(0.25t3) m,
12–78.
A rocket is fired from rest at and travels along a
parabolic trajectory described by . If the
xcomponent of acceleration is , where tis
in seconds, determine the magnitude of the rocket’s velocity
and acceleration when t= 10 s.
ax=a1
4 t2b m>s2
y2=[120(103)x] m
x=0
SOLUTION
12–79.
The particle travels along the path defined by the parabola
If the component of velocity along the xaxis is
where tis in seconds, determine the particle’s
distance from the origin Oand the magnitude of its
acceleration when When y=0.x=0,t=0,t=1s.
vx=15t2ft>s,
y=0.5x2.
dna, tA.,suhT
.The particle’s distance from the origin at this moment is
Ans.
Acceleration:Taking the first derivative of the path ,we have .
The second derivative of the path gives
(1)
However,, and .Thus, Eq. (1) becomes
(2)
ay=vx
2+xax
y
$=ay
x
$=ax
x
#=vx
y
$=x
#2+xx
$
y
#=xx
#
y=0.5x2
d=2(2.50 –0)2+(3.125 –0)2=4.00 ft
y=3.125
A
14
B
=3.125 ft
x=2.5
A
12
B
=2.50 ftt=1sy=0.5
A
2.50t2
B
2=
A
3.125t4
B
ft
x=
A
2.50t2
B
ft
Lx
0
dx =Lt
0
5tdt
dx =vxdt
dt
y
y 0.5x2
8 5
*12–80.
The motorcycle travels with constant speed along the
path that, for a short distance, takes the form of a sine curve.
Determine the xand ycomponents of its velocity at any
instant on the curve.
v0
SOLUTION
y=csin ap
Lxb
LL
cc
x
y
v0
y c sin ( x)––
L
π
vy=
L
a
cosp
L
x
b
c
1+
a
L
c
b
cos2
a
L
x
bd
12–81.
45
30
30 m
y
C
.Thus,
Average Velocity: The displacement from point Bto Cis
.
Ans.(vBC)avg =¢rBC
¢t=7.765i+13.45j
3–1={3.88i+6.72j}m>s
=(28.98i–7.765j)–(21.21i–21.21j)={7.765i+13.45j}m
¢rBC =rC–rB
={28.98i–7.765j}m
rC=(30 sin 75° –0)i+[(30 –30 cos 75°) –30]j
={21.21i–21.21j}m
rB=(30 sin 45° –0)i+[(30 –30 cos 45°) –30]j
[30 sin 75°, 30 –30 cos 75°]
A particle travels along the circular path from Ato Bin 1s.
If it takes 3 s for it to go from Ato C, determine its average
velocity when it goes from Bto C.
12–82.
x
y
SOLUTION
Ans.
Ans.a=(–ck2 sin kt)2+(–ck2 cos kt)2+0=ck2
v=(ck cos kt)2+(–ck sin kt)2+(–b)2=c2k2+b2
z
$=0z
#=-bz =h–bt
y
$=-ck2 cos kty
#=-ck sin kty =c cos kt
x
$=-ck2 sin ktx
#=ck cos ktx =c sin kt
The roller coaster car travels down the helical path at
constant speed such that the parametric equations that define
its position are x = c sin kt, y = c cos kt, z = h − bt, where c, h,
and b are constants. Determine the magnitudes of its velocity
and acceleration.
z
12–83.
Pegs Aand Bare restricted to move in the elliptical slots
due to the motion of the slotted link. If the link moves with
a constant speed of , determine the magnitude of the
velocity and acceleration of peg Awhen .x=1m
10 m/s
SOLUTION
A
CD
y
x
*12–84.
The van travels over the hill described by
. If it has a constant speed of
, determine the xand ycomponents of the van’s
velocity and acceleration when .x=50 ft
75 ft>s
y=(–1.5(10–3)x2+15) ft
x
y( 1.5 (10 3)x215) f
t
y
100 ft
15 ft
SOLUTION
12–85.
T
h
e f
li
g
h
t pat
h
of t
h
e
h
e
li
copter as
i
t ta
k
es off from A
i
s
defined by the parametric equations and
where tis the time in seconds. Determine
the distance the helicopter is from point Aand the
magnitudes of its velocity and acceleration when t=10 s.
y=10.04t32m,
x=12t22m
SOLUTION
y
A
9 1
12–86.
Determine the minimum initial velocity and the
corresponding angle at which the ball must be kicked in
order for it to just cross over the 3-m high fence.
u0
v
0
SOLUTION
coincides with the ball’s initial position.
x-Motion: Here,,, and .Thus,
(1)
y-Motion: Here,, , and .Thus,
(2)
Substituting Eq. (1) into Eq. (2) yields
(3)
From Eq. (3), we notice that is minimum when is
maximum. This requires
Ans.
Substituting the result of into Eq. (2), we have
Ans.(v0)min =
B
58.86
sin 116.57° –cos
2
58.28°
=9.76 m>s
u
u=58.28° =58.3°
2u=116.57°
tan 2u=-2
df(u)
du=2 cos 2u+sin 2u=0
df(u)
du=0
f(u)=sin 2u–cos2 uv0
v0=B58.86
sin 2u–cos2 u
3=v0 (sin u) t–4.905t2
3=0+v0 (sin u) t+1
2 (–9.81)t2
y=y0+(v0)y t+1
2 ay t2
A
+c
B
y0=0ay=-g=-9.81 m > s
2
(v0)x=v0 sin u
t=6
v0 cos u
6=0+(v0 cos u)t
x=x0+(v0)x
t
A
+
:
B
x=6 mx0=0(v0)x=v0 cos u
v03 m
6 m
u0
Ans:
u=58.3°
(
v
0
)
min
=
9.76
m>s
9 2
SOLUTION
1
+
S
2
s=v0t
18 =vA cos
u
(1.5)
(1)
1
+
c
2
v
2=
v
2
0+2ac (s–s0)
0
=
(v
A
sin u)
2+
2(
–
32.2)(h
–
3.5)
1
+
c
2
v=v0+act
0=vA sin
u
–32.2(1.5)
(2)
u
12–87.
The catapult is used to launch a ball such that it strikes the
wall of the building at the maximum height of its trajectory.
If it takes 1.5 s to travel from A to B, determine the velocity
vA at which it was launched, the angle of release
u
, and the
height h.
Ans:
u=76.0°
v
A
=49.8 ft>s
h=39.7 ft
18 ft
3.5 ft
h
A
vA
B
u
9.81
SOLUTION
1
+
S
2
s=s0+v0t
R=0+(10 cos
u
)t
1
+
c
2
v=v0+act
–10 sin u=10 sin u–9.81t
t=
20
9.81
sin u
Thus, R=
200
9.81
sin u cos u
R=
100
9.81
sin 2u (1)
Since the function
y=sin 2
u is symmetric with respect to
u=45°
as indicated,
Eq. (1) will be satisfied if
|
f
1 | =|
f
2 |
Choosing f
=15° or
u
1=45°–15°=30°
and u
2=45°+15°=60°,
and
substituting into Eq. (1) yields
R=8.83 m
Ans.
12–90.
Show that the girl at A can throw the ball to the boy at B
by launching it at equal angles measured up or down
from a 45° inclination. If vA = 10 m
>
s, determine the range
R if this value is 15°, i.e.,
u
1 = 45° − 15° = 30° and
u
2 = 45° +
15° = 60°. Assume the ball is caught at the same elevation
from which it is thrown.
R
B
A
vA fi 10 m/s
u
SOLUTION
(vA)x=80 cos 30°=69.28 ft>s
(v
A
)
y
=80 sin 30°=40 ft>s
1
+
S
2
s=s0+v0t
x=0+69.28t
(1)
1
+
c
2
s=s0+v0t+
1
2
act2
–y=0+40t+
1
2
(–32.2)t2 (2)
y
=–
0.04x
2
From Eqs. (1) and (2):
–
y
=
0.5774x
–
0.003354x
2
0.04x2=0.5774x–0.003354x2
0.04335x2=0.5774x
x=13.3 ft
Ans.
Thus
y
=–
0.04 (13.3)
2=–
7.09 ft Ans.
12–91.
The ball at A is kicked with a speed vA = 80 ft
>
s and at an
angle
u
A = 30°. Determine the point (x, –y) where it strikes
the ground. Assume the ground has the shape of a parabola
as shown.
x
y
B
A
vA
y fi 0.04x2
y
x
uA
SOLUTION
1
+
S
2
s=s0+v0t
15 =0+vA cos 30° t
1
+
c
2
s=s0+v0t+
1
2
act2
–9=0+vA sin 30° t+
1
2
(–32.2)t2
vA=16.5 ft>s
Ans.
t=1.047 s
1
+
S
2
(vB)x=16.54 cos 30°=14.32 ft>s
1
+
c
2
v=v0+act
(
vB
)
y
=16.54 sin 30°+(–32.2)(1.047)
=–25.45 ft>s
v
B=2
(14.32)2
+
(
–
25.45)2
=
29.2 ft
>
s Ans.
*12–92.
The ball at A is kicked such that u
A=30°.
If it strikes the
ground at B having coordinates
x=15 ft,
y=–9 ft,
determine the speed at which it is kicked and the speed at
which it strikes the ground.
x
y
B
A
vA
y fi 0.04x2
y
x
uA
SOLUTION
1
+
S
2
s=s0+v0t
d cos 10°=0+80 cos 55° t
1
+
c
2
s=s0+v0 t+
1
2
act2
d sin 10°=0+80 sin 55° t–
1
2
(32.2)
(
t2
)
Solving
t=3.568 s
d=166 ft
Ans.
12–93.
A golf ball is struck with a velocity of 80 ft
>
s as shown.
Determine the distance d to where it will land.
d
B
A10
45
vA fi 80 ft/s
SOLUTION
(vA)x=80 cos 55°=44.886
(v
A
)
y
=80 sin 55°=65.532
1
+
S
2
s=s0+v0t
d cos 10°=0+45.886t
1
+
c
2
s=s0+v0t+
1
2
act2
d sin 10°=0+65.532 (t)+
1
2
(–32.2)
(
t2
)
d=166 ft
t=3.568 =3.57 s
Ans.
(vB)x=(vA)x=45.886
1
+
c
2
v=v0+act
(v
B
)
y
=65.532 –32.2(3.568)
(
vB
)
y
=–49.357
v
B=2
(45.886)2
+
(
–
49.357)2
vB=67.4 ft>s
Ans.
12–94.
A golf ball is struck with a velocity of 80 ft
>
s as shown.
Determine the speed at which it strikes the ground at B and
the time of flight from A to B.
d
B
A10
45
vA fi 80 ft/s
100
12–95.
SOLUTION
Solving
Ans.
Solving
Ans.h=11.5 ft
tAB =0.786 s
h=7+36.73 sin 30°tAB –1
2(32.2)(tAB
2)
(+c)s=s0+v0t+1
2act2
25 =0+36.73 cos 30°tAB
(:
+)s=s0+v0t
tAC =0.943 s
vA=36.73 =36.7 ft>s
10 =7+vAsin 30°tAC –1
2(32.2)(tAC
2)
(+c)s=s0+v0t+1
2act2
30 =0+vAcos 30°tAC
(:
+)s=s0+v0t
10 f
h
C
B
A
v
A
30
5ft25 ft
7ft
The basketball passed through the hoop even though it barely
cleared the hands of the player B who attempted to block it.
Neglecting the size of the ball, determine the magnitude vA of
its initial velocity and the height h of the ball when it passes
over player B.
Ans:
vA=36.7 ft>s
h=11.5 ft