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SOLUTION
r=4(t-sin t) i+(2 t2-3)j
v=
dr
dt
=4(1 -cos t)i+(4 t)j
v
0
t=1=
1.83879i
+
4j
v
=2
(1.83879)2
+
(4)2
=
4.40 m
>
s Ans.
u=tan-1
a
4
1.83879 b
=65.312° a
a=4 sin ri +4j
at=1=3.3659i+4j
a
=2
(3.3659)2
+
(4)2
=
5.22773 m
>
s2
f=tan-1
a4
3.3659
b
=49.920° a
d
=
u
-
f
=15.392°
a
2=5.22773 cos 15.392°=5.04 m>s2
Ans.
a
n=5.22773 sin 15.392°=1.39 m>s2
Ans.
12–113.
The position of a particle is defined by r = {4(t - sin t)i
+ (2t2 - 3)j} m, where t is in seconds and the argument for
the sine is in radians. Determine the speed of the particle and
its normal and tangential components of acceleration when
t = 1 s.
u
u
122
12–114.
The automobile has a speed of 80 ft
>
s at point A and an
acceleration a having a magnitude of 10 ft>s2, acting in
the direction shown. Determine the radius of curvature
of the path at point A and the tangential component of
acceleration.
SOLUTION
12–115.
The automobile is originally at rest at If its speed is
increased by where tis in seconds,
determine the magnitudes of its velocity and acceleration
when t=18 s.
v
#=10.05t22ft>s2,
s=0.
When
Therefore the car is on a curved path.
Ans.
Ans.a=42.6 ft>s2
a=2(39.37)2+(16.2)2
at=0.05(182)=16.2 ft/s2
an=(97.2)2
240 =39.37 ft>s2
v=0.0167(183)=97.2 ft>s
t=18 s,
s=437.4 ft
s=4.167(10-3)t4
Ls
0
ds =Lt
0
0.0167 t3dt
v=0.0167 t3
Lv
0
dv=Lt
0
0.05 t2dt
s
240 ft
300 ft
*12–116.
SOLUTION
So that
Ans.
Ans.a=2(55.51)2+(18.17)2=58.4 ft>s2
at=0.05(19.06)2=18.17 ft>s2
an=(115.4)2
240 =55.51 ft>s2
v=115 ft>s
v=0.0167(19.06)3=115.4
t=19.06 s
550 =4.167(10-3)t4
s=4.167(10-3)t4
Ls
0
ds =Lt
0
0.0167 t3dt
v=0.0167 t3
Lv
0
dv=Lt
0
0.05 t2dt
The automobile is originally at rest If it then starts to
increase its speed at where tis in seconds,
determine the magnitudes of its velocity and acceleration at
s=550 ft.
v
#=10.05t22ft>s2,
s
=0.
s
240 ft
300 ft
125
12–117.
The two cars A and B travel along the circular path at
constant speeds vA = 80 ft
>
s and vB = 100 ft
>
s, respectively.
If they are at the positions shown when t = 0, determine the
time when the cars are side by side, and the time when they
are 90° apart.
A
B
v
A
vB
rB 390 ft
r
A 400 ft
SOLUTION
126
12–118.
Cars A and B are traveling around the circular race track. At
the instant shown, A has a speed of 60 ft
>
s and is increasing
its speed at the rate of 15 ft
>
s2 until it travels for a distance of
100p ft, after which it maintains a constant speed. Car B has
a speed of 120 ft
>
s and is decreasing its speed at 15 ft
>
s2 until
it travels a distance of 65p ft, after which it maintains a
constant speed. Determine the time when they come side by
side.
A
B
v
A
vB
rB 390 ft
r
A 400 ft
Ans:
t=66.4 s
12–119.
SOLUTION
dn
n=20 Mm>h=20(106)
3600 =5.56(103)m>s
The satellite Stravels around the earth in a circular path
with a constant speed of If the acceleration is
determine the altitude h. Assume the earth’s
diameter to be 12 713 km.
2.5 m>s2,
20 Mm>h.
h
S
*12–120.
The car travels along the circular path such that its speed is
increased by , where tis in seconds.
Determine the magnitudes of its velocity and acceleration
after the car has traveled starting from rest.
N
eglect the size of the car.
s=18 m
at=(0.5et)m>s2
SOLUTION
Lv
0
dv =Lt
0
0.5etdt
s18 m
12–121.
Radius of Curvature:
Acceleration:
When the car is at
Thus, the magnitude of the car’s acceleration at Bis
Ans.a=2at
2+an
2=2(-2.591)2+0.91942=2.75 m>s2
at=
C
0.225
A
51.5
B
-3.75
D
=-2.591 m>s2
B
A
s=51.5 m
B
at=vdv
ds =
A
25 -0.15s
BA
-0.15
B
=
A
0.225s-3.75
B
m>s2
an=vB2
r
=17.282
324.58 =0.9194 m>s2
r=
B
1+ady
dx b2
B
3>2
2d2y
dx22=c1+a-3.2
A
10-3
B
xb2d3>2
2-3.2
A
10-3
B
24x=50 m
=324.58 m
d2y
dx2=-3.2
A
10-3
B
dy
dx =-3.2
A
10-3
B
x
y=16 -1
625 x2
The car passes point Awith a speed of after which its
speed is defined by . Determine the
magnitude of the car’s acceleration when it reaches point B,
where and x = 50 m.s=51.5 m
v=(25 -0.15s)m>s
25 m
>
s
y16 x2
1
625
y
s
x
16 m
B
A
12–122.
If the car passes point Awith a speed of and begins
to increase its speed at a constant rate of ,
determine the magnitude of the car’s acceleration when
s=101.68 m and x = 0.
at=0.5 m>s2
20 m
>
s
SOLUTION
Radius of Curvature:
Acceleration:
The magnitude of the car’s acceleration at Cis
Ans.a=2at
2+an
2=20.52+1.602=1.68 m>s2
an=vC2
r
=22.3612
312.5 =1.60 m>s2
at=v
#=0.5 m>s
r=
B
1+ady
dx b2
R
3>2
2d2y
dx22=c1+a-3.2
A
10-3
B
xb2d3>2
-3.2
A
10-3
B
4x=0
=312.5 m
d2y
dx2=-3.2
A
10-3
B
dy
dx =-3.2
A
10-3
B
x
y=16 -1
625 x2
vC=22.361 m>s
vC2=202+2(0.5)(100 -0)
y16 x2
1
625
y
s
x
16 m
B
A
12–123.
The motorcycle is traveling at when it is at A. If the
speed is then increased at determine its speed
and acceleration at the instant t=5s.
v
#=0.1 m>s2,
1m>s
x
s
y
y0.5x
2
SOLUTION
x31+x2+1n ax+31+x2b=12.5
6.25 =1
2cx31+x2+1n ax+31+x2bdx
0
6.25 =Lx
0
31+x2dx
d2y
dx2=1
dy
dx =x
y=0.5x2
L6.25
0
ds =Lx
0
A
1+ady
dx b2
dx
s=0+1(5) +1
2(0.1)(5)2=6.25 m
s=s0+n0t+1
2act2
at=n=0.1
*12–124.
SOLUTION
Ans.a=2a2
t+a2
n=2(4)2+(7.622)2=8.61 m>s2
an=yB2
r
=82
8.396 =7.622 m>s2
r=
C
1+(dy
dx)2
D
3>2
`
d2y
dx2
`
4x=2m
=
C
1+(1.6)2
D
3>2
|0.8| =8.396 m
d2y
dx22x=2m
=0.8
dy
dx 2x=2m
=0.8x2x=2m
=1.6
y=0.4 x2
The box of negligible size is sliding down along a curved
path defined by the parabola .When it is at A
(, ), the speed is and the
increase in speed is . Determine the
magnitude of the acceleration of the box at this instant.
dv >dt =4m>s2
v=8m>syA=1.6 mxA=2m
y=0.4x2
y
x
y0.4x
2
A
SOLUTION
at=v=0.06t
dv=at dt
Lv
s
dv=
Lt
0
0.06t dt
v=0.03t2+5
ds =v dt
Ls
0
ds =
Lt
0
(0.03t2+5) dt
s=0.01t3+5t
s=
1
3
(2p(300)) =628.3185
0.01t3+5t-628.3185 =0
Solve for the positive root,
t=35.58 s
v
=
0.03(35.58)
2+
5
=
42.978 m
>
s
=
43.0 m
>
s Ans.
an=
v2
r
=
(42.978)
2
300 =6.157 m
>
s2
a
t=
0.06(35.58)
=
2.135 m
>
s
2
a
=2
(6.157)2
+
(2.135)2
=
6.52 m
>
s2 Ans.
12–125.
The car travels around the circular track having a radius of r= 300 m
such that when it is at point A it has a velocity of 5 m
>
s, which is
increasing at the rate of
v
#
= (0.06t) m
>
s2, where t is in seconds.
Determine the magnitudes of its velocity and acceleration when it
has traveled one-third the way around the track.
y
x
r
A
12–126.
The car travels around the portion of a circular track having
a radius of r = 500 ft such that when it is at point A it has a
velocity of 2 ft
>
s, which is increasing at the rate of
v
#
=
(0.002t) ft
>
s2, where t is in seconds. Determine the
magnitudes of its velocity and acceleration when it has
traveled three-fourths the way around the track.
y
x
r
A
SOLUTION
at=0.002 s
at ds =v dv
Ls
0
0.002s ds =
Lv
2
v dv
0.001s2=
1
2
v2-
1
2
(2)2
v2=0.002s2+4
s=
3
4
[2p(500)] =2356.194 ft
v
2=
0.002(2356.194)
2+
4
v=105.39 ft>s=105 ft>s
Ans.
an=
v
2
r
=
(105.39)2
500 =22.21 ft
>
s2
a
t=
0.002(2356.194)
=
4.712 ft
>
s
2
a
=2
(22.21)2
+
(4.712)2
=
22.7 ft
>
s2 Ans.
12–127.
At a given instant the train engine at Ehas a speed of
and an acceleration of acting in the
direction shown. Determine the rate of increase in the
train’s speed and the radius of curvature of the path.r
14 m>s2
20 m>s
v20 m/s
a14 m/s
2
E
75
r
SOLUTION
SOLUTION
*12–128.
The car has an initial speed v0 = 20 m
>
s. If it increases its
speed along the circular track at s = 0, a
t=
(0.8s) m
>
s
2
,
where s is in meters, determine the time needed for the car
to travel s = 25 m.
s
s
12–129.
The car starts from rest at s = 0 and increases its speed at
at = 4 m
>
s2. Determine the time when the magnitude of
acceleration becomes 20 m
>
s2. At what position s does this
occur?
SOLUTION
12–130.
At
Ans.a=a2
t+a2
n=42+6.252=7.42 ft s2
t=5s,at=v
#=0.8(5) =4ft>s2
an=v2
r
=252
100 =6.25 ft>s2
v=25 ft>s
Lv
15
0
A boat is traveling along a circular curve having a radius of
100 ft. If its speed at is 15 ft/s and is increasing at
determine the magnitude of its acceleration
at the instant t=5s.
v
#=10.8t2ft>s2,
t=0
12–131.
SOLUTION
Ans.a
=2
a2
t+
a2
n=2
22
+
1.252
=
2.36 m
>
s2
an=y2
r
=52
20 =1.25 m>s2
at=2m>s2
A boat is traveling along a circular path having a radius of
20 m. Determine the magnitude of the boat’s acceleration
when the speed is and the rate of increase in the
speed is v
#
=2m>s2.
v=5m>s
140
*12–132.
Starting from rest, a bicyclist travels around a horizontal circular
path, at a speed of
where tis in seconds.Determine the magnitudes of his velocity
and acceleration when he has traveled s=3m.
v=10.09t2+0.1t2m>s,r=10 m,
SOLUTION
When
Solving,
Ans.
Ans
.
a=2a2
t+a2
n=2(0.8465)2+(0.3852)2=0.930 m>s2
an=v2
r
=1.962
10 =0.3852 m>s2
at=dv
dt =0.18t+0.1
`
t=4.147 s
=0.8465 m>s2
v=0.09(4.147)2+0.1(4.147) =1.96 m>s
v=ds
dt =0.09t2+0.1t
t=4.147 s
s=3m,
3=0.03t3+0.05t2
s=0.03t3+0.05t2
Ls
0
ds =Lt
010.09t2+0.1t2dt
Ans:
v=1.96
m>s
a
=
0.930 m
>
s
2
