PROBLEM 11.1
The motion of a particle is defined by the relation
x = 8t3 - 8 + 30 sin iz t, where x and t are expressed in millimeters and
seconds, respectively. Determine the position, the velocity, and the
acceleration of the particle when / = 5 s.
SOLUTION
Position:
f3 - 8 + 30sin(#t) mm
Velocity:
dx = 24r + 30# cos (#r) mm/'s
V ~~dt
Acceleration:
dv
a - —
dt
48t - 30#2sin(#t) mm/s2
At t = 5 s,
8 + 30 sin (5#)
x5 = 992 mm A
v5 = (24)(5): + 30#cos(5/r)
v5 = 506 mm/s A
= M O ) - 30#2 sin(5#)
a5 = 240 mm/s2 A
*5 = (8)(5)J
PROBLEM 11.2
The
motion of a particle is defined by the relation
10 * 5 ,
x = — t - —r - 20r + 10, where x and t are expressed in meters and
seconds, respectively. Determine the time, the position, and the
acceleration o f the particle when u - 0.
SOLUTION
x = i10
V ? _ i5 o _ 20t + 10 m
3
2
Position:
.,2
Velocity:
V
= — = lor2 - 5t - 20 m/s
dt
a = — = 20t - 5
dt
Acceleration:
1012 - 5r - 20 = 0
When v = 0,
Solving the quadratic equation for t,
t = 1.6861 s
and
t = -1.1861 s
t = 1.686 s <
When t = 1.6861,
S
o
x = y ( l.6 8 6 l ) - ^(1.6861) - 20(1.6861) + 10
1 0
x = -14.85 m -4
o
II
>
a = 20(1.6861) - 5
a = 28.7 m/s2 4
PROBLEM 11.3
The motion of the slider^ is defined by the relation x = 20 sin fa, where
x and t are expressed in inches and seconds, respectively, and k is a
constant. Knowing that k = 10 rad/s, determine the position, the velocity,
and the acceleration of slider A when t - 0.05 s.
SOLUTION
x =
Position:
sin Art in.
v = — = 20Arcosfa in./s
dt
Velocity:
a = — = — Ar sin At in./s
dt
Acceleration:
When / = 0.05 s,
2 0
2 0
and
k =
1 0
2
2
rad/s
kt = (10)(0.05) = 0.5 rad
x = 9.59 in. A
x = 20sin(0.5)
v = (20)(l0)cos(0.5)
a = - ( 2 0 ) ( l0
) 2
sin(0.5)
v = 175.5 in./s A
a = -959 in./s <
2
PROBLEM 11.4
The
motion
of
the
slider
A
is
defined
by
the
relation x =
sin(£,r - k2t 2), where x and t are expressed in inches and
seconds, respectively. The constants kx and k2 are known to be 1 rad/s
and 0.5 rad/s2, respectively. Consider the range 0 < / < 2 s and
determine the position and acceleration of slider A when v = 0.
2 0
SOLUTION
Position:
x =
Where
2 0
s in ^ ,r - k2t 2) in.
kx = 1 rad/s
Let
k2 = 0.5 rad/s
and
2
# = kxt - k2t 2 = t - 0.512 rad
— = (l - / ) rad/s
dt
and
’
v
= -1 rad/s2
dr
J *2
Position:
x =
2 0
sin# in.
Velocity:
dx
v=— =
dt
2 0
dG
cos#— in./s
dt
Acceleration:
a =
a =
When v = 0,
2 0
cosG^-^r dt2
2 0
s in # f —
\ dt )
dG ,
— = —r =
dt
1
in./s
1
either
or
dv
~dt
cos 6 =
0
t =
0
2
1
s
Over 0 S / S 2 s, values of cosG are:
r(s)
0
0.5
1 . 0
1.5
2 . 0
#(rad)
0
0.375
0.5
0.375
0
cos#
1 . 0
0.931
0.878
0.981
1 . 0
No solutions cos# = 0 in this range.
For / = 1 s,
# = 1 - (0.5)(1
) 2
= 0.5 rad
x = 20sin(0.5)
a = 20co s(0 .5 )(-l) - 20sin(0.5)(0)
x = 9.59 in. A
a = -17.55 in./s «
2
PROBLEM 11.5
The motion of a particle is defined by the relation x = 5t4 - 413 + 3t - 2,
where x and t are expressed in feet and seconds, respectively. Determine
the position, the velocity, and the acceleration of the particle when
t = 2 s.
SOLUTION
Position:
x = 5r4 - 4t3 + 3t - 2 ft
Velocity:
v = — = 20t3 - 1212 + 3 ft/s
dt
a = — = 60t2 - 241 ft/s2
dt
Acceleration:
When t = 2 s,
x = (5)(2)4 - ( 4 ) ( 2 ) 3 - ( 3 ) ( 2) - 2
v
= (20)(2)3 - (1 2 )(2 )2 + 3
* = (60)(2)2 - ( 2 4 ) (2 )
x = 52 ft <
v = 115 ft/s
a = 192 ft/s <
PROBLEM 11.6
The motion of a particle is defined by the relation
x = r + 13 - 14r2 - 1Or + 1 , where x and t are expressed in inches
and seconds, respectively. Determine the position, the velocity, and the
acceleration o f the particle when t - 3 s.
6
4
8
6
SOLUTION
Position:
X
= 6r + r - 14r2 - lor + 16 in.
4
8
3
V = * = 24r3 + 24r2 - 281 - 10 in./s
dt
Velocity:
a = — = l i t 2 + 48/ - 28 in./s
dt
Acceleration:
2
When t = 3 s,
x
= (6)(3)4 + (8 )(3 )3 -(1 4 )(3 )2 -(1 0 )(3 ) + 16
v = (24)(3)3 + (24)(3)2 - (28)(3) - 10
a = (72)(3)2 + (48)(3) - 28
x = 562 in. M
v = 770 in./s -4
a = 764 in./s 4
2
PROBLEM 11.7
The motion of a particle is defined by the relation
x = 2/3 - 1212 - l i t - 80, where x and / are expressed hi meters and
seconds, respectively. Determine (a) when the velocity is zero, (b) the
velocity, the acceleration, and the total distance traveled when x = 0.
SOLUTION
Position:
x = 2? - 1212 - 72t - 80 m
Velocity:
v = — = 612 - 24/ - 72 m/s
dt
a = — = 12/ - 24 m/s2
dt
Acceleration:
611 - 24/ - 72 = 0
(a) When v = 0,
/ = -2 s
Solving the quadratic equation:
and
/ = 6s
Reject the negative time. Then
For 0 < / < 6 s,
v is negative.
x is decreasing,
For / > 6 s,
v is positive.
x is increasing.
Minimum value of x occurs when / = 6 s.
*mi„ = (2)(6)3 - (12)(6)2 - (72)(6) - 80 - -512 m
x0 = -80 m
when / = 0,
(b) Distance traveled over 0 < / < 6 s.
4 = kin - *o H
2P
When x = 0,
Roots of the cubic equation are:
"512 - (-80) I= 432 m
-
1212 - 72/ - 80 = 0
/ = 10 s
and
/ = - 2 s (twice).
Reject the negative roots.
When / = 10 s,
v - (6)(10) - (24)(10) —72
v = 288 m/s A
a = (12)(10) - 24
a = 96 m/s2 A
Distance traveled over 6 s < / < 1 0 s .
d2 = | x / - x min| = |0 - ( - 5 1 2 ) | = 512 m
Total distance traveled:
dx + d 2 = 432 + 512
d = 944 m A
PROBLEM 11.8
The motion of a particle is defined by the relation
x = 2t3 - 12 + 48/ -1 6 , where x and / are expressed in millimeters
and seconds, respectively. Determine (a) when the velocity is zero,
(b) the position and the total distance traveled when the acceleration is
zero.
1 8
SOLUTION
Position:
x = 2t3 - 1812 + 48f - 16 mm
Velocity:
v = — = f - 36t + 48 mm/s
6
dt
2
a = — = 12/ - 36 mm/s
dt
Acceleration:
(a) When v = 0,
6
2
12 - 36/ + 48 = 0
/ = 2 s and / = 4 s A
Solving the quadratic equation,
For 0 < t < 2 s,
v is positive,
and
x is increasing.
2 s < t < 4 s,
v is negative,
and
x is decreasing.
/ > 4 s,
v is positive,
and
x is increasing.
(ib) When a = 0,
When / = 3 s,
12/ - 36 - 0
/ =3s
* J = ( 2 ) ( 3 ) 3 - ( 18)(3)2 + ( « ) ( 3 ) - 1 6
When / = 0,
x3 =
2 0
mm A
Xq = -1 6 mm
When t = 2 s,
x2
= (2 )(2 )3 -(1 8 )(2 )2 + ( 4 8 ) ( 2 ) - 1 6 = 24 mm
Distance traveled over 0 < t < 2 s:
c/, = | x2 - x0 | = 124 - (-1 6 ) | = 40 mm
Distance traveled over 2 s < t < 3 s:
d2 - | x3 - x2 | = | 20 - 2 4 1= 4 mm
Total distance traveled:
dj + d 2 —40 + 4
d = 44 mm A
PROBLEM 11.9
The acceleration of point A is defined by the relation a = -1.8sinfa,
where a and t are expressed in m/s and seconds, respectively, and
k = 3 rad/s. Knowing that x = 0 and u = 0.6 m/s when / = 0, determine
the velocity and position of point A when t = 0.5 s.
—
2
iHa
1
\c
Ly
[
B
|
1
i
r
-------
D
SOLUTION
Given:
a = -1.8sin/rf m/s2,
x0 = 0,
v = 0.6 m/s,
0
k = 3 rad/s
s
i/
J^sinfar/f = -£-cosfa|
1
v —v = £ adt = 0
1
1 .8
8
v - 0.6 = -^ -(co skt - l) = 0.6cosA? - 0.6
v=
Velocity:
x - x = £ vdt =
0
x Position:
When t = 0.5 s,
0
0 . 6
0 . 6
cos kt m/s
£ cosktdt = ^ s i n f o |
= ^ (s in fa -
0
) = . sin£r
0
2
x = . sinfa m
0
2
kt = (3)(0.5) = 1.5 rad
v = 0.6 cos 1.5 = 0.0424 m/s
v = 42.4 mm/s A
x = 0.2sinl.5 = 0.1995 m
x = 199.5 mm A
PROBLEM 11.10
The acceleration of point A is defined by the relation
a = -1.08 sin kt - 1.44 cos kt, where a and t are expressed in m /s ' and
seconds, respectively, and k = 3 rad/s. Knowing that x - 0.16 m and
v = 0.36 m/s when t = 0, determine the velocity and position of point A
when t = 0.5 s.
SOLUTION
Given:
a = -1.08sinfa - 1.44cosfa m /s',
x„ = 0.16 m,
k = 3 rad/s
v = 0.36 m/s
0
v - v = £ adt = -1.08 £ sin ktdt - 1.44 £ co sktdt
0
v - 0.36 =
1.08
, |' 1.44 . . i'
cosx/
sm w
k
>o
k
'0
1.08 /
,
,»1.44/ . .
= —— (coskt - ) --- — (sin fa 1
0
)
= 0.36cos£f - 0.36 - 0.48sinfa
Velocity:
v = 0.36cosfa - 0.48sin£f m/s
x - x = £ vdt = 0.36 £ co sktdt - 0.48 £ sin kt dt
0
x - 0.16 =
0.36 . , i» 0.48
-sinfa +
-cos/:/
k
lo
k
0.36/ . .
= —— (sm kt -
When t - 0.5 s,
0.48/
.
) + —— (coskt - )
= .
1 2
sinfa + 0.16cosfo - 0.16
x = .
1 2
sinfa + 0.16cosfa m
0
Position:
0
0
1
kt = (3)(0.5) = 1.5 rad
v = 0.36cosl.5 - 0.48sinl.5 = -0.453 m/s
x = 0.12sinl.5 + 0.16cosl.5 = 0.1310 m
v = -453 mm/s 4
x = 131.0 mm 4
PROBLEM 11.11
The acceleration of a particle is directly proportional to the square of the
time t. When t = 0, the particle is at x = 36 ft. Knowing that at t - 9 s,
x = 144 ft and v = 27 ft/s, express x and v in terms o f t.
SOLUTION
Given:
a = kt2 ft/s2,
Xg = 144 ft,
x0 - 36 ft,
v9 = 27 ft/s
v - vo = l adt = l ^ 2 dt = j* # 3
Velocity:
V
=
Vn
0
+
1
,
~kt
3
x ~ xo = f\vdt = v0t + j ^ k t 4
Position:
x = xn + v0t + — kt4 = 36 + v0f + — kt4
0
0
12
0
12
When t = 9 s,
x = 144 ft
and
v = 27 ft/s
36 + v0(9) + ^ * ( 9 ) 4 = 144
or
9v0 + 546.15k = 108
v
0
(l)
+ j * ( 9 ) 3 = 27
v0 + 243& = 27
^2)
Solving equations (1) and (2) simultaneously yields:
v0 = 7 ft/s
Then,
and
k = 0.082305 ft/s4
x = 36 +
It
+ 0.00686/4 ft <
v = 7 + 0.0274r3 ft/s <
PROBLEM 11.12
It is known that from t = 2 s to ? = 10 s the acceleration o f a particle is
inversely proportional to the cube of the time t. When t - 2s,
v - -15 ft/s, and when t = 10 s, v = 0.36 ft/s. Knowing that the particle
is twice as far from the origin when t = s as it is when t s,
determine (a) the position of the particle when t = s and when I = s,
(ib) the total distance traveled by the particle from f = s to f =
s.
2
1 0
2
1 0
2
SOLUTION
For 2 s < f < 10 s, a = —r where £ is a constant.
r
At t = 2 s,
v = -15 ft/s
dv
a =—
dt
At t = 10 s,
or
dv = adt
v - 0.36 ft/s
/
-15 - 0.5k
1
v (10)2
1
= 0.36
k = 128 ft -s
from which
4j
Then,
t2 = 64
When v = 0,
For 2 s < t <
8
s<t <
8
1 0
s,
s,
or
t =
8
v< ,
x is decreasing.
v> ,
x is increasing.
0
0
s
1 0
PROBLEM 11.12 CONTINUED
Position:
At t — 2 s,
At t = 8 s,
At f = 10 s,
Given:
Using the plus sign:
Distance traveled:
Using the minus sign:
x = jdx = jv d t = J
—2 +
Xg = 8 4
f
.
64
dt — t "\
hC
t
64
(■C = 34 + C ft
2
64
8
h C = 16 + C ft
x10 = 10 h
10
64
1 0
^ = 2|x10| = ±2x10
h C = 16.4 + C ft
or
34 + C = ± (l6.4 + C)
C = 1.2 ft, which gives xg = 17.2 ft,
x2 = 35.2 ft and x10 = 17.60 ft A
d = | xg - x2| + | x10 - Xg| = 18 + 0.4
d - 18.40 ft A
C - -22.27 ft, which gives xg = -6.27 ft,
x2 = 11.73 ft and x10 = -5.87 ft A
Distance traveled:
d = Ixg - x2| + I x10 - xg| = 18 + 0.4
d - 18.40 ft A
PROBLEM 11.13
The acceleration of a particle is directly proportional to the time /. At
/ = 0, the velocity of the particle is 400 mm's. Knowing that
v = 370 mm/s and x = 500 mm when t = 1 s, determine the velocity,
the position, and the total distance traveled when t = 7 s.
SOLUTION
a = kt mm/s
= 400 mm/s;
II
©
>
Given:
where & is a constant.
at t = 1 s,
v = 370 mm/s,
L * = i adt =
v - 400 = —kt2
2
or
=\
t2
v = 400 + —kt2
2
v = 400 + ~ A (l )2 = 370,
At / = 1 s,
x = 500 mm
k = -60 mm/s3
v = 400 - 30?2 mm/s
Thus
v7 = 400 - (30)(7)2
At t - 7 s,
When v = 0,
400 - 30/2 = 0. Then t2 = 13.333 s2,
v7 = -1070 mm/s -4j
t = 3.651 s
For 0 < t < 3.651 s,
v>0
and
x is increasing.
For t > 3.651 s,
v<0
and
x is decreasing.
£ o* = f vr f < = f ( 4 O 0 - 3 0
? )d t
x - 500 = (400t - 10r2)|| = 400/ - 10/3 - 390
II
©
£
Position:
At / = 3.651 s,
At / = 7 s,
x = 400/ - 10/3 +110 mm
x = x0 =110 mm
x = Xmax —(400)(3.651) - (10)(3.651)3 + 110 = 1083.7 mm
x = Xl = (4G0)(7) - (10)(7)3 +110
x7 = -520 mm A
Distances traveled:
Over 0 < t < 3.651 s,
dl = Xmax - *0 = 973-7 mm
Over 3.651 < / < 7 s,
d2 = xmax - x7 = 1603.7 mm
Total distance traveled:
d = dx + d2 = 2511A mm
d = 2580 mm A
PROBLEM 11.14
The acceleration of a particle is defined by the relation a - 0.15 m/s'1.
Knowing that x = -1 0 m When t = 0 and v = -0.15 m/s when / = 2 s,
determine the velocity, the position, and the total distance traveled when
/ = 5 s.
SOLUTION
Determine velocity.
to ,,* =
v - (-15) = 0 .1 5 /- (0.15)(2)
v = 0.15/ - 0.45 m/s
At / = 5 s,
v = 0.300 m/s M
v = (0 .1 5 )(5 )-0 .4 5
5
When v = 0,
0.15/ - 0.45 = 0
5
/ = 3.00 s
For 0 < / < 3.00 s,
v< ,
x is decreasing.
For 3.00 < / < 5 s,
v> ,
x is increasing.
0
0
£<0.15, - 0 .4 5 ) *
Determine position.
* -(-io )-ja m * 1X
= o m t2- m s
- 0.075/2 - 0.45/ - 10 m
x = (0.075)(5)2 - (0.45)(5) - 10 - -10.375 m
At t = 5 s,
5
x = -10.38 m M
5
At t - 0,
At / - 3.00 s,
x =
0
- 1 0
m (given)
x = xmin = (0.075)(3.00)2 - (0.45)(3.00) -TO = -10.675 mm
3
PROBLEM 11.14 CONTINUED
Distances traveled:
Over 0 < t <, 3.00 s,
d\ = xo ~ xmin = 0.675 m
Over 3.00 s < t < 5 s,
d 2 = *5 - *min = ° - 3 0 0 m
Total distance traveled:
d = dx + d2 = 0.975 m M
PROBLEM 11.15
The acceleration of point A is defined by the relation a = 200x(l + kx2),
where a and x are expressed in m/s and meters, respectively, and k is a
constant. Knowing that the velocity of A is 2.5 m/s when x = 0 and 5 m/s
when x = 0.15 m, determine the value of k.
2
SOLUTION
Note that a is a given function of x
Use
vdv = a dx =
2 0 0
* (l + kx2)dx = (
v = 2.5 m/s
Using the limits
v = 5 m/s
and
J^vdv =
2 0 0
x+
2 0 0
fcc )r&
3
when x = 0,
when jc = 0.15 m,
3^dx
l5^200x + 200fcc
0.15
5
L
5
2
2
J 2.5
j
-
o
2.5
= (100)(0.15)2 + (50)*(0.15)4
2
9.375 = 2.25 + 0.0253125*
Solving for k ,
k = 281 m
" 2
<
PROBLEM 11.16
The acceleration of point A is defined by the relation a = 200x + 3200x ,
where a and x are expressed in m/s and meters, respectively. Knowing
that the velocity o f A is 2.5 m/s and x = 0 when t = 0, determine the
velocity and position of A when t = 0.05 s.
2
SOLUTION
Note that a is a given function of x.
vdv = adx = (200x + 3200x^^dx
Use
Using the limits v = 2.5 m/s when x = 0,
£ svrfv = £(200* + 3200x ^ d x
j c
— - — =
2
2
u 2
2
v = 1600*4 + 200
2
Then
where
v =
2
m,
1 6 0 0 m2
+
1 0 0
* 2
200m
and u2 are the roots o f
x + 800*4
2
+ 6.25
+ 6.25 =
Let u = x
1 6 0 0 (m
1 6 0 0 m2
+
2
- m,)(m - m2),
200m
+ 6.25 = 0
Solving the quadratic equation,
-200 ± 7(200 - (4)(1600)(6.25)
i
m, = ---------- vv
v . -----= -------------- = -0.0625 ± 0
(2)(1600)
3200
) 2
- 2 0 0
0
2
1 ,2
U] = m = -0.0625 ft
2
So
Taking square roots,
v =
2
1 6 0 0 (m
2
+ 0.0625)2 = 1600(*2 + 0.252)2 ft /s
2
v = ± 40(x2 + 0.252 jft/s
2
PROBLEM 11.16 CONTINUED
Use
,
,
dx = vdt
40dt = ±
,
dx
,
dx
dt = — - ± —
v
v
4 0 |x + 0.25 j
or
—— tx + 0.25
x = 0
Use limit
when
t = 0
2
40 td t = ± F
—— j- = ± — tan-1—
*
* x + 0.25
0.25
0.25
2
40t = ±4.0tan- (4x)
or
1
4x = ± tan(l0f)
v =^
or
tan- (4x) = ±10/
1
x = ±0.25tan(l0f)
= ±0.25[sec2(10/)](10) = ±2.5 sec (l0r)
2
At t - 0, v = ±2.5 m/s, which agrees with the given data if the minus sign is rejected.
Thus,
v = 2.5 sec (l0t)m /s,
2
At t = 0.05 s,
x = 0.25tan(l0f)m
and
10/ = 0.5 rad
v = 2.5 sec (0.5) =
2f
v ’ cos 0.5
2
v = 3.25 m/s A
2
x = 0.25tan(0.5)
x = 0.1366 m -4
PROBLEM 11.17
Point A oscillates with an acceleration a = 2880 - 144x, where a and x
are expressed in in./s and inches, respectively. The magnitude of the
velocity is 11 in./s when x = 20.4 in. Determine (a) the maximum
velocity of A, (b) the two positions at which the velocity of A is zero.
2
1
m
m
m
SOLUTION
Note that a is a given function o f x.
a = 2880 - 144x = 144(20 - x)
(a) Note that v is maximum when a = 0,
or
x = 20 in.
Use vdv = adx = 144(20 - x)dx with the limits
v = 11 in./s when x = 20.4 in.
and
v = vmax when x = 20 in.
r20
f r v* = C |4 4 (2 0 - * ) *
20
2
2
= -144
2
0
-
(0.4
) 2
= 11.52
20.4
vmax =
1 4 4 -0 4
in /s
2
Vmax =
2
1 2 . 0 0
in./s <
(b) Note that x is maximum or minimum when v = 0.
Use vdv = adx = 144(20 - x) with the limits
v = 11 in./s
when x = 20.4 in.,
J
0 - —— = -144
1 1
2
( 2 0
-
* ) 2
and
v= 0
when x = xm
= - 7 2 ( 2 0 - x m)2 + (7 2 )(0 .4 )2
2
20.4
(20 - x
m
) 2
= 1.00028
20 - xm = ±1.00014 in.
xm = 19.00 in. and
2 1 . 0
in. -4
PROBLEM 11.18
Point A oscillates with an acceleration a = 144(20 - x), where a and x
are expressed in in./s and inches, respectively. Knowing that the system
starts at time / = with v = and x = 19 in., determine the position and
the velocity of A when / = 0.2 s.
2
I v,
M
j w m m m
M
0
fl *
0
SOLUTION
Note that a is a given function of x.
a = 144(20 - x)in./s
2
Use vdv = adx = 144(20 - x) with the limits v = 0 when x = 19 in.
£ vdv = £^144(20 - x)dx
—— 0 = -144
= 7 2 - 7 2 ( 2 0 - jc )
2
19
v2 = 144^1 - (20 - x)2J
Use
.
.
ax = vat,
or
or
v = ±12^1 - (20 - x
) 2
in./s
,
dx
dx
at = — = -------,
v
±12^1 - (20 - x f
with the limits t = 0 when x = 19 in.
dx
or
y > - ±-
/ =
12 9 J l - ( 2 0 - x f
12^1 - (20 - x f
Let u = 20 - x.
u = 1 in. when x = 19 in. dx = -du
du
J ,
*
1 2
sin
u =
Then,
and
At t = 0.2 s,
dx
± - {
1 0 Jl
u
-
' • -l
n\
= T — sin u ----)
.
l
1
1 2
n
m= —T
1
1 2
2
/
T 12/j = cos(±12/) = cos(l2/) = 20 - x
x =
v=
2 0
1 2
- cos(l /) in.
2
s in (l /) in./s
2
12/ = 2.4 rad
x = 20 - cos(2.4)
v = 12sin(2.4)
x: = 20.7 in. <
v= .
8
1 1
in./s A
PROBLEM 11.19
The acceleration of a particle is defined by the relation a = 12jc - 28,
where a and x are expressed in m/s and meters, respectively. Knowing
that v = m/s when x = , determine (a) the maximum value of x,
(b) the velocity when the particle has traveled a total distance of 3 m.
2
8
0
SOLUTION
f
a =
Note that a is a given function of x.
1 2
m/s when x = .
0
V
8 2
+
n
= --- r x ----7 ? 1
I
3J _
1 2
—
x ---- \dx
3J
V2
2
2
8
12 ' L - l Y .
,3,
{
i)
82
1 2
x ---- m/s
3,
i
8
1
I
2
1 2
v
7^
Use vdv - adx = 12 x ---- dx with the limits v =
3y
v =
x - 28 =
2~
-
1 (
l\
v = ±,|
x ---I
3J
1 2
Reject minus sign to get v =
{a) Maximum value of x.
8
m/s at x = 0.
v = 0 when x = xn
12 M l 2 - ± = o
3
l
3J
x
7
3
= +—
3
+
1
•^max
^m
or
f x ----7 )
I
3J
9
and
Now observe that the particle starts at x = 0 with v > 0 and reaches x = 2 m . A tx = 2 m , v = 0 and
2
a < 0, so that v becomes negative and x decreases. Thus, x = 2 — m is never reached.
xmav =
2
m ^
PROBLEM 11.19 CONTINUED
(b) Velocity when total distance traveled is 3 m.
The particle will have traveled total distance d = 3 m when d - xmax = xmax - r or 3 -
2
=
2
- x
or x = m.
1
Using v = —
4 / 1 2
x —
----- , which applies when x is decreasing, we get
v=
- . 1
2
V
i '2
3.
v = -4.47 m/s -4
PROBLEM 11.20
The acceleration o f a particle is defined by the relation a = k { \ - e~ j ,
where k is a constant. Knowing that the velocity o f the particle is
v = +9 m/s when x = -3 m and that the particle comes to rest at the
origin, determine (a) the value of k, (b) the velocity o f the particle when
x =
m.
- 2
SOLUTION
Note that a is a function of x.
a = Ar^l - e~x j
Use vdv = a dx = k { 1 - e~x )dx with the limits v = 9 m/s w hen* = -3 m, and v = 0 w hen* = 0.
Cvdv = s ^ j l - e~x )dx
v
T
= k[x + e -* l
0 - t - = i [ 0 + 1 - (-3 ) - e3] = -16.0855*
(a)
k = 2.5178
Use vdv = adx = A 1 - e~xjdx = 2.5178|l -
k = 2.52 m/s A
2
ix with the limit v = 0 w henx = 0.
- e ~ ‘ )dx
f 24178(1
y
= [2.5178(* + c- ') ]* = 2.5178(* +
V2 =
5.0356(x
+
- l)
v
=
- l)
±2.2440(x
+
- l)
1/2
Letting x = - 2 m ,
v = ±2.2440(-2 ■‘r e 2 - l ) ^ = ±4.70 m/s
Since x begins at x =
- 2
m and ends at x = , v > .
0
0
Reject the minus sign.
v = 4.70 m/s A
PROBLEM 11.21
The acceleration o f a particle is defined by the relation a = - k 4 v , where
& is a constant. Knowing that x = 0 and v - 25 ft/s at / = 0, and that
v = ft/s when x = ft, determine (a) the velocity of the particle when
x = ft, (b) the time required for the particle to come to rest.
1 2
6
8
SOLUTION
vdv = adx = -ky/vdx,
x0 = 0 ,
v = 25 ft/s
0
dx = - ~ v V2dv
k
r dx =
f yjvdv - —- v * 2!
/fc Jl’°
3 k -lv
0
—
Noting that x =
6
I K 1-**)
-
ft when v = 12 ft/s,
6
Then,
= — [ 125 - 12 ^1 =
3/tL
J
or
k
k = 9.27 Jm /s
V
x = ^ | ^ y [ l 2 5 - v3'2] = 0.071916(l25 - v*2)
vm = 125 - 13.905x
(a) When x =
8
ft,
v
,3/2
3/2
= 125 - (13.905)(8) = 13.759 (ffc/s)
v = 5.74 ft/s <
(b)
At rest, v = 0
dv =
a dt - -kyfvdt
,
dt =
dv
r-pr
kV
1
2vq
(2)(25)
t = —— = — —— k
9.27
2
1/2
t = 1.079 s <
PROBLEM 11.22
Starting from x = Owith no initial velocity, a particle is given an
acceleration a = O.S-Jv2 + 49, where a and v are expressed in ft/s and
2
ft/s, respectively. Determine (a) the position of the particle when
v = 24 ft/s, (b) the speed of the particle when x = 40 ft.
SOLUTION
a = 0.8 Vv + 49
2
,
.
vav = a ax
.
vdv
vdv
ax ------- = ------ ,■ ■ a
0.8 Vv + 49
2
Integrating using x = 0 when v = 0,
F dxb
1
® -8
- Vv + 49
2
\Jv2 + 491
1
JQ
0 . 8
x = 1.25 (Vv + 4 9 - 7 )
2
( )
1
(a) When v = 24 ft/s,
x - 1.25(V242 + 49 - 7j
x = 22.5 ft A
(b) Solving equation (1) for v2,
yjv2 + 49 = 7 + 0.8x
v = (7 + 0.8x
2
) 2
- 49
When x = 40 ft,
v = [7 + (0.8)(40)]2 - 49 = 1472 ft /s
2
2
2
v = 38.4 ft/s A
PROBLEM 11.23
The acceleration of slider A is defined by the relation a = -2 k\]k 2 - v 2,
where a and v are expressed in m/s2 and m/s, respectively, and k is a
constant. The system starts at time t = 0 with x = 0.5 m and v - 0.
Knowing that x = 0.4 m when t = 0.2 s, determine the value o f k.
SOLUTION
dv
a = — = -2 k \Jk - v2,
dt
-2 k dt =
v = 0 when t = 0
and
-
dv
=d
• -1
sm
\lk 2 -
(V ~
—
{kj
-
-2 k ^ d t = sin 1
V
U ,
sm
u.
= -2kt
v = &sin(-2&/) = -ksin (2 k t)
dx = vdt = -k s m (2 k t)d t
Integrating, using x = 0.5 m at / = 0, and x = 0.4 m at / = 0.2 s,
£. * = - T f t s i n f 2* ! ) ] *
5
I* C = i cos(2i<)jo = °-4 - °-5 = ^COS[(2)*(0.2)] - i
cos(0.4&) = 0.8
0.4k = cos-1 (0.8) = 0.6435 rad
k =
0.6435
0.4
k = 1.609 m/s 4
B
PROBLEM 11.24
j t
The acceleration of slider A is defined by the relation a = —2Vl - v 2,
w here a and vv are expressed in m/s
m /s2 and m
/s. respectively.
resnectivelv. The system
svstem
where
m/s,
starts at time / = 0 with x = 0.5 m and v = 0. Determine (a) the position
of A when v = - 0.8m/s, (b) the position of A when t = 0.2 s.
m
SOLUTION
Acceleration is a given function of velocity.
(a) Position when v = -0.8 m/s.
vdv
From vdv = adx, we get dx = ----a
vdv
- \/l - v
2
2
Integrating, using x = 0.5 m when v = 0.
vdv
or
i/T -
- VF“
2
2
When v = -0.8 m/s,
i r.
r
X= -V l-V
or
- 0 .5 = —(Vl - v - VT)
x .
x = 0.300 m 4
^ - ( - . )!
1
1
0
( 1)
8
( ) Position when t = 0.2 s.
6
From dv = a dt we get
dv
dt = — =
a
dv
-2V 1 - V 2
Integrating, using / = 0 when v = 0.
dv
£<* = £
tSolving for v,
When r = 0.2/ s,
0
-2 V1 - V 2 *
= —-s in
2
-1
or
v-
H> = - { [ sin~lv I
t = —5-sin v
2
0
" 1
v = s in (- /) - - s i n ( r)
2
2
21 = 0.4 rad
v = - s i n 0.4 = -0.3894 m/s
Using equation (1),
x = —^1 - (-0.38941)2
x = 0.461 m 4
PROBLEM 11.25
The acceleration of a particle is defined by the relation a = - k v 25,
where k is a constant. The particle starts at x = 0 with a velocity of
16 mm/s, and when x = 6 mm the velocity is observed to be 4 mm/s.
Determine (a) the velocity of the particle when x = 5 mm, (b) the time
at which the velocity of the particle is 9 mm/s.
SOLUTION
Acceleration is a given function of velocity,
a = -k v 2 5
(a) Velocity when x = 5 mm.
c
From
,
vdv
vdv
1 _] 5
dx = ----- = ------ ^ - — v
a
kv
k
,
,
vdv = adx
Integrating, using the condition x = 0 when v = 16 mm/s.
f * =- tk L v~l$dv or
x - 0 = - ( v -0-5 - 16“°-5)
k
or
M
o =f[v
L J0
k L “°'5J
x = | (v“0-5 - 0.25)mm
k
( 1)
When x - 6 mm, v = 4 mm, hence
6 .. | , 4 ‘0-5 -0 .2 5 )
k - 0.08333mm_1-5s0-5
or
Solving (1) for v,
v-0'5 = 0.25 + Q.Skx
v = (0.25 + 0.5kx) 2
When x = 6 mm,
v = [0.25 + (0.5)(0.08333)(5)]"2
v = 4.76 mm/s <
(6) Time when v = 9 mm/s.
dv —------dv
1 -ir2 5
dt = — =
— -fr- ==---- —v~1
2>dv
a
kv25
k
From dv = a dt,
Integrating, using the condition t - 0 when v = 16 mm/s
td t Jo
f v~25dv
k Xe
or
t - 0 = Y ^ [ v_1'5 _ (16)_1'5 j
When v = 9 mm,
L Jo
or
i ' S k ik 1-5!Jie
t = 8(v" '’5 " 0.015625)
t = 8(9"15 - 0.015625)
t = 0.1713 s <
PROBLEM 11.26
The acceleration of a particle is defined by the relation a = 0.6(1 - k v ),
where k is a constant. Knowing that at f = 0 the particle starts from rest
at x =
m and that v =
m/s when t =
s, determine (a) the
constant k, (b) the position of the particle when v = 7.5 m/s, (c) the
maximum velocity o f the particle.
6
6
2 0
SOLUTION
(a) Determination of k.
From
dv = adt,
dt =
dv
a
0
.
6 ( 1
dv
- kv)
Integrating, using the condition v = 0 when t = 0,
dv
0
.
6 ( 1
( 1)
- *v)
Using t = 20 s when v =
6
mm/s,
20 - - ^ - ^ - ln ( l - 6k)
Solving by trial,
k = 0.1328 s/m A
(b) Position when v = 7.5 m/s.
From vdv = adx,
dx =
vdv
vdv
a
0
.
6 ( 1
- kv)
vdv
Integrating, using the condition x = 6 m when v = 0,
Now
1
- kv
So
f *
k
0 . 6
x - 6-
0.6k
v+
0
6 ( 1
- kv
1
dv —
= £ .
k
“ ^v)
- v - -^ln(l - kv)
-10
- jfcv)
PROBLEM 11.26 CONTINUED
Using v = 7.5 m/s and the determined value o f k :
x =6
1
(0.6)(0.1328) 7-5 + s k i h (1- ( 0-1328)(7'5))
x = 434 m A
(c) Maximum velocity occurs when a = 0.
Then,
1
1
k
0.1328
v max = 7 - 5 3 m / s A
PROBLEM 11.27
Experimental data indicate that in a region downstream of a given
louvered supply vent the velocity of the emitted air is defined by
v = 0.18u0/x, where v and x are expressed in ft/s and feet, respectively,
and v0 is the initial discharge velocity of the air. For v 0 = 12 ft/s,
determine (a) the acceleration of the air at x = ft, (b) the time required
for the air to flow from x = 3 ft to x = 10 ft.
6
SOLUTION
(a) Acceleration at x =
6
ft.
v _ (0.18)(12) _ 2.16
v = 0.18—
ft/s
x
0
dv
dx
2.16
x2
dv
2.16 2.16
4.6656
a = v— = ----- ---------- -- -------- t—
dx
x
x
x
5
When x =
6
4.6656
a = —
ft,
a = -0.0216 ft/s <
2
( ) Required time for flow from x = 3 f t t o x = 10ft.
6
Use — = v from which dt = — = —
dt
dt
v
0.18v0
dt =
dx
v
xdx
0.18v0
xdx
= 0.46296xo!x
(0.18)(12)
Integrating, using t = 0 when x = 3 ft,
j^dt = 0.46296 %xd x
or
[r]J, = 0.46296
t = 0.23148(x2 - 9 )
When x = 10 ft,
= (0.23 1 48)( 10 2 - 9 )
t =
2 1 . 1
s4
PROBLEM 11.28
t;
Based on observations, the speed of a jogger can be approximated by the
relation v = 7.5(1 - 0.04x)° , where v and x are expressed in km/h and
kilometers, respectively. Knowing that x = 0 at t = 0, determine
(a) the distance the jogger has run when t = h, (b) the jogger’s
acceleration in m/s at t = , (c) the time require for the jogger to run
km.
i
1
2
0
6
SOLUTION
Given: v = 7.5(1 - 0.04x)°3 with units km and km/h.
(a) Distance at t = 1 hr.
. . . . .
.
dx
dx
Using dx = vdt, we get dt = — = ------------------ ttv
7.5(1 - 0.04x)
Integrating, using t = 0 when x = 0,
rd t = — r —
h
7.5 * (i _ 0.04)
or
3
[,r= JL .
I i - o . m * 07)*
(7.5) (0.7)(0.04)l
io
7
0
t = 4.7619 jl - (1 - 0.04x)0 7}
Solving for x,
When f =
x = 25 jl - ( l - 0.2 lOr J170'71
h,
1
(1)
x = 25 jl - [ - ( .
0
2 1 0
)(l)]‘/° 7J
x = 7.15km^
(b) Acceleration when t = 0.
^
= (7.5)(0.3)(-0.04)(l - 0.04x)^>'7 = -0.0900(1 - 0.04x)"°’7
When t = 0 and x = 0,
v = 1.5 km/h,
a =
— - 0.0900 h
dx
-1
= (7.5)(-0.0900) = -0.675 km/h
2
- ( 0 .6 7 5 ) ( 1 0 0 0 ) ^
(3600)“
(c) Time to run
Using x =
6
6
a = -52.1 x 10
- 6
m/s A
2
km.
km in equation (1),
= 4.7619 jl - [l - (0.04)(6)]° 7J = 0.8323 h
t = 49.9 min ^
PROBLEM 11.29
The acceleration due to gravity of a particle falling toward the earth is
a = -g R 2 / r 2, where r is the distance from the center o f the earth to the
particle, R is the radius of the earth, and g is the acceleration due to
gravity at the surface of the earth. If R = 3960 mi, calculate the escape
velocity, that is, the minimum velocity with which a particle must be
projected upward from the surface of the earth if it is not to return to
earth. (Hint, v = 0 for r = oo.)
SOLUTION
dv
gR
v— = a = —
dr
r~
The acceleration is given by
Then,
vdv = -
gR2dr
Integrating, using the conditions v = 0 at r = oo, and v = vesc at r = R
£ v d v
= -g R 2% { j
0
Iv
2
= g*2
2
V
vesc
r
'e s c = > / 2 g ^
Now, R = 3960 mi = 20.909 x 10 ft and g = 32.2 ft/s2.
6
Then,
v,^ = J(2)(32.2)(20.909 x 106)
= 36.7 x 10 ft/s <
3
PROBLEM 11.30
The acceleration due to gravity at an altitude y above the surface of the
earth can be expressed as
-32.2
a -
[ l + ( y / 2 0 . 9 x l 0 6)]
2
where a and y are expressed in ft/s and feet, respectively. Using this
expression, compute the height reached by a projectile fired vertically
upward from the surface o f the earth if its initial velocity is
(a) v = 2400 ft/s, (b) v = 4000 ft/s, (c) v = 40,000 ft/s.
SOLUTION
-32.2
The acceleration is given by a =
-32.2dy
vdv = ady =
[ u K ) T
/J
\ 20.9x10
\ 20.9x10 ) \
Integrate, using the conditions v = v a ty = 0 and v = 0 a ty = ymax. Also, use g = 32.2 ft/s' and
0
R = 20.9 x 10 ft.
6
n
( 1
+
i
_ , . „
Solving forymax,
ymax =
1
= gR2
7
[ r
) 2
1
1
R + Tmax
R
0 - ~ vo = gR2
>'mmx
dy
(° vdv = - g C — ——z- = - g R 2
gRya
R + yn
v0
R+y
Vo{R + Tmax) = gRy„
2
Rva
u
2 gR - v
2
0
20.9
Using the given numerical data,
Tmax
x
10 v
6
_
02
(2)(32.2)(20.9 x 106) - v
02
(
(a) v = 2400 ft/s,
2 0 . 9
20.9
x 1Q6 Vq
1.34596 x 10 9
x 106)(2400)'
ymax = 89.8 x 10 ft <
3
0
'Vmax ~ 1.34596 x 10 - 2 4 0 0 2
9
(20.9 x 106)(4000)2
Cb) v = 4000 ft/s,
ymax = 251 x 10 ft ^
3
0
■Vm“ ~ 1.34596 x 10 - 40002
9
(c) v = 40000 ft/s,
Tmax =
0
(20.9 x 106)(40000)2
------------- 'q------------T = negative
1.34596 x 10 - 400002
9
Negative value indicates that v is greater than the escape velocity.
0
Tmax =
0 0
<
PROBLEM 11.31
The velocity o f a slider is defined by the relation v - t>'sin (cont + <f>J.
Denoting the velocity and the position o f the slider at / = 0 by v 0 and
x0, respectively, and knowing that the maximum displacement o f the
slider is 2x0, show that (a) -u' =
+ XqCO2 )/2 x 0co„ , (b) the maximum
value of the velocity occurs when x = x013 - ( i >0 / xQw„} / 2.
SOLUTION
(a) Given: v = v'sin(ft>„/ + tp)
At / = 0,
v = v0 = v sm <p
or
sm p = —
(1)
v'
Let x be maximum at t = tx when v = 0.
Then,
sin(ffl„f1 + cp) = 0
dx
— =v
dt
Using
At / = 0,
Then,
x =x = C
V
0
X = x0 +
c»„
(Or,
cos<z>
v
V
cos (cot +
V
C = x0 + — cos^
or
C O S< p
co„
*max = *0 +
,
,
dx - vdt
or
x = C
Integrating,
V
/
COSlfi>„/ + .
(3)
(0„
V
COS^ + —
co
.
U S ing COS<0„/, + <p = - 1
cos cp =
Solving for cos cp,
(2)
cos(<a„/s + cp) = ± 1
and
_ 1
v
With xmax = 2x0,
Using
Solving for v' gives
X0(On
cos cp ~
sin cp + cos 9? = 1 ,
or
v'
|
- - 1
, | +
(4)
* 0 «>n
-1
= 1
(vp + xW n )
,
V
=
2 x06)n
(5 )<
PROBLEM 11.31 CONTINUED
(ib) Acceleration:
a = — = v'con cos(tw„f + <p)
dt
Let v be maximum at t = t2 when a - 0.
cos{o)nt2 + <p) - 0
Then,
From equation (3), the corresponding value of x is
v'
v' f
x = Xq + — cos q> = x H-----
- 1
0
’n \
y
] = 2xq - —
y
°>n
= 2x
V() + XWn _ 3
1 Vq
- ZX° - ( 2 w „ ) a > n - 2 0 ~ 2 x0a)2
xo
PROBLEM 11.32
The velocity of a particle is v = u0[ l - s in (/rt/7 )]. Knowing that the
particle starts from the origin with an initial velocity v 0, determine (a) its
position and its acceleration at t = 37, (b) its average velocity during the
interval t to t = 7,
0
SOLUTION
W
dx
-
1
= V = V„
1
. nt
- sm —
7
Integrating, using x = x0 = 0 when t = 0,
,
£ dx = { v d t = Jjv
1
0
. nt
- sm — dt
7
,'nj
nt
v0r + - y—cos—
n
v0r
n t v0T
H cos —
x = v0t + ——
- _
n
n
( 1)
x = 3v0T + - '—cos ( 3^ ; 7
When t - 37,
n
= 3-
v j
x = 2.36VcJ A
dv
a =— =
dt
nvr
nt
-co s—
7
7
a
When t = 37,
7
a =
-cos3;r
(b) Using equation (1) with t = 7,
xx = v0T +
cos n -
n
n
= v07
'.-A'
V
ttj
Average velocity is
Ax _ x, - x _
'l-A '
a7“
7
- v
nj
0
vave = 0.363v0 A
PROBLEM 11.33
An airplane begins its take-off run at A with zero velocity and a constant
acceleration a. Knowing that it becomes airborne 30 s later at B and that
the distance AB is 900 m, determine (a) the acceleration a, (b) the take­
off velocity V g .
SOLUTION
Constant acceleration.
v = vA = 0,
x0 - xA = 0
0
v = vr,
at
’o +
+ at -= ai
1
2
x = x(t+ v0t + -1a t 2 = —at
At point B,
Solving (2) for a,
Then,
x = xB = 900 m
and
( )
2
t = 30 s
a = % = W -t 9? )
t
(30)
a = 2 m/s 4
vB = at = (2)(30)
vs = 60 m/s ^
2
PROBLEM 11.34
Steep safety ramps are built beside mountain highways to enable vehicles
with defective brakes to stop safely. A truck enters a 250-m ramp at a
high speed v0 and travels 180 m in s at constant deceleration before its
speed is reduced to v0/2. Assuming the same constant deceleration,
determine (a) the additional time required for the truck to stop, (b) the
additional distance traveled by the truck.
6
SOLUTION
x0 = 0
Constant acceleration.
v = v + at
0
(1)
1 2
X = Xq + V + 2 a r
(
(3)
x = x0 + v0r + 1 v -( Vn >'? = x0 + ^(v0 + v)t = ^ (v + v)t
Then,
0
At t = 6 s,
v = -^-v
1
1
0
v = —vn =
2
0
2 0
6
x6 = 180 m
and
0
(
^
180 = - v + —v ( ) = 4.5v
L V.
2. y
180
v = — = 40 m/s
4.5
or
0
0
m/s
0
a = ---------- = ------ m/s = -3.333 m/s
Then, from (3),
2
Substituting into (1) and (2),
v = 40 - 3.333f
x = 40 + 40/ - ^(3.333
or
2
3
6
40 - 3.333ts = 0
(a) Additional time for stopping = 12 s -
6
) / 2
ts = 12 s
x = 0 + (40)(12) - -j(3.333)(12
s
( ) Additional distance for stopping = 240 m - 180 m
6
2
„ - v " vo
Solving ( 1) for a,
At stopping, v = 0
( )
) 2
= 240 m
At -
6
sA
Ad = 60 m -4
PROBLEM 11.35
A truck travels 540 ft in s while being decelerated at a constant rate of
1.5 ft/s2. Determine (a) its initial velocity, (b) its final velocity, (c) the
distance traveled during the first
s.
8
a ■ 1.5 ft/s"
0 . 6
SOLUTION
a = -1.5 ft/s
x =
(a) Solving for v using x = 540 ft when t =
0
x ~ xo ~ 2 atf
V° “
(b) At t =
8
t
*0
8
x = 0
2
0
1 2
+ Vq/ + —at'
s,
540 - 0 - (0.5)(-1.5)(8
“
) 2
8
Vn = 73.5 ft/s 4
s,
v = 61.5 ft/s 4
v = v0 + at = 73.5 + (—1.5)(8)
(c) At t = 0.6 s,
x = 0 + (73.5)(0.6) + i( - 1 .5 ) ( 0 .6
) 2
x = 43.8
ft 4
PROBLEM 11.36
%*25ini/ta
A motorist enters a freeway at 25 mi/h and accelerates uniformly to
65 mi/h. From the odometer in the car, the motorist knows that she
traveled 0.1 mi while accelerating. Determine (a) the acceleration of the
car, (b) the time required to reach 65 mi/h.
SOLUTION
Constant acceleration.
v = 25 mi/h = 36.667 fit/s
0
Vy —65 mi/h = 95.333 ft/s
x =0
and
0
Xy = 0.1 mi = 528 ft
Vy = v + 2a[xf - x0)
02
(a) a =
V/ - Vo
95.333 - 36.6672 „ , , , ,
r = ------;
r
= 7.3333 ft/s
2 ( x y - x 0)
2(528-0)
2
2
— 7
a = 7.33 ft/s ^
2
(b) Vy = v + atf
0
_
Vf
vf
- vn
95.333 - 36.667
7.3333
tf =
8 . 0 0
s
PROBLEM 11.37
it
A group of students launches a model rocket in the vertical direction.
Based on tracking data, they determine that the altitude of the rocket was
27.5 m at the end of the powered portion of the flight and that the rocket
landed 16 s later. Knowing that the descent parachute failed to deploy so
that the rocket fell freely to the ground after reaching its maximum
altitude and assuming that g = 9.81 m/s2, determine (a) the speed t>, of
the rocket at the end of powered flight, (b) the maximum altitude reached
by the rocket.
27.5 m
SOLUTION
Constant acceleration.
Choose t = 0 at end of powered flight.
y t = 27.5 m
Then,
(а) When y reaches the ground, y f = 0
and
a = - g = -9.81 m/s
2
/ = 16 s.
y f = y x + y + -1a t 2 = y, + y - -1g r2
J-/ - r , ^ r ; = o - 2 7 . 5 ^ ( 9 . 8 i)(i6): =767 6[n /s
t
16
v. = 76.8 m/s A
( ) When the rocket reaches its maximum altitude ymax,
б
v=
0
v = v + a ( y - y,) = v - 2 g ( y - y })
2
,2
2
,2
v -v
2
y -y \-
2
ymax = 27.5 -
g
2
-
0 - 76.762
...
—
(2)(9.81)
ymax = 328 m A
max
PROBLEM 11.38
A sprinter in a 400-m race accelerates uniformly for the first 130 m and
then runs with constant velocity. If the sprinter’s time for the first 130 m
is 25 s, determine (a) his acceleration, (b) his final velocity, (c) his time
for the race.
SOLUTION
1 2
(a) During the acceleration phase x = x0 + v0t + —at
Using x0 = 0, and v0 ==0, and solving for a gives
2x
a = ~tT
Noting that x = 130 m when t = 25 s,
„ . (4 (1 3 0 )
a = 0.416 m/s M
(25)J
(b) Final velocity is reached at t = 25 s.
Vf = v0 + at = 0 + (0.416)(25)
Vf = 10.40 m/s 4
(c) The remaining distance for the constant speed phase is
Ax = 400 - 1 3 0 - 270 m
For constant velocity,
Total time for run:
At - — = _?Z2_ _ 25.96 s
v
10.40
t = 25 + 25.96
t = 51.0 s 4
PROBLEM 11.39
In a close harness race, horse 2 passes horse 1 at point
where the two
velocities are v 2 - 1 m/s and v x = 6.8 m/s. Horse 1 later passes horse 2
at point B and goes on to win the race at point C, 400 m from A. The
elapsed times from A to C for horse 1 and horse 2 are /j = 61.5 s and
t2 - 62.0 s, respectively. Assuming uniform accelerations for both horses
between A and C, determine (a) the distance from A to B, (b) the position
of horse 1 relative to horse 2 when horse 1 reaches the finish line C.
. W i ,
SOLUTION
Constant acceleration (oj and a2) for horses 1 and 2.
Let x - 0 and t = 0 when the horses are at point/I.
Then,
x = v0t + —at
0
2
Solving for a,
2 (x - vn/)
a = — —g— ~
2
Using x - 400 m and the initial velocities and elapsed times for each horse,
'?
_^
(61.5)2
_ 2 [ 4 0 0 -( 7 .0 )( 62.0)] _
t
(62.0)
Calculating x{ - x2,
xl - x2 = (vt - v2)t + —(aj - a2) t 2
xI - x 2 = (6.8 - 7.0)/ + -[(-9 .6 2 3 9 x 10 3 1- (-17.6899 x 10~3'. jt2
= -0.2/ + 8.066 x 10“3/ 2
At point B,
xi - x2 - 0
/
-0 .2 tB + 4.033 x 10l_3/ | = 0
— -_ = 49.59 s
4.033 x 10"3
Calculating xB using data for either horse,
Horse 1:
Horse 2:
xB = (6.8)(49.59) + ^-(-9.6239 x 10“3)(49.59)2
xB = 325 m 4
xB = (7.0)(49.59) + ^-(-17.6899 x 10_3)(49.59)2 = 325 m
When horse 1 crosses the finish line at / = 61.5 s,
X j - x 2 = -(0.2)(61.5) + ( 4.033 x 10~3)(61.5)2
Ax = 2.95 m 4
PROBLEM 11.40
Two rockets are launched at a fireworks performance. Rocket A is
launched with an initial velocity v 0 and rocket B is launched 4 seconds
later with the same initial velocity. The two rockets are timed to explode
simultaneously at a height of 80 m, as A is falling and B is rising.
Assuming a constant acceleration g = 9.81 m/s2, determine (a) the initial
velocity v 0, (b) the velocity of B relative to A at the time of the
explosion.
SOLUTION
Choose x positive upward.
Constant acceleration a = - g
Rocket launch data:
Velocities:
Rocket A:
x = 0, v = v0, t = 0
Rocket B:
x = 0, v = v0, / = tB = 4 s
Rocket A:
vA= v -
Rocket B:
vB= v - g ( t - tB)
0
gt
0
Rocket A: xA = v</ - ^ g t 2
Positions:
Rocket B: xB = v0(t - tB) - - g ( t - tB)~,
t > tB
For simultaneous explosions at x A = xB = 80 m when t = tE,
v(Je
~ ~ % rE -
vo ( fE ~ h )
~
~
{h )
= V a
“
V fl
”
+ fpEh
^
~
(1)
Solving for v0,
Then, when t = tE,
Solving for tE,
2
°r
{ E
2x,
~
t R t F.
(4)- + (4)(l)(2)(80)
9.81
g
—_ 0
6.507 s
PROBLEM 11.40 CONTINUED
v = (9.8l)(6.507) -
(a) From equation (1),
At time
0
tE ,
VA = V0 ~ g t E
v B
~
v A
=
g t B
=
(9.81)(4)
_ 4^ 2 m/s j
A
39.2 m/s f
<
VB = V0 - g ( t E - t B )
v
Bia
=
PROBLEM 11.41
A police officer in a patrol car parked in a 65 mi/h speed zone observes a
passing automobile traveling at a slow, constant speed. Believing that the
driver of the automobile might be intoxicated, the officer starts his car,
accelerates uniformly to 85 mi/h in
s, and, maintaining a constant
velocity of 85 mi/h, overtakes the motorist 42 s after the automobile
passed him. Knowing that 18 s elapsed before the officer began pursuing
the motorist, determine (a) the distance the officer traveled before
overtaking the motorist, ( ) the motorist’s speed.
8
6
SOLUTION
Motion o f motorist: vm = vm = constant,
xm = vmt
Motion o f patrol car: Starts at t = 18 s with constant acceleration
At t = 18 +
8
= 26 s,
vp = 85 mi/h = 124.67 ft/s
v = a (t - 18)
p
v
'
a =—
t - 18
26-8
= 15.583 ft/s
2
*„ = ! « (< - 1 8 )!
At t = 26 s,
For t > 26 s,
xp = ^(15.583)(26 - 18)2 = 498.67 ft
vp = 124.67 ft/s
xp = 498.67 + 124.67(t - 26)
(a) At t = 42 s,
xp = 498.67 + 124.67(42 - 26) = 2493.4 ft
xp = 0.472 mi 4
( ) At t = 42 s,
6
xm = vm(42) = xp = 2493.4 ft
v = 2493.4 =
42
5 9
3? M
vm = 40.5 mi/h 4
PROBLEM 11.42
As relay runner A enters the 65-ft-long exchange zone with a speed of
42 ft/s, he begins to slow down. He hands the baton to runner B 1.82 s
later as they leave the exchange zone with the same velocity. Determine
(a) the uniform acceleration of each of the runners, (b) when runner B
should begin to run.
SOLUTION
Let x = 0 at the start of the exchange zone, and t = 0 as runner A enters the exchange zone.
Then,
( v ^ = 42 ft/s.
Motion of runner A:
vA = (v^)Q + aAt
xA = { v A)0t + \ a/
(a) Solving for aA, and noting that xA = 65 ft when t = 1.82 s
2 ^ -(v ,V l
-=
t2
At t =
1 .8
2[65 -(42 )(l .82 )]
=
s,
(v ^ =
Motion of runner B: (vB)
and
= 0
Then,
vb
6 5 ft,
4 2
0
~ (
)1 = 2
vb
- 6 .9 1
ft/s A
2
( vfl)2 “ ( v« )o
2x
s
) 0
( - 6 . 9 l ) ( l . 8 2 ) = 2 9 .4 3
ft/s
at the starting time tB of runner B.
2 9 .4 3
aB = -----= (v
+
xB =
Solving for aB, and noting that vB = {vA)^ =
W henxB =
aA =
1 .8 2
o b x b
ft/s
2 9 .4 3 2 - 0
_
aB =
= -
6 .6 6
ft/s A
( 2 )(6 5 )
b
+ aB(t - tB) = aB(t - tB)
Vr
t~ lB=—
°B
(b) Solving for tB, and noting that vB =
tB =
B
2 9 .4 3
r - —
aB
ft/s when t =
= 1 .8 2 -
6.66
Runner B should begin
2 .6 0
1 .8 2
s
= - 2 .6 0
s
s before runner A reaches the exchange zone. A
PROBLEM 11.43
In a boat race, boat A is leading boat B by 38 m and both boats are
traveling at a constant speed of 168 km/h. At t = 0 , the boats accelerate
at constant rates. Knowing that when B passes A, t - 8 s and
vA = 228 km/h, determine (a) tire acceleration o f A, (b) the acceleration
of S.
SOLUTION
(a) Acceleration of A.
)0 = 168 km/h
VA = M o + aA*’
CO
00
II
•k*
3
vA = 228 km/h =
=
46.67 m/s
63.33 m/s
va ~ M 0 63.33- 46.67
fli = ------—
al -------------t
8
(b)
^ = ( * 4 )0 +
W o' + \
af
*s = (**)„ +
xa ~ xb = M o - (xb ) 0 + [Wo -
= 0,
When t
= 8 s,
M 0-
{xb)0
= 38 m
W o] '
and
W o' +
2
ai f 2
+ j ( a A - aB)t2
(v ,)0 -
(v ^ = 0
0
11
>?
1
H
When t
„ AOm/s, 2^
aA = 2.08
-4
0 = 38 +
Hence,
aB
;(aA - aB)(8)2,
= aA + 1.1875 = 2.08 +1.1875
or
aA - a B =
- 1.1875
aB
I
=*3.27 m/s2 -4
PROBLEM 11.44
Car A is parked along the northbound lane of a highway, and car B is
traveling in the southbound lane at a constant speed of 96 km/h. At
t = 0, A starts and accelerates at a constant rate aA, while at t = 5 s, B
begins to slow down with a constant deceleration of magnitude aA / .
Knowing that when the cars pass each other x = 90 m and v A = v B,
determine (a) the acceleration aA, (b) when the vehicles pass each other,
(c) the distance between the vehicles at t = .
11
6
0
SOLUTION
(a)
Acceleration of A.
va
= W
and
+ a At
o
Using
xA =
(vA = 0
va
(xA\
j
and
(vA)„ t
* I a
/
{xA)0 = Ogives
and
= a At
+
1
Xj = - a J
2
When cars pass at t - tp xA - 90 m
(2 = 2f i = (3 M
aA
= M
md
aA
For 0 < t < 5 s,
vB = (vs ) = -96 km/h = -26.667 m/s
For t > 5 s,
vb
= ( vs ) o
+
- 5) = -26.667 +
^ a A (t
Va = -VB
When vehicles pass,
aJ \ = 26.667 - ~ a A(tx - 5 )
7
— a At]
Using
Let u -
or
-
5
— ctA
= 26.667
or
_
ltx
160
- 5 = -----
[ m , 7-n/180
. 160
gives —
- 5=
h = I
l M 6 u - 5 = 160u2
160«2 - 7 ^ 8 0 « + 5 = 0
- 5)
PROBLEM 11.44 CONTINUED
Solving the quadratic equation,
u =
7^180 ± V(49)(180) - (4)(160)(5) = 93.915 ± 74.967
(2)(160)
= 0.0592125
aA =
and
u
320
0.52776
= 285.2 m/s
and
3.590 m/s
The corresponding values for /, are
t
1
=
V285.2
=
0.794 s,
and
tt =
1
V3.590
=
7.08 s
Reject 0.794 s since it is less than 5 s.
aA = 3.59 m/s M
Thus,
t = /j = 7.08 s "4
(b) Time of passing.
(c) Distanced.
0 ^ t < 5 s,
xB = (xB)0 - (v*)01 = d - 26.667/
At t = 5 s,
xB = d - (22.667)(5) = d - 133.33
F o r / > 5 s,
xB — d —133.33 + (v8 )0(/ - 5) + i « s (/ - 5)2
xB = d - 133.33 - 26.667(/ - 5) +
When / = /, = 7.08 s,
- 5)2
xB = xA = 90
90 = d - 133.33 - (26.667)(2.08) + ^
d =90 + 133.33 + 55.47 + 1.29
j ~
d = 278 m -4
PROBLEM 11.45
Two automobiles A and B traveling in the same direction in adjacent
lanes are stopped at a traffic signal. As the signal turns green, automobile
A accelerates at a constant rate of 6.5 ft/s2. Two seconds later,
automobile B starts and accelerates at a constant rate o f 11.7 ft/s2.
Determine (a) when and where B will overtake A, (b) the speed of each
automobile at that time.
SOLUTION
For
t
> 0,
For
t
> 2 s,
xa
=
XB
( * a )o
=M o
or
For
+ W o ' + \ aAf2 = 0 + 0 + -|(6 .5 )r
xB
xA
=
+ (
vb
)o( ‘
-
2 )
2
or
xA
= 3.25/2
+ : % (' - 5)2 = 0 + 0 + | ( » . 7 ) ( » - 2
) 2
= 5.85(t - i f = 5.85r2 - 23.4r + 23.4
3.25/2 = 5.8512 - 23.4 1 + 23.4,
x B,
2.60?2 - 23.41 + 23.4 = 0
or
Solving the quadratic equation,
t =
1.1459 and t
=
7.8541 s
Reject the smaller value since it is less than 5 s.
(a)
t =
xA = x B =
v^=(v
(b)
vB =
(v
* ) 0
^ ) 0
(3.25)(7.8541)2
+V
+ a B {t
- 2)
=
=
O + (6.5)(7.854l)
0
+
(11.7)(7.8541 - 2)
x =
vA =
vB
7.85 s
<
ft
<
2 0 0
51.1 ft/s <
= 68.5 ft/s A
PROBLEM 11.46
f Two automobiles A and B are approaching each other in adjacent
highway lanes. At / = 0, A and B are 0.62 mi apart, their speeds are
v A = 68 mi/h and v B = 39 mi/h, and they are at points P and Q,
respectively. Knowing that A passes point Q 40 s after B was there and
that B passes point P 42 s after A was there, determine (a) the uniform
accelerations o f A and B, (b) when the vehicles pass each other, (c) the
speed of B at that time.
4^ I s f
>l-
I?
SOLUTION
Let x be the position relative to point P.
Then,
(x^)0 = 0
Also,
(v^, )0 = 6.8 mi/h = 99.73 ft/s
(a)
and
(xs )Q = 0.62 mi = 3273.6 ft
and
(vg )Q= -39 mi/h = -57.2 ft/s
Uniform accelerations.
I
\
*a = ( x a ) o
/
+
\
2 [ X A ~ { X a ) 0 - { Va ) 0 A
~2------------------------ i
2
1
t + 2 a Af
(v a )0
a A -=
o r
—
,
7
aA = 0.895 ft/s2 — 4
-
2 f 3 2 7 3 .6 -0 -(9 9 .7 3 )(4 0 )l
= -0.895 ft/s2
aA = -L \
(40)
l
\
XB = { XB ) 0
,
+
v
1
i VB ) 0
+
2[ ^ ~
.2
2 a Bf
0 f
°B =
(
^
)
0
~2---------- *
2 f o ~ 3273.6 - (-57.2)(42)1
,
aB = -I*
\
^
= -0.988 ft/s2
(42)
(b)
,
aB = 0.988 fl/s2 — 4
When vehicles pass each other x A = xB.
( * a ) o +
( va )
0t
+
\
af
= M
o
+ M
0t +
0 + 99.73/ + -—(-0.895)/2 = 3273.6 - 57.2/ + i( - 0 .9 8 8 ) /2
0.0465/2 - 156.93/ + 3273.6 = 0
Solving the quadratic equation,
/ = 20.7 s and -3390 s
Reject the negative value. Then,
/ = 20.7 s 4
(c) Speed of B.
vB = (vB)o + aBt = -57.2 + (-0.988)(20.7) = -77.7 ft/s
Iv J = 77.7 ft/s 4
PROBLEM 11.47
¥
Block A moves down with a constant velocity o f 2 ft/s. Determine (a) the
velocity of block C, (b) the velocity of collar B relative to block A, (c) the
relative velocity of portion D of the cable with respect to block A.
¥
D
SOLUTION
Let x be positive downward for all blocks and for point D.
vA =
2
ft/s
Constraint of cable supporting A: xA + (xA - xB) = constant
2
va
-
v
b
or
= 0
vB = 2
va
=
(2)(2)
4 ft/s
=
Constraint of cable supporting B: 2xB + xc = constant
vc + 2vfi = 0
or
vc = -2vg = - (2 )(4 ) =
(c)
ft/s
vc =
(a)
(b)
- 8
vB/A = vB - vA = 4 - 2
8
ft/s | A
\ B/A = 2 ft/s i <
xD + xc = constant, vD + vc =
vD = - vc =
8
8
0
ft/s
Vv a = 6 ft/s { <
PROBLEM 11.48
Block C starts from rest and moves down with a constant acceleration.
Knowing that after block A has moved 1.5 ft its velocity is 0.6 ft/s,
determine (a) the accelerations o f A and C, (h) the velocity and the
change in position o f block B after 2 s.
SOLUTION
Let x be positive downward for all blocks.
Constraint of cable supporting^: xA + (xA - xB) = constant
2va - vg =
or
0
vs = v^
aB = 2aA
and
2
Constraint o f cable supporting B 2xB + xc = constant
2vg + vc = 0,
or
vc = -2vg,
and
ac - - 2 a B = - 4 a A
Since vc and ac are down, vA and aA are up, i.e. negative.
v 2a -
{
v a )]
= 2 a A [ x A - ( x A )0 ]
= 0-12 ftJs2 f M
ac = - 4 a A
(b) aB = 2aA = (2)(-0.12) = -0.24 ft/s
ac = 0.48 ft/s | -4
2
2
Avfl = aBt = (-0.24)(2) = -0.48 ft/s
Axg = - iBt 2 = - ( - 0 ,2 4 )( 2 ) 2 = -0.48 ft
Avs = 0.48 ft/s ( <
AxB = 0.48 ft ( <
PROBLEM 11.49
Block C starts from rest and moves downward with a constant
acceleration. Knowing that after 12 s the velocity of b lo ck s is 456 mm/s,
determine (a) the accelerations of A, B, and C, (b) the velocity and the
change in position of block B after s.
8
SOLUTION
Let x be position relative to the fixed supports, taken positive if downward.
2xA + 3xB = constant
Constraint of cable on left:
2va
+
3vfl
=
0
,
2
or
vB
=
- ~ v A,
2
and
aB
=
-
- a A
xB + 2xc = constant
Constraint of cable on right:
vB + 2vc = ,
0
vc = - ~1 v B1= - v A,
or
andA
ac = -1 a A
Block C moves downward; hence, block A also moves downward.
(a)
Accelerations.
va
= ( VA
+
or
°aT
aA
= VA
= 4-56{2 ° = 38.0 mm/s
2
aA = 38.0 mm/s | -4
2
= — a . = -I -
(38.0) = -25.3 mm/s
ac = j a A = ^-jj(38.0) = 12.67 mm/s
(b) Velocity and change in position o f B after
vB = (v
f l) 0
8
aB = 25.3 mm/s { ^4
2
2
ac = 12.67 mm/s j
2
2
^ 4
s.
+ aBt = 0 + (-25.3)(8) = -203 mm/s
\ B = 203 mm/s | -4
- ('» )» = M „ '
*- 0 + j(- 2 5 .3 )(mm
8 ) ! = -811
&xB =811 mm | 4
PROBLEM 11.50
W = W W 8 = &
IB
til
Block B moves downward with a constant velocity of 610 mm/s.
Determine (a) the velocity of block A, (b) the velocity of block C, (c) the
velocity of portion D of the cable, (d) the relative velocity of portion D of
the cable with respect to block B.
ffl
SOLUTION
Let x be position relative to the fixed supports, taken positive if downward.
Constraint of cable on left:
2xA + 3xB = constant
2
va
+ 3
vb
= 0,
Constraint of cable on right:
or
vA = - - v B
xB + 2xc = constant
vB
+ vc
2
=
0
or
Constraint of point on D on cable:
vc = — v
+ xp = constant
2 * 4
2 va +
vd
= °
o r
vd
= ~ 2 va
(a) Velocity o f block A.
vA = ~ ~ vb
= -^-(610) = -915 mm/s
v^ = 915 mm/s { A
(b) Velocity of block C.
vc = —- v B = —-(610) = -305 mm/s
(c) Velocity of point D.
vD = —(2 )(—915)
(d) Velocity relative to B.
v ixb = vd ~ v b
vc = 305 mm/s |
A
v D = 1830 mm/s i
A
= 1830 - 610
\ [yB =
1 2 2 0
mm/s { A
PROBLEM 11.51
In the position shown, collar B moves to the left with a constant velocity
of 300 mm/s. Determine (a) the velocity of collar A, (b) the velocity of
portion C of the cable, (c) the relative velocity of portion C of the cable
with respect to collar B.
SOLUTION
Let x be position relative to the right supports, increasing to the left.
2xA + xB + (xB - x A) = constant
Constraint of entire cable:
2 v b + vA =
vA = - 2 v b
0
2xA + xc = constant
Constraint of point C of cable:
2
V;+ vc =
0
vc = - v;
2
(а) Velocity of collar A.
V; = - vs = -(2 )(3 0 0 ) = -600 mm/s
2
vA
= 600 mm/s —►A
( ) Velocity of point C of cable.
б
vc = — v; = -(2 )(-6 0 0 ) =
2
1 2 0 0
mm/s
vc =
1 2 0 0
mm/s -— A
(c) Velocity of point C relative to collar B.
vc,b = v c - v g
- 1200 - 300 = 900 mm/s
va B
= 900 mm/s -— A
PROBLEM 11.52
Collar A starts from rest and moves to the right with a constant
acceleration. Knowing that after s the relative velocity o f collar B with
respect to collar A is 610 mm/s, determine (a) the accelerations of A and
B, (b) the velocity and the change in position o f B after s.
8
6
SOLUTION
Let x be position relative to the right supports, increasing to the left.
2xA + xB + {xB - xA) = constant,
Constraint of entire cable:
2
vg + vA = ,
1
vB = ~ ~ v A,
or
0
and,
aB = -~aA
1
{a) Accelerations o f A and B.
1
Vb /a = V
b
~
va
2
= ~ T vA ~
va
va
=
“ T
VB/A
vA = —“ (610) = -406.67 mm/s
/ X
#
~ (v^)o = aAt,
or
aA =
- (v^)o
-406.67 - 0
= ------= -50.8 mm/s
2
aA = 50.8 mm/s —►A
2
aB = ~ a A = “
(b) Velocity and change in position o f B after
6
(-114.4)
%B = 25.4 mm/s ■*— M
2
s.
Vg = (vg)Q + aBt = 0 + (50.8)(6)
XB ~ {xb )0 = K V + | aBt2 = |( 2 5 .4 ) ( 6 ) i
v g = 152.5 mm/s - — -4
Axg = 458 mm — <
PROBLEM 11.53
At the instant shown, slider block B is moving to the right with a constant
acceleration, and its speed is in./s. Knowing that after slider b lo ck s has
moved 10 in. to the right its velocity is 2.4 in./s, determine (a) the
accelerations of A and B, (b) the acceleration o f portion D of the cable,
(c) the velocity and the change in position of slider block B after 4 s.
6
SOLUTION
Let x be position relative to left anchor. At the right anchor, x = d.
Constraint o f cable:
+ 2 ( d - x A)
Xg +
<N | ro
II
II
©
2 vb -
or
0
=
or
0
and
aA ~ "JaB
ii
v
i
{d~xA
I
ii
VA +
2
and
+
Constraint of point D o f cable:
constant
=
=
constant
aD = ~aA
o
II
On
(a) Accelerations o f A and B.
in./s
M
M o
l
=
2
= |(
aA[xA ~
aB = | aA = |(0 .5 1 2 ) = -0.768 in./s
(ib) Acceleration of point D.
6
) = 4 in./s
( x a )o ]
2
aD = - a A = -(-0 .5 1 2 )
aB = 0.768 in./s — <
2
aD = 0.512 in./s —-*■ A
2
(c) Velocity of block B after 4 s.
Vg = ( vb )q + a Bt =
6
+ (-0.768)(4)
vB = 2.93 in./s — <
2
Change in position of block B.
*b ~ ( * ,) o = M 0 t + \ a / = (6)(4) + 1 (-0 .7 6 8 )(4 )2
Axfi = 17.86 in. — 4
PROBLEM 11.54
j"
M K
i
j
-J
Slider block B moves to the right with a constant velocity of 12 in./s.
Determine (a) the velocity o f slider block A, (b) the velocity o f portion C
of the cable, (c) the velocity o f portion D o f the cable, (d) the relative
velocity of portion C o f the cable with respect to slider block A.
L
SOLUTION
Let x be position relative to left anchor. At right anchor x = d.
xB + (xB - xA) + 2 (d - xA)
Constraint of entire cable:
(a) Velocity of A:
xB + xB - xc
(d) Relative velocity.
2vB - 3vA =
0
yA =
constant
vD
=
d - xA + d - xc
-vA
vOA
=
=
-8.00 in./s
vc - vA = 2 4 - 8
vc
=
constant
8 .0 0
in./s —- 4
2vB - vc = 0
vc = lv B = 2(12)
Constraint of point D of cable:
(c) Velocity of D:
constant
vA = ^ v B = y (l2 )
Constraint of point C of cable:
(b) Velocity of C:
-
vA + vD
-
=
24 in./s
— •-
-4
•* —
4
—
4
0,
vD =
y OA
=
8 .0 0
in./s
16.00 in./s
►
PROBLEM 11.55
M M M kk
Collars A and B start from rest, and collar A moves upward with an
(w
acceleration of 3t2 mm/s2. Knowing that collar B moves downward with
a constant acceleration and that its velocity is
mm/s after moving
800 mm, determine (a) the acceleration of block C, (b) the distance
through which block C will have moved after 3 s.
I
i—L i ■
%
2 0 0
SOLUTION
Let x be position relative to upper support, positive downward.
Let d = value of x at lower support.
{ d - x A) + (xc
Constraint of cable:
4vc - vfi - 2 va =
0
and
) + xc + (xc - xB) = constant
2
4ac - aB - 2 a A=
0
Accelerations:
a
(b) Velocity and position:
vc = (v
c
) 0
+ ^ a c dt = 0 + 6.25/ - 0.5
/ 3
* c - ( * c ) o = £vc <// = 3.125/2 - 0.125r4
At f = 3 s,
^ c - ( ^ c ) o = (3 - 1 2 5 ) (3 ) 2 -(0 .1 2 5 )(3 )4
Axc = 18.00 mm { A
PROBLEM 11.56
Collar A starts from rest at t = 0 and moves downward with a constant
acceleration of 180mm/s2. Collar B moves upward with a constant
acceleration, and its initial velocity is 200 mm/s. Knowing that collar B
moves through 500 mm between t = 0 and t = 2 s, determine (a) the
accelerations of collar B and block C, (b) the time at which the velocity of
block C is zero, (c) the distance through which block C will have moved
at that time.
SOLUTION
Let x be position relative to upper support, positive downward.
Let d = value of x at lower support.
(d - xA) + (xc - xA) + 2xc + (xc - xB) = constant
Constraint of cable:
4vc - vB -
2
va
= 0,
4ac - aB - 2aA = 0
and
(a) Accelerations of B and C.
aA = 180 mm/s2,
VB = { VB ) 0 +
( v /< )0
=
0
* B ~ { * B ) 0 = ( VB ) 0 * + \ a B*2
__ 2 [ » . - f a k - f a ) . ' ] . 2 [-5 0 0 - (-2001(2)] _
t
( )
2
'
2
ac -
+ 2aA) = -“ (-5 0 + (2)(l80))
ac = 77.5 mm/s { ^
2
(b) When vc = 0.
(vc
) 0
= t [ ( vb
) 0
+ 2 (v .)0] = ~(~200 + 0) = -50 mm/s
vc = (v c ) + act
c
vc
c
t = Vc
/0
*c- (*c), -
ac
(vc )„< +
= -16.13 mm
= °
^ 5°^
77.5
/ = 0.645 s**
= (-50)(0.645) + i(77.5)(0.645)2
Axc = 16.13 mm { 4
PROBLEM 11.57
Slider block B moves to the left with a constant velocity of 2 in./s. At
t = 0, slider block A is moving to the right with a constant acceleration
and a velocity of 4 in./s. Knowing that at / = 2 s slider block C has moved
1.5 in. to the right, determine (a) the velocity o f slider block C at t = 0,
(b) the velocity of portion D o f the cable at t = 0, (c) the accelerations of
A and C.
SOLUTION
Let x be position relative to the anchor, positive to the right.
- x B + [xc - xB) + 3(xc - xA) = constant
Constraint of cable:
4vc -
2
vb
vB = - 2 in./s
When / = 0,
<*>
4ac - 2 aB - 3aA - 0
- 3v^ = 0
and
(v
a ) 0
= 4 in./s
(v c), = j [ 2 v i, + 3 ( v , ) 0] = i [ ( 2 ) ( - 2 ) + (3)(4)]
Constraint o f point D:
(v
c
= 2 in./s —
-4
{xD - xA) + {xc - xA) + (xc - xB) - xB = constant
vD
+ vc 2
2va
-
2vb
=0
M o =2( va )0 + 2 vb - 2(vc )0 = (2)(4)+ (2)(-2) - (2)(2)
(b)
) 0
(vD)0 = 0 A
xc-(*c)0 =(vc)ot + J act2
(c)
k - ( U - ( v c ) „ 'l
r i. - ( )( )l
,
ac = V
=-i
V
;J = -1.250 in./s
C
t2
(2)
2
^
2
5
2
2
~ 2aB) = i[(4 )(-1 .2 5 0 ) - (2 )(0 )]
= -1.667 in./s
ac = 1.250 in./s
2
-*— A
aA = 1.667 in./s
2
•— A
2
PROBLEM 11.58
Slider block A starts with an initial velocity at / = 0 and a constant
acceleration of 9 in./s to the right. Slider block C starts from rest at / = 0
and moves to the right with constant acceleration. Knowing that at t = 2 s,
the velocities of A and B are 14 in./s to the right and 1 in./s to the left,
respectively, determine (a) the accelerations of B and C, (b) the initial
velocities of A and B, (c) the initial velocity of portion E of the cable.
2
SOLUTION
Let x be position relative to the anchor, positive to the right.
Constraint of cable:
- x B + (xc - x B) + 3(xc - x A) = constant
4vc -
2
vb
-
3
va
= 0
and
4ac - 2aB - 3aA = 0
(a) Accelerations of B and C.
At t = 2 s,
vA = 14 in./s
and
vB = -1 in./s
vc = ^ (2 v s + 3V,) = ~ [ ( 2 ) ( - l ) + (3)(14)] = 10 in./s
(vc)0 = 0
vc -S-(=vc)0
10-0
——
vc )o + ac t ac =
«« -
»C =
5 m./s" —
<
} [ ( 4 ) ( 5 ) - (3 )(9 )] = -3 .5 in./s2
afl = 3.5 in./s2 -— A
(b) Initial velocities o f A and B.
va
= ( va )o “ aAt
(va)o = VA~ aAt = 14 - (9 )(2 ) = - 4 in./s
= -4 in ./s —
<
( y s )0 = 6 in./s —
A
( VA
vB = ( vs ) 0 - V
(vs)o = VB —Ogt = - l - ( - 3 . 5 ) ( 2 )
Constraint o f point E :
2(xc - x A) + ( x E - x A) = constant
v£ -
(c>
i.vE)0 = 3( v^)o -
3v
a
+ 2vc = 0
2(vc)0 = I3) ! " 4 ) -
( 2 )(0 ) = -1 2 in./s
(v £ )0 = 12 in./s — A
PROBLEM 11.59
The system shown starts from rest, and the length of the upper cord is
adjusted so that A, B, and C are initially at the same level. Each
component moves with a constant acceleration, and after s the relative
change in position of block C with respect to block ^4 is 140 mm upward.
Knowing that when the relative velocity of collar B with respect to block
A is 40 mm/s downward, the displacements o f A and B are 80 mm
downward and 160 mm downward, respectively, determine (a) the
1
accelerations of A and B if aB > 5 mm/s2, (b) the change in position of
block D when the velocity of block C is 300 mm/s upward.
SOLUTION
Let x be position relative to the support taken positive if downward.
Constraint of cable connecting blocks A, B, and C:
2xA + xB + xc = constant,
2vA + vfl + vc = ,
2
( VA
2
= ( v*)o = ( vc)o = ° ’
( vsm )
-
( v b i a )0 =
( XA
2aP/A^ * b i a
=
0
~ ( x b /a
)0]
2 (xB - x A)
aA + 2aB + ac =
( x bia )
) 0
VB/A
o= ,
0
(vBM)o =
0
0
“ 0 = 2aB/A( x B - xA - 0)
402
=
2(160 - 8 0 )
VB/A
a B/A ~
= (* c
( x b )o
2
1 0
mm/s
2
XB/A = (XB/a )0 + ( VB/a )0‘ + - aB l f = 0 + 0 + - a B, f
t2 = K bU
I2 ( x b
, =
or
xa
~ {
xa
x
a)
12(160 - 8 0 )
_
V
° B !A
V
B/A
-
)o = M
4s
10
o 1 + 2 ° ^
2 [ * 4 - f o ) o - ( vA ' ] - 2(8° - ° )
(a)
=
aB = aA + aB/A =
1 0
+
ac = - ( aB + 2aA) = - [ ( )(
2
2
t =
vc = ( vc)o + M
aA
=
1
0
mm/s | 4
2
(4)2
aB =
1 0
2 0
2 0
mm/s { 4
2
) + ( ) ( l )] = -6 0 mm/s
2
0
'c ~ ( vc)o _ "300 - 0
= 5s
-60
Constraint of cable supporting block D:
( XD ~ x a ) + { x d ~
2aD - aA - aB = 0,
(*)
xd -
M o
=
x b
) =
constant,
2vD -
vA
- vs = 0
aD = —(u^ + aB) = —(10 + 20) = 15 mm/s
M o ‘ + \ aDt2 = 0 + ^ (1 5 )(5 )2
AxD = 187.5 mm } 4
PROBLEM 11.60
The system shown starts from rest, and each component moves with a
constant acceleration. If the relative acceleration of block C with respect
k
n
m
f
to collar B is 120 mm/s upward and the relative acceleration of block D
n
2
with respect to block A is 220 mm/s downward, determine (a) the
velocity of block C after s, (b) the change in position of block D after
s.
2
r
i
6
1 0
SOLUTION
Let x be position relative to the support taken positive if downward.
Constraint of cable connecting blocks A, B, and C:
2xa
+ 2 x B + xc = constant
2aA
+ 2 v B + vc =
2vA
+ 2aB + ac =
0
( )
0
1
Constraint of cable supporting block D:
(xd
- * * ) + (*/>- xb) = constant,
a c/B
=
ac
- aB = -120
Given:
aaA
=
aD - aA
or
= 220
v - vA - vB =
0
0
=0
2aD — aB — aA
Given:
2
( )
2
- 120
ac - a B
or
aD
=
aA
(3)
+ 220
(4)
Substituting (3) and (4) into (1) and (2),
2
aA
+ 2 a B + (a B -
2 [ a A + 220) -
1 2 0
aA
-
)=
aB
or
0
=0
2
or
aA
+ 3a B =
aA
-
aB
(5)
1 2 0
= -440
( )
6
Solving (5) and ( ) simultaneously,
6
aA = -240 mm/s
From (3) and (4),
ac
(a) Velocity of C after
6
= 80 mm/s
aB = 200 mm/s
and
2
and
2
aD
2
= -2 0 mm/s
2
s.
vc = 480 mm/s { -4
vc = ( vc ) o + act = 0 + (80)(6)
(b) Change in position of D after 10 s.
xd
- M o =M o* +
=
0
+ ~ ( — )(
2 0
1 0 ) 2
=
- 1 0 0 0
mm
AxD = 1.000 m | 4
PROBLEM 11.61
A particle moves in a straight line with a constant acceleration of - 4 ft/s
for s, zero acceleration for the next 4 s, and a constant acceleration of
2
j
6
10
u
6
««) +4 ft/s" for the next 4 s. Knowing that the particle starts from the origin
and that its velocity is
ft/s during the zero acceleration time interval,
(a) construct the v - 1 and x - t curves for < t <, 14 s, (b) determine
the position and the velocity o f the particle and the total distance traveled
when t = 14 s.
- 8
0
PROBLEM 11.62
A particle moves in a straight line with a constant acceleration of - 4 ft/s
J
for s, zero acceleration for the next 4 s, and a constant acceleration of
...
-y
Ki) +4 ft/s"
for the next 4 s. Knowing that the particle starts from the origin
2
6
—
10
1
M
with v = 16 ft/s, (a) construct the v - f and x - t curves for
0
0
< t <
14 s, (b) determine the amount of time during which the particle is further
than 16 ft from the origin.
PROBLEM 11.62 CONTINUED
Integrating, using limits x = 0 when t = 0 and x = 16 ft when
t = ti
[x
] ’6
= [l r 6
or
/ , 2
2
r2]'o
or
- /, +
8
8
=
16 = 16/, -
2 /,2
0
Solving the quadratic equation,
± J(
- (4)(1)(8)
t, = ---------- [ w \ /V - ’ = 4 ±2.828 = 1.172 s
8
8 ) 2
and
6.828 s
(2)0 )
The larger root is out o f range, thus /, = 1.172 s
Over
6
< t < 10,
Setting x = 16,
x = 24 16 = 72 -
Required time interval:
8 / 2
8
(/ -
or
6
) = 72 -
8
/
t2 = 7 s
( / 2
- /i) = 5.83 s A
PROBLEM 11.63
*
A particle moves in a straight line with the velocity shown in the figure.
Knowing that x = -540 m at t = 0, (a) construct the a - t and x - t
curves for 0 < t < 50 s, and determine (b) the total distance traveled by
the particle when t = 50 s, (c) the two times at which x = 0.
,i „
SOLUTION
(a) Construction of the curves.
a = slope of v - t curve
Construct the a - t curve.
s:
A/ = 10 s,
Av = 0
Av = n
a =—
At
s < t < 26 s:
At = 16 s,
Av = -80 m/s
Av
a = — = -5 m/s
At
26 s < t < 41 s:
At = 15 s,
Av = 0
Av
a =— =
At
41 s < / < 46 s:
At = 5 s,
Av = 15 m/s
a = — = 3 m/s
At
46 s < t < 50 s:
At = 4 s,
Av = 0
Av
a =— =
At
0
1 0
<f <
1 0
0
2
0
2
0
CL (y* / S* ^
Construct the x - t curve.
Ax = area of v - 1 curve.
x is maximum or minimum where v = .
0
For 10 s ^ f < 26 s,
v=
0
v = 60 - 5(r - 10)
60 - 5t + 50 = 0
when
or
t = 22 s
x0 = -540 m
Also
0 to 10 s
Ax = (10)(60) = 600 m
*io = -540 + 600 = 60 m
10 s to 22 s
Ax = -(1 2 )(6 0 ) = 360 m
x22 = 60 + 360 = 420 m
22 s to 26 s
Ax = ■^■(4)(-20) = -40 m
x
26
= 420 - 40 = 380 m
PROBLEM 11.63 CONTINUED
26 s to 41 s
Ax = (l5 )(-2 0 ) = -300 m
= -62.5 m
41 s to 46 s
46 s to 50 s
x41 = 380 - 300 = 80 m
Ax = (4 )(-5 ) = -2 0 m
x
= 80 - 62.5 = 17.5 m
x
= 17.5 - 20 = -2.5 m
4 6
5 0
£ (s i
(b) Total distance traveled.
Total:
0 < t < 22 s,
dx = |x
22 s < t < 50 s,
d2 = |x
22
50
- x0| = |420 - ( —540)| = 960 m
- x22| = |—2.5 - 420| = 422.5
d = dx + d 2 = 1382.5 m
d = 1383 m <
(c) Times when x = 0
For 0 < t < 10 s,
A tx = 0,
For 46 s S t < 50,
A tx = 0,
x = -540 + 60f m
-5 4 0 + 60r = 0
x = 17.5 - 5(/ - 46) m
17.5 - 5(/ - 46) = 0
/ - 46 = 3.5
t = 49.5 s <
PROBLEM 11.64
A particle moves in a straight line with the velocity shown in the figure.
Knowing that x = -540 m at t = 0, (a) construct the a - t and x - t
curves for 0 < I < 50 s, and determine (b) the maximum value of the
position coordinate of the particle, (c) the values of t for which the
particle is at x =
m.
1 0 0
SOLUTION
(a) Construction of the curves.
Construct the a - t curve.
a = slope of v - t curve
s:
At = 10 s,
Av = 0
Av
a =— =
At
s < / < 26 s:
At = 16 s,
Av = -80 m/s
Av
i
a = — = -5 m/s
At
26 s < t < 41 s:
At = 15 s,
Av = 0
Av
a =— =
At
41 s < t < 46 s:
At = 5 s,
Av = 15 m/s
Av
a = — = 3 m/s
At
46 s < t < 50 s:
At = 4 s,
Av = 0
Av
a =— =
At
0
1 0
<t <
1 0
0
0
2
0
t(s^ i
Construct the x - t curve.
Ax = area of v - / curve.
x is maximum or minimum where v = .
0
For 10 s £ t <, 26 s,
v=
0
v = 60 - 5(r - 10)
60 - 5t + 50 = 0
when
Also
or
t = 22 s
x = -540 m
0
0 to 10 s
Ax = (10)(60) = 600 m
x,o = -540 + 600 = 60 m
10 s to 22 s
Ax = ^(12)(60) = 360 m
x
22 s to 26 s
Ax = -^-(4)(-20) = -40 m
x
22
2 6
= 60 + 360 = 420 m
= 420 - 40 = 380 m
PROBLEM 11.64 CONTINUED
26 s to 41 s
Ax = (l5 )(-2 0 ) = -300 m
x41 = 380 - 300 = 80 m
41 s to 46 s
.
.( -5 ^
Ax = (5) -----------
x
= 80 - 62.5 = 17.5 m
x
= 17.5 - 20 = -2.5 m
2
v
0
2
=-62.5 m
4 6
J
i
Ax = (4 )(-5 ) = -2 0 m
46 s to 50 s
50
X (W 'l
(b) Maximum value of position coordinate,
v=
0
when t =
2 2
*max = *22 = 420 m 4
s
(c) Values of t when x = 100 m.
For 10 s < f < 26 s,
x = 60 + 60(/ -
A tx = 100 m,
) 2
1 0 ) 2
= 100
2.5 r - 60r + 40 = 0
6 0 t ^ ( 6 0 ) i -(4 )(2 .5 )(4 0 )
(2)(2.5)
1 = 10.69 s,
A tx = 100 m,
) - j(5 )(t -
60 + 60 (r - 10) - 2.5 (t - 10
L e tr = t - 10,
For 26 s <, t <, 41 s,
1 0
and
.695
and
33.3 s (out of range)
23.3 s
t = 10.69 s
A
x - 380 + (-2 0 )(f - 26)
100 = 380 - 20(t - 26)
t - 26 = 14
t = 40.0 s A
PROBLEM 11.65
A parachutist is in free fall at a rate of 180 ft/s when he opens his
parachute at an altitude of 1900 ft. Following a rapid and constant
deceleration, he then descends at a constant rate of 44 ft/s from 1800 ft to
ft, where he maneuvers the parachute into the wind to further slow
his descent. Knowing that the parachutist lands with a negligible
downward velocity, determine (a) the time required for the parachutist to
land after opening his parachute, (b) the initial deceleration.
1 0 0
vf
SOLUTION
Let x be the altitude. Then v is negative for decent and a is positive for
deceleration.
Sketch the v - / and x - t curves using times r,, t2 and
as shown.
Use constant slopes in the v - / curve for the constant acceleration stages.
/ 3
Areas of v - / curve:
A, = - - ( 1 8 0 + 144)/, = -112/, ft
O
A2 = -44/,
4 =
- | ID
I
y (ft'i i
HO"
1
Changes in position:
2
= -22/3
Ax, = 1 800- 1900 = -100 ft
Ax2 = 1 00-1800 = -1700 ft
Ax3 = 0 - 100 = -1 0 0 ft
Using Ax, = Aj gives
t, =
/ 7
=
-112
-1700
= 38.64 s
-44
-100
h =
-22
(a) Total time:
(b) Initial acceleration.
= 0.893 s
= 4.55 s
/, +
a =
/ 2
+
/ 3
= 44.1 s M
Av _ (-4 4 ) - ( - 1 8 0 )
At
0.893
a = 152.3 ft/s A
2
PROBLEM 11.66
A machine component is spray-painted while it is mounted on a pallet
that travels 12 ft in 20 s. The pallet has an initial velocity o f 3 in./s and
can be accelerated at a maximum rate of 2in./sx . Knowing that the
painting process requires 15 s to complete and is performed as the pallet
moves with a constant speed, determine the smallest possible value of the
maximum speed of the pallet.
SOLUTION
Sketch v - 1 curve as shown. Label areas A{, A2, and A3
A,
o
it ,
A, = (3) (20) - 60 in.
Av = at, = 2t, in./s
A3 = ! Av)(20 ~ tj) = 2r,(20 - tj) in.
Distance traveled:
Ax = 12 ft = 144 in.
Ax = total area, 144 = 60 + 1? + 2r, (20 - f,)
or
t} - 40r, + 84 = 0
and
Reject the larger root.
37.8 s
t, = 2.224 s
Av = 2r, = 4.45 in./s
vmax = 3 + Av = 3 + 4.45
W
= 7.45 in./s <
PROBLEM 11.67
In a 400-meter race, runner A reaches her maximum velocity vA in 4 s
with constant acceleration and maintains that velocity until she reaches
the half-way point with a split time of 25 s. Runner B reaches her
maximum velocity vB in 5 s with constant acceleration and maintains
that velocity until she reaches the half-way point with a split time of
25.2 s. Both runners then run the second half of the race with the same
constant deceleration of 0.1 m/s2. Determine (a) the race times for both
runners, (b) the position of the winner relative to the loser when the
winner reaches the finish line.
SOLUTION
Sketch v - t curves forfirst 200 m.
rr(tv./s'i
/0„
Runner/I:
/, = 4 s, t2 = 25 - 4 = 21 s
A,
> 1
1, L-
Us
ti
4
=
=
2 ( VJ n u x
1
A 2 =
tS (mVs )
21K )m a x
A, + A2 = Ax = 200 m
2 3(v.)
= 200
V A /m a x
Runner B:
or
(v
.)
= 8.6957 m/s
' A /m ax
t2 = 25.2 - 5 = 20.2 s
/, = 5 s,
^ = j ( 5K l,« L , = 2 -5(v» L .
^
A, + A2 = Ax = 200 m
22.7(Vo)
=
V a /m a x
Sketch v - t curve for second 200 m.
Ai = v
m ax / 3
2 0 0
Av = |a
- -jAv/j = 200
or
|/ 3
(v»)
=8.8106 m/s
V ° /m ax
= 0. l
/ 3
or
0.05/32 - vmaxr + 200 = 0
3
Vmax±>/(Vmax)2 -(4)(0-05)(200)
10
(2)(0.05)
Runner/4: (vmax)^ = 8.6957,
(t3)A = 146.64 s
Reject the larger root. Then total time
and
V max —
+
\ livmax)2 “ 40
27.279 s
rA = 25 + 27.279 = 52.279 s
t A = 52.2 s <
PROBLEM 11.67 CONTINUED
Runner B: (vmax)g = 8.8106, (t3)fl = 149.45 s
Reject the larger root. Then total time
and
26.765 s
tB - 25.2 + 26.765 = 51.965 s
tB = 52.0 s 4
Velocity o f A a t t = 51.965 s:
^ = 8.6957 - (0.1)(51.965 - 25) = 5.999 m/s
Velocity o f A at t = 51.279 s:
v2 = 8.6957 - (0.1)(52.279 - 25) = 5.968 m/s
Over 51.965 s < t < 52.965 s, runner 4 covers a distance Ax
Ax = vave(Ai) = ^(5.999 + 5.968)(52.279 - 51.965)
Ax = 1.879 m 4
PROBLEM 11.68
A commuter train traveling at 64 km/h is 4.8 km from a station. The train
then decelerates so that its speed is 32 km/h when it is 800 m from the
station. Knowing that the train arrives at the station 7.5 min after
beginning to decelerate and assuming constant decelerations, determine
(a) the time required to travel the first 4 km, (b) the speed of the train as it
arrives at the station, (c) the final constant deceleration of the train.
SOLUTION
Sketch the v - t curve
fiX
Data:
x2 = 4.8 km = 4.8 x 10 m
Vj = 32 km/hr = 8.889 m/s
x, - 4.8 x 103 - 800 = 4.0 x 103 m
t2 = 450 s
(a) Time /, to travel first 4 km.
fj = 300 s A
X! .= 4.0 x 103 = 4 = i ( v 0 + v,) r, = | (17.778 + 8.889)^
(b) Velocity v2.
X2 ~
X,
= 800 = A2 = '
(v ,
v2
+ v )(f - /,) = ~(vj + v2)(450 - 300)
2
2
+ vi “ 10.667 m
v2 = 10.667 - 8.889
v, = 1.778 m/s
A
(c) Final deceleration.
_ v2 -‘12
t2 -
Vi
_ 1.778 - 8.889 _
= -0.0474 m/s2
450 - 300
au I = 0.0474 m/s2 A
PROBLEM 11.69
Two road rally checkpoints A and B are located on the same highway and
are mi apart. The speed limits for the first 5 mi and the last 3 mi are
60 mi/h and 35 mi/h, respectively. Drivers must stop at each checkpoint,
and the specified time between points A and B is 10 min 20 s. Knowing
that a driver accelerates and decelerates at the same constant rate,
determine the magnitude of her acceleration if she travels at the speed
limit as much as possible.
8
SOLUTION
10 min 20 s = — +
= 0.1722 h
60 3600
Sketch the v - t curve
AT
60
t= —
i/h i
tu —
25
a
35
t= —
a
Al = 60tx - -j(60)(ia ) - |( 2 5 ) t i = 60 tx - 1 8 0 0 - - 312.5—
A, - 5 mi
But
1
60^ - 2112.5— = 5
a
( 1)
A2 = 35(0.1722 - /,) - 35tc = 6.0278 - 351, - 612.5—
A2 -
But
8
- 5 = 3 mi
(2)
35tx + 6 1 2 .5 - = 3.0278
a
tx = 85.45 x 10
Solving equations (1) and (2) for tx and —,
a
- = 60.23 x l 0
a
2
a = 16.616 x
1 0
“ 6
(16.616
x
3
h = 5.13 min
h2/m i
103)(5280)
a = 6.77 ft/s 4
mi/h =
2
(3600)2
PROBLEM 11.70
In a water-tank test involving the launching of a small model boat, the
model’s initial horizontal velocity is 20 ft/s and its horizontal acceleration
varies linearly from -4 0 ft/s
at / = 0 to - 6 ft/s* at / = /, and then
remains equal to - 6 ft/s2 until t = 1.4 s. Knowing that v = 6 ft/s when
/ = /,, determine (a) the value o f % (b) the velocity and position of the
model at t = 1.4 s.
SOLUTION
Sketch the a - / curve as shown
a ( f t / s 1)
tc*>
A = ~6h
4
= -1 ( 4 0 - 6 )/,= -1 7 /,
V, = v0 +
A + A2
—6 = 20 —6/, - 17/,
/, = 0.6087 s
(«)
/, = 0.609 s ^
h = 1-4 s
t2 —/, = 0.7913 s
4 + A = "(6)(1.4) = -8.4 ft/s
A2 = -(17)(0.6087) = -0.348 ft/s
V = Vq + Aj +
+ A2 — 20 ~ 8.4 —10.348
2
(ft)
x2 = x0 + V0/ 2 +
= 0 + v0/2 +
( 4
( 4
+ 4 ) % + 4*2
+
4
v, = 1.252 ft/s 4
by moment-area method
)
= 0 + (20)(1.4) - (8.4 - 1 (1.4) - (10.348)^1.4 -
x, = 9.73 ft 4
PROBLEM 11.71
A bus is parked along the side of a highway when it is passed by a truck
traveling at a constant speed o f 70 km/h. Two minutes later, the bus starts
and accelerates until it reaches a speed of 100 km/h, which it then
maintains. Knowing that 12 min after the truck passed the bus, the bus is
1.2 km ahead of the truck, determine (a) when and where the bus passed
the truck, (b) the uniform acceleration of the bus.
SOLUTION
Sketch the v - 1 curves.
at f
21.1
/s'}
bos
1
V=l.*+M
/
//
/
1 2 .0
/
\
\
1
1
1
1
1
\
t,
i
\
tx
'
I
!i
\
72o
At t = 12 min = 720 s,
W
=(19.44)(720) = 14000 m
xbus - 14000 + 1200 = 15200 m
cbus ~
under v - t curve
- ( / , - 120)(27.78) + (720 - ^ (2 7 .7 8 ) = 15200
tx = 225.8 s
(u) When xbus —-^truck* areas under the v - 1 curves are equal.
^(27.78)(f, - 120) + 27.78(f2 - f,) = 19.44t2
With f, = 225.8 s,
t2 = 576 s <
■^truck
(b)
(19,44)(576) a 11200 m
v -v
27.78 - 0
f, - 120 ~ 225.8 - 120
0
° bus
W
= H-20 km <
abus = 0.262 m/s2 M
PROBLEM 11.72
Cars A and B are d = 60 m apart and are traveling respectively at the
S
S
h
constant speeds o f (®^)0 = 32 km/h and (vB)0 = 24 km/h on an icecovered road. Knowing that 45 s after driver A applies his brakes to avoid
overtaking car B the two cars collide, determine (a) the uniform
deceleration o f car A, (b) the relative velocity o f car A with respect to car
B when they collide.
SOLUTION
(vA)0 = 32 km/h = 8.889 m/s
vB
= 24 km/h = 6.667 m/s
Sketch the v - t curves.
Ax = (6.667)(45) = 300 m
/tf'On/s')
_
4 = i( 2 .2 2 2 ) (4 5 ) + i v ^ ( 4 5 )
= 50 -(- 22i.5vA/B
c. an
XA = f o ) „
+ A + A 2
XB = { X b ) 0 +
X B/A = { X B!a \
~ 4
0 = 60 - 50 - 22.5<vm
(b)
va
=
vb
V, - ( v ^ ) 0
(«)
4
aa =
+ VA!B ~
6.667
v a/ b
=
7.111 - 8.889
45
= 0-444 m/s
4
7.111 m/s
a a = -0.0395 m/s2 4
PROBLEM 11.73
Cars A and B are traveling respectively at the constant speeds of
( v A)0 = 22 mi/h and (v B)0 = 13 mi/h on an ice-covered road. To avoid
overtaking car B, the driver of car A applies his brakes so that his car
decelerates at a constant rate of 0.14 ft/s2. Determine the distance d
between the cars at which the driver of car A must apply his brakes to just
avoid colliding with car B.
SOLUTION
(v^ )Q = 22 mi/h = 32.267 ft/s
(v
B ) 0
= 13 mi/h = 19.067 ft/s
Sketch the v - t curves.
Slope of v - t curve for car A.
rr (-R /s ;
7 ✓>/ n
a =
= -0.14 ft/s
2
h
A
A2 = ^(13.2)(94.29) = 622.3 m
0
0
t,
XB = {
XA
xb
)0 +
4
= ( Xa \ + A +A2
XB/A = X B ~ X A = { X b ) 0 - (
xa
d - A2
) 0 - A 2>
°r
0 = d - A2
d = 622 m ^
PROBLEM 11.74
An elevator starts from rest and moves upward, accelerating at a rate of
4 ft/s until it reaches a speed of 24 ft/s, which it then maintains. Two
seconds after the elevator begins to move, a man standing 40 ft above the
initial position of the top of the elevator throws a ball upward with an
initial velocity of 64 ft/s. Determine when the ball will hit the elevator.
2
40 ft
SOLUTION
Construct the a - t curves for the elevator and the ball.
a(ft/ s ’ )
Limit on A, is 24 ft/s. Using A, = 41
c)t
t(i '
4t2 = 24
t2 =
s
6
Motion of elevator.
For 0 £
<
6
s,
(*fc ) o
=
(v£
0
=
) 0
0
Moment of A, about t = t,:
xe = (*£ )0 + W o ' i + 2 t \ = 2 r \
f*—t1-2 -»i
Motion of ball. At t = 2,
(* b
For /( > 2 s,
A2 = -32.2(/| - 2)ft/s
Moment of A2 about t = t2:
- 3 2 . 2 - 2)
)0
= 40 ft
*» = (* « )o +
( vb
't\ -
2
) 0
^
= —16. l(r, -
2
):
2f
- 2 ) - 1 6 1 ( 'i -
= 40 + 64(/, - 2) - 16.1(7, - 2
= 04 ft/s
) 2
xB = xE
When ball hits elevator,
40 + 6 4 ( r , - 2 ) - 1 6 . l ( / , - 2 ) 2 = 2/f
or
18.1/,2 - 128.4r, + 152.4 = 0
Solving the quadratic equation,
r, = 1.507 s
and
The smaller root is out of range, hence
Since this is less than
6
s, the solution is within range.
5.59 s
/. = 5.59 s <
PROBLEM 11.75
Car A is traveling at 70 km/h when it enters a 50 km/h speed zone. The
driver of car A decelerates at a rate of 5 m/s2 until reaching a speed of
50 km/h, which she then maintains. When car B, which was initially 19 m
behind car A and traveling with a constant speed of 72 km/h, enters the
speed zone, its driver decelerates at a rate of 6 m/s2 until reaching a
speed of 48 km/h. Knowing that the driver of car B maintains a speed of
48 km/h, determine (a) the closest that car B comes to car A, (b) the time
at which car A is 22 m in front of car B.
SOLUTION
At / = 0,
( ^ ) 0 = 19 m,
Final speeds,
( va ) q
= 70 km/h = 19.444 m/s
(vA) , = 50 km/h - 13.889 m/s,
xb
= ( x a\
= 19 m
{tB\ =
(XB)0
and
= 0,
(vs )Q = 72 km/h - 20.000 m/s
(vg )y = 48 km/h t= 13.333 m/s
ivA)f - ( VA
Deceleration of car A: a , = -5 m/s
At time (tB\ ,
( * s ) 0
_ 13.889 - 19.444
= 1.1111s
xB = {xB\ + (vB\ ( t B),
= 1 2 -1 ° = 0.9500 s
20
Deceleration of car B: aR = - 6 m/s
r. \
\
K ) / - K ) 0 13.333-20.000
l1t1„
{ h ) 2 - { h \ = — ^ ----------=
7
= 1.111 s
aB
-o
(tB)2 = 0.9500 + 1.111 = 2.0611s
Sketch the v - t curves for the two cars.
\ x ai b ) .
V
/m in
(a)
For
occurs when
vA = vB
(tB\ < t < { t B)v
vb
= ( v g )<, + % [ / - ( ? * ) , ]
VA = { va )j -
Equating, setting t = tm, and solving for tm,
{vA) f - { v B)0
UB
= 0.950 +
= 1.9685 s
13.889 - 20
PROBLEM 11.75 CONTINUED
At time t = tm,
(x A)
= (x/t )0
,
\
- ( x a)o +
=
1 9
+ area under v - t curve for car A.
(v4)0 + (vi ) / ,
r w ,
2
A +^ A m ~
+ 19.444 + 13.88 9 ^
1H ^ +
^
x
3
3 3 9
)^
9 6 8 5
_ i.n n ) =
4 9 . 4 2 6
m
( x* L = (xB)0 + area under v - t curve for car B.
-
(*b)o + ( v« ) o ( ^ ) i +
-
9 0
= 0 + (20)(0.950) + —
+
1 ^ OOQ
,
(1.9685 - 0.950) = 36.258 m
( x ai b )
V
At time t = (tB)1,
(xa
)2 = (x^
,
.
('* )J
/m in
~
xa ~
xb
~ 49.426 - 36.258 - 13.17 m 4
)o + 3163 un<ler v - 1
curve for car A.
(v^)0 + (v^)/
r
=(^)o +---- 2---
A+
~*A
= 19 + 19-444 * 13-889(1.1 111) + (13.889)(2.0611 - 0.950) = 50.712m
( xb )2 = (xB)Q + area under v - t curve for car B.
- ( * , ) „ + (vs )0( » j) i + (v‘ x \ (VA)f[ { h \ - (<«),]
?0
= 0 + (20)(0.950) + —
(*4 ) 2
=
-(2.0611 - 0.950) = 37.517 m
" ( xs ) 2 = 50.712 - 37.517 = 13.195 m < 22 m
For xA/B = 22 m,
XA
4- 1 3
1 > (?s )2
(*
<)2 + {vA ) f [ t - { h ) 2]
and
xB = ( x s )2 + (v g )/ [ t - ( t s )2]
Subtracting,
X A ~ X B ~ X A1B = [ { x a ) 2 ~ ( % ) 2] + [ ( V ^
Solving for t,
= (> )
- (v fl) / ] [ f - { t B ) 2 ]
, 2.0611 +
2
M
/ - ( VJ»)/
2 2 - 13' 195
13.889- 13.333
i = 17.91 s 4
PROBLEM 11.76
Car A is traveling on a highway at a constant speed ( v ^ = 100 km/h
and is 120 m from the entrance of an access ramp when car B enters the
acceleration lane at that point at a speed (vB)0 = 25 km/h. Car B
accelerates uniformly and enters the main traffic lane after traveling
70 m in 5 s. It then continues to accelerate at the same rate until it reaches
a speed of 100 km/h, which it then maintains. Determine the final
distance between the two cars.
SOLUTION
(
)Q = 100 km/h = 27.778 m/s
( v«)0 = 25 km/h = 6.944 m/s
Sketch acceleration curve for car B over 0 < t < 5 s.
Using moment-area formula at t = 5 s.
* b -(* « )„ = (v° ) ' + (a fl)(5)( 2-5)
70 = (6.944)(5) + 12.5aB
aB = 2.822 m/s2
Determine when B reaches 100 km/h.
a*
K ) , = K ) 0 + A2
tc o
27.778 = 6.944 + 2.822tB
2.1TI
tB = 7.38 s
A2 = (2.822)(7.38) = 20.83 m/s
Then,
xB = (xB)Q + ( vB)QtB + A2^ -
* a = { x A )0 + ( v A )0 t B
an d
Subtracting,
Then,
by moment-area formula
xB - x A = (xB)0 - ( x ,) Q + [(vB)0 - (v^)0]rB + A2
f 7 38'
x B - x A = 120 + (6.944 - 27.778)(7.38) + (20.83)1 '
xB!A = 43.1 m
Car B is ahead of car A. A
PROBLEM 11.77
During a manufacturing process, a conveyor belt starts from rest and
travels a total of 0.36 m before temporarily coming to rest. Knowing that
the jerk, or rate of change of acceleration, is limited to ±1.5 m/s2 per
second, determine (a) the shortest time required for the belt to move
0.36 m, (b) the maximum and average values of the velocity of the belt
during that time.
SOLUTION
Sketch acceleration curve.
da
dt
Let
j = jerk
Then,
amta. ~ J ( ^ )
A = |«m ax(2A t) = amax(Ar)
- /(A t)2
V /
=
V0 +
0 —0 +
Ax - A2
—A.2
A2 = Ax
Ax = v0(4At) + (/4j)(3Ar) - A2(At)
= 0 + 3j(A t f - j ( A t ) = 2y (At)3
Ax
0.36
At = 3 — = 3
= 0.4932
V
V(2)(L5)
(a) Shortest time:
(b) Maximum velocity:
4At = (4)(0.4932) = 1.973 s A
vmax = V0 + A = 0 + j ( A t f
= (1.5)(0.4932)2 = 0.365 m/s A
Average velocity:
Ax
4 At
0.36
= 0.1825 m/s A
1.973
PROBLEM 11.78
An airport shuttle train travels between two terminals that are 5 km apart.
To maintain passenger comfort, the acceleration of the train is limited to
±1.25 m/s2, and the jerk, or rate o f change of acceleration, is limited to
±0.25 m/s2per second. If the shuttle has a maximum speed of 32 km/h,
determine (a) the shortest time for the shuttle to travel between the two
terminals, (b) the corresponding average velocity o f the shuttle.
SOLUTION
Sketch the a - t curve.
4 = i(5 )(1 .2 5 ) = 3.125 m/s
vmax = 52 km/hr = 8.889 m/s = 2Ax + A2
4 = vmax - 2 4 - 8.889 - (2)(3.125) = 2.639 m/s
A2
2.639
AU = - 2 - = -------- = 2.111 s
*ma*
1-25
Total distance is 5 km = 5000 m. Use moment-area formula.
/
x = x0+ v0r + (24 +
4)
1
\
'l - A f l - T A ?2
V
**
—
J
+^ )
"I---
2
V
2
= 0 + 0 + vmax( t - 2 A t 1 - At2)
(a) t = 2Atl + At2 + —
v
= (2)(5) + 2.111 +
= 10 + 2.111 + 562.5 = 575 s
8.889
t = 9.58 min A
...
s 5000
„
(b) vavs = - -= ------ 8.70 m/s
ave
t 575
= 31.3 km/h 4
PROBLEM 11.79
An elevator starts from rest and rises 125 ft to its maximum velocity in
T seconds with the acceleration record shown in the figure. Determine
(a) the required time T, (b) the maximum velocity, (c) the velocity and
position of the elevator at / = 772.
r/lft/s2)
2
/
0/
/
1
|
1
773
r t(s)
PROBLEM 11.80
An accelerometer record for the motion of a given part of a mechanism is
approximated by an arc of a parabola for 0.2 s and a straight line for the
next 0.2 s as shown in the figure. Knowing that v = 0 when t = 0 and
x = 0.8 ft when t = 0.4 s, (a) construct the v - I curve for 0 £ t <, 0.4 s,
(b) determine the position of the part at t = 0.3 s and t = 0.2 s.
SOLUTION
Divide the area of the a - t curve into the four areas A,, A2, A3 and A4.
A, = |( 8 ) ( 0 .2 ) = 1.0667 ft/s
A2 = (16)(0.2) = 3.2 ft/s
A3 = i ( l 6 + 8)(0.l) = 1.2 ft/s
A4 = |(8 )(0 .1 ) = 0.4 ft/s
Velocities: v0 = 0
O-i
v0.2 = v0 + A + A 2
0.2
= 4.27 ft/s A
v 0.3 ” v0.2 + A 3
0.3
= 5.47 ft/s <
OM
V0.4 = V0.3 + A 4
V0.4 = 5 -8 7 f t / s *
Sketch the v - t curve and divide its area into As, A^, and A7 as shown.
£ 8dx = 0.8 - x = f Avdt
A t / = 0.3 s,
or
x = 0.8 - ^* v d t
x03 = 0 .8 - As -(5.47)(0.1)
With /<5 = y (0 .4 )(0 .l) = 0.0267 ft,
At / = 0.2 s,
x0 3 = 0.227 ft <
■*0.2 —ft-8 —(-^5 + Aft) ~ Ai
With Af + Af, = —(l.6)(0.2) = 0.2133 ft,
and
A7 = (4.27)(0.2) = 0.8533 ft
Xq2 = 0.8 - 0.2133 - 0.8533
x0 2 = -0.267 ft <
PROBLEM 11.81
nfcV)
Two seconds are required to bring the piston rod of an air cylinder to rest;
the acceleration record of the piston rod during the 2 s is as shown.
Determine by approximate means (a) the initial velocity of the piston rod,
(b) the distance traveled by the piston rod as it is brought to rest.
4 U
•VO\
20
10
II
\
o*V os o n
i n i !V 15 1T
20
t s1
SOLUTION
Approximate the a - I curve by a series of rectangles of height ah each with its centroid at t =
When
equal widths of At = 0.25 s are used, the values of /, and a, are those shown in the first two columns of the
table below.
At / = 2 s,
',
ai
2 -t,
* ,( 2 - 0
(s)
{U S)
(s )
(ft/s)
0.125
-3.215
1.875
-6.028
0.375
-1.915
1.625
-3.112
0.625
-1.125
1.375
-1.547
0.875
-0.675
1.125
-0.759
1.125
-0.390
0.875
-0.341
1.375
-0.205
0.625
-0.128
1.625
-0.095
0.375
-0.036
1.875
-0.030
0.125
-0.004
Z
-7.650(ft/s2)
v = v0 + j^adt
a
- 1 1.955(ft/s)
v0 + la , (A/)
a v0 + (S a / )(A/)
0 a v0 -(7.650)(0.25)
v0 = 1.913 ft/s <
Using moment-area formula,
* = *o + V + £ « , ( ' “ ti ) dt ~ xo + V + £«*/(2 - t,)(At)
a j r 0 + V + (Ea(( 2 -f ,)) ( A /)
a 0 + (l.9 1 3 )(2 )-(ll.9 5 5 )(0 .2 5 )
x = 0.836 ft A
PROBLEM 11.82
The acceleration record shown was obtained during the speed trials of a
sports car. Knowing that the car starts from rest, determine by
approximate means (a) the velocity of the car at t = 8 s, (b) the distance
the car traveled at t = 20 s.
SOLUTION
Approximate the a - I curve by a series of rectangles of height a,, each with its centroid at t =
When
equal widths of A/ = 2 s are used, the values of /, and a, are those shown in the first two columns of table
below.
a,
t,
(s)
K
)
2 0 - /,.
a, ( 2 0 - / , )
(s)
(ft/s)
1
17.58
19
334.0
3
13.41
17
228.0
5
10.14
15
152.1
7
7.74
13
100.6
9
6.18
11
68.0
11
5.13
9
46.2
13
4.26
7
29.8
15
3.69
5
18.5
17
3.30
3
9.9
19
3.00
1
3.0
990.1 (ft/s)
Z
(a) At / = 8 s,
vg = v0 + fcadt » 0 + la , (A/)
« (lfl,)(A r)
Since t = 8 s, only the first four values in the second column are summed:
la , = 17.58 + 13.41 + 10.14 + 7.74 = 48.87 ft/s2
vg = (48.87)(2)
(b) At I = 20 s,
v8 = 97.7 ft/s <
jc20 = v„r + £ °a (2 0 - t)dt = 0 + la , (20 - t)(At)
= (990.1)(2)
x20 = 1980 ft 4
PROBLEM 11.83
A training airplane has a velocity of 32 m/s when it lands on an aircraft
carrier. As the arresting gear of the carrier brings the airplane to rest, the
velocity and the acceleration of the airplane are recorded; the results are
shown (solid curve) in the figure. Determine by approximate means
(a) the time required for the airplane to come to rest, (b) the distance
traveled in that time.
i
SOLUTION
The given curve is approximated by a series of uniformly accelerated motions.
- Ct ( m o /s M
l5 H L
II -S
f
© ! ©
i8
©
®
r i 3©
I -E -—
-4-
S
lo
2o
15
25
Jo i?
tC s 'i
For uniformly accelerated motion,
2
2 - , /
\
v2 - v, = 2a(x2 - x,)
or
v2 - v, = a(t2 - f|)
or
.
V2 -
Ax = —
V,
2a
L
At = v2 ~ vi
For the regions shown above,
Region
V| (m/s)
v2 (m/s)
a(m /s2j
Ax(m)
A/(s)
1
32
30
-3
20.67
0.667
2
30
25
-8
17.19
0.625
3
25
20
-11.5
9.78
0.435
4
20
10
-13
11.54
0.769
5
10
0
-14.5
3.45
62.63
0.690
3.186
I
t = I(A /) = 3.19 s <
(«)
(b) Assuming
x0 = 0,
x = Xq + E(Ax) = 62.6 m A
SOLUTION
Use a = v-—~ noting that ~
= slope of the given curve.
Slope is calculated by drawing a tangent line at the required point, and using two points on this line to
determine Ax and Av. Then, — = — .
dx
Ax
(a) When x = 0.25,
v = 1.4 m/s
Av = 1 m/s
and
from the curve
Ax = 0.25 m
— =— =4s'
d x ' 2.5 ~ S
a = (l.4 )(4 )
x = 0.5 m
(b) When v = 2.0 m/s,
Av = 1 m/s
and
Ax = 0.6 m
— = — = 1.667 s"1,
dx
0.6
from the tangent line
a = 5.6 m/s: A
from the curve.
from the tangent line.
a = (2)(l.667)
a = 3.33 m/s2
A
PROBLEM 11.85
Using the method of Sec. 11.8, derive the formula x = x0 + v0f + ^ a t 2
for the position coordinate of a particle in uniformly accelerated
rectilinear motion.
SOLUTION
,
The a - t curve for uniformly accelerated motion is shown. The area of the rectangle is
A = at.
-
Its centroid lies at
t =
1
2
By moment-area formula,
X
= X0 + v0 + A (t - t ) =
* 0
f \ N
+ V + (a t) —t
v2 ,
1 2
= x0 + v + - a t
PROBLEM 11.86
Using the method of Sec. 11.8, determine the position o f the particle of
Prob. 11.61 when t = 12 s.
SOLUTION
to
iz
m
t on
-H
From the a - 1 curve, Ax = (-4 )(6 ) = -2 4 ft/s
A2 = (4)(2) = 8 ft/s
Over 6 s < / < 10 s,
v = -8 ft/s
v = v0 + Au
or
- 8 = v0 - 24,
or
v0 = 16 ft/s
By moment-area formula,
xi = xo + vol + moment of shaded area about t = 12 s
2
xl 2 = 0 + (16)(12) + (-24)(12 - 3) + (8)(12 - 11)
x12 = -1 6 ft 4
PROBLEM 11.87
~
An automobile begins a braking test with a velocity of 30 m/s at / = 0 and
comes to a stop at / = /, with the acceleration record shown. Knowing
that the area under the a - t curve from / = 0 to / = T is a semiparabolic
area, use the method of Sec. 11.8 to determine the distance traveled by
the automobile before coming to a stop if (a) T = 0.2 s, (b) T = 0.8 s.
SOLUTION
a (»»>/&*■)
(a) T = 0.2 s.
A, = —(-8.4)(0.2) = -1.12 m/s
A2 = (-8.4)(/, - 0.2)
= -8.4/, + 1.68
Vj- = v0 + £/l
0 = 3 0 - 1 .1 2 - 8 .4 /, +1.68
/, = 3.638 s
A-, = -28.88 m/s
/, - T = 3.438 s
By moment-area formula,
x, = x0 + v0/, + moment of area
x, = 0 + (30)(3.638) + (-1.12) g ( ° . 2 ) + 3.438) + ( - 2 8 .8 8 ) [ i £ i '
x, = 55.6 m A
(6) T = 0.8 s.
A, = |( - 8 .4 ) ( 0 .8 ) = -4.48 m/s,
A2 = ( —8.4)(/, - 0.8) = -8.4/, + 6.72
Vy = v0 + LA
or
0 = 30 —4.48 - 8.4/, + 6.72, /, - 3.838 s
/, - T = 3.038 s
A2 = -25.52 m/s
By moment-area formula,
x, = 0 + (30)(3.838) + (-4.48) ^(0.8) + 3.038
( - ...( ( M i)
x, = 61.4 m A
PROBLEM 11.88
For the particle of Prob. 11.63, draw the a - 1 curve and determine, using
the method of Sec. 11.8, (a) the position of the particle when
t = 52 s, (b) the maximum value of its position coordinate.
SOLUTION
a = slope of v - 1 curve.
Construct the a - 1 curve.
0 < t < 10 s,
At = 10 s,
Av = 0,
Av = n0
a =—
At
10 s < t < 26 s,
At = 16 s,
Av = -80 m/s,
Av
o
a - — = -5 m/s
At
26 s < t < 41 s,
At = 15 s,
Av = 0,
AV = A
a =—
0
At
41 s < t < 46 s,
At = 5 s,
Av = 15 m/s,
Av
2
a = — = 3 m/s
At
46 s < t < 52 s,
At = 6 s,
Av = 0,
a =0
O - i r n / s * -)
—
S /p f/s
1©
t
t
45
4 s.s
HI J
a
-S
,\
S
~£
Ax - Av = -80 m/s
A2 - Av - 15 m/s
(a) x at t = 52 s.
x = *o + v0f + I 4 ( f - f,)
= -540 + (60)(52) + (-80)(52 - 18) + 15(52 - 43.5)
(b) Calculation of xmax.
10
“
W - k f r y
'■■>7 /
-v
1
-12.5 m <
PROBLEM 11.88 CONTINUED
v = v0 + A3 - 0
60 - 5(fj - 10) = 0
- 10 = 12 s, tx - 22 s
—+ — = 16 s, A = -6 0 m/s
*max = * 0 + V
+ A
= -540 + (60)(22) + (-60)(22 - 16)
PROBLEM 11.89
The
motion
of
a
particle
is
defined
by
the
equations
x = 17.6 - 0.8t(t2 - 9 1 + 24) and y - -13.8 + 0,6f(Y2 - 9 1 + 24), where
x and y are expressed in feet and t is expressed in seconds. Show that the
path of the particle is a portion of a straight line, and determine the
velocity and acceleration when (a)t = 2 s, ( b ) t - 3 s, (c) t = 4 s.
SOLUTION
u = t ( t 2 - 9 1 + 24] = t3 - 912 + 24t
Let
— = 3t2 - 18f + 24,
dt
Then,
and
x = 17.6 - 0.8m ft
*
dt
=
-
y = -13.8 +0.6m ft
0.8 *
dt
±
=
dx
— = dL =
dx
§
^ 4 - 6/-18
dt2
0.8
+
0. 6 —
dt
= -0.75 = constant
Since — does not change, the path is straight.
dx
(a) At t = 2 s,
w
— = 0,
dt
v, = f
and
^ 4 = _6.
dt2
- (-0.8)(0) = 0,
v, = ^
,2
fljc = £ - £ = (-0 .8 )(-6 ) = 4.8 ft/s2,
= (0.6)(0) - 0
ay = (0 .6 )(-6 ) = -3.6 ft/s2
v = 0, a = 6.0 ft/s2 ^
du
— = -3,
dt
.
(b) At t = 3 s,
36.9°
d u „
— j- = 0
dt1
,
and
vx = (-0 .8 )(-3 ) = 2.4 ft/s,
vy - (0.6)(-3) = -1.8 ft/s
ax = 0,
ay = 0
v = 3.0 ft/s ^ 3 6 .9 ° , a - 0
du „
— - 0,
dt
, . i
(c) Att - 4 s,
v* = 0 ,
ax - (-0.8)(6) = -4.8 ft/s2,
,
and
d 2u ,
—T= 6
dt2
vy = 0,
= (0.6)(6) = 3.6 ft/s2
v = 0, a = 6.0 ft/s2 3 6 . 9 ° ^ <
PROBLEM 11.90
The motion of a particle is defined by the equations
x = (4 cos nt - 2)1(2 - cos nt) and y - 3 sin nt 1(2 - cos nt), where x
and y are expressed in feet and t is expressed in seconds. Show that the
path of the particle is part of the ellipse shown, and determine the
velocity when (a) t = 0, (b) t = 1/3 s, (c)t= 1 s.
SOLUTION
Substitute the given expressions for x and y into the given equation of the ellipse, and note that the equation is
satisfied.
x 2 v2
il6cos2;r? - 16cos;rf + 4)
9 sin2nt
— + 2— = 3
L+
—
—_
4
3
4(2 - cos^-r)
3(2 - cos-m )
4cos“ nt - 4cos^r + 1 + 3sin* nt
4 - 4cos/zt + cos1nt
(2 - cos nt)
(2 -
COS n t ) '
Calculate x and y by differentiation.
.
-An sin nt
(4 cos nt - 2 )(^ sin nt) _
(2 - cos nt)
(2 - cos^-r)2
(c) When t = 1 s,
(2 - cos n t)'
3n cos nt
'ism.ntinstn.nt)
3n)2cosnt - l)
(2-cos;rt)
(2 - cos;rt)
(2 - cos nt)'
x =0
(a) When t = 0 s,
(b) When t = — s,
3
-6nsw .nt
x -
and
-6n ( # ) _
| A
y = 3n,
v - 9.42 ft/s
-n - v 3, y — 0
v = 7.26 ft/s -— ^
(2 - 4 )
• A
x =0
,
and
3n(-3)
y =—
= -n,
(3)
v = 3.14 ft/s { A
PROBLEM 11.91
The
motion
of
a
particle
is
defined
by
the
equations
x = (/ - 4)3/ 6^ + t2 and y = (t3/ 6) - (t - l)2 / 4, where x and y are
expressed in meters and t is expressed in seconds. Determine (a) the
magnitude o f the smallest velocity reached by the particle, (b) the
corresponding time, position, and direction o f the velocity.
SOLUTION
(a) Given:
1 / .
*\3
. ,2
x = -6 ( t - 4 y + t
and
6
4
By differentiation,
vx vy
dx - I ,
~dt
2
1
dy
dt ~ 2
-1 r 2 - - 1t + -1
2
dvx
=t - 2
dt
and
2
2
2
dv,y _ t _ i
dt
Magnitude of velocity is
- f t
3 + v y2
Note that v is minimum when v is minimum.
v2 = v2
x + v2
y
Differentiating and setting equal to zero,
, dv
„ dvr
dv
2v— = 2vr —— + 2vv—— = 0
dt
x dt
y dt
2 1
(<2 - < + 1 6 )(> -2 ) + vr — 2 t
+
1
2
-
2 f - 1.5t2 + 25.51 - 32.5 = 0
The only real root of the cubic equation is t = 1.757 s.
The corresponding values of vx and vy are
vx = ^ (l-7 5 7 )2 - 2(1.757) + 8 = 6.03 m/s
vy = \ ( ^ f - j ( l . 7 5 7 ) + | = 1.165 m/s
mn = J (6 .0 3 )2 +(1.165)2
vmin = 6.14 m/s <
PROBLEM 11.91 CONTINUED
!>) Time, position, and direction o f velocity.
t = 1.757 s A
x = i( l.7 5 7 —4)3 + (1.757)2
^ = 6 (L757)
4 (L757 _1)2
tan<9 = ^L =
= 0.1932,
v„
6.03
x = 1.206 m A
y = 0.761 m A
e = 10.9° ^
a
PROBLEM 11.92
The motion of a particle is defined by the equations x = 6/ - 3 sin t and
y = 6 - 3 cos t, where x and y are expressed in meters and t is expressed
in seconds. Sketch the path of the particle for the time interval
0 < t <, 271, and determine (a) the magnitudes of the smallest and largest
velocities reached by the particle, (6) the corresponding times, positions,
and directions of the velocities.
SOLUTION
Sketch the path of the particle, i.e. plot ofy versus x.
Using x = 6/ - sinr, and y = 6 - 3cosr obtain the values in the table below. Plot as shown.
x(m )
y(m )
0
3
6.42
6
71
18.85
9
3—
2
31.27
6
2 71
37.70
3
'(* )
0
71
2
JC * 1
(a) Differentiate with respect to t to obtain velocity components.
v = — = 6 - 3cos/
x
dt
and
v., = 3sin/
y
v2 = v2 + v2 = (6 - 3cos/)~ + 9sin2/ = 45 - 36cos/(m /s)'
d(v)~
—L -2 -= 3 6 s in /= 0
dt
/ = 0, ± n, ± 2n, ... ± N n
When / = 2Nn,
cos / = 1,
and
When / = (2N + l)/r,
cost = -1,
and
(v2)
'
'm in
v2 is minimum.
v2 is maximum.
= 45 - 36 = 9 (m/s)2,
vmin = 3 m/s 4
PROBLEM 11.92 CONTINUED
(v 2 )
V -'max
- 45 + 36 = 81 (m/s),
•
(b) When v = VKmin>
= 9 m/s
A
t - 2 n N s,
A
v max
Where N - 0, 1, 2,.
x - 6(2 n N ) - 3sin(2^iV)
y = 6 - 3cos(2^A^)
vx = 6 - 3cos(2;rAf)
vy = 3sin(2^yV)
tan 0 = — = 0,
vv
When v
x = \2n~N m A
}' = 3 m ^
vr = 3 m/s
vy =0 A
0
=
0A
t = (2N + l ) x s
V
Kmax>
x = 6[2^(7V - 1)] - 3sin[2;r(Ar + l)]
y = 6 - 3 c o s [ 2 ;r ( W + l)]
vx = 6 - 3cos[2;r(Ar + l)]
vy = 3sin[2/r(./V + l)]
tan# = — = 0,
vv
A
A
x = \ 2 ti( N + \)
y = 9m ^
vx = 9 m/s
A
v„ = 0
A
0
=
0A
PROBLEM 11.93
The motion of a particle is defined by the position vector
r = yf(cost + t sin ?)i + (sin t - t cos t)\, where ? is expressed in
seconds. Determine the values of ? for which the position vector and the
acceleration vector are (a) perpendicular, (6) parallel.
SOLUTION
Given:
r = ^4(cos? + ?sin?)i + A{saxt - ?cos?)j
\ =Y
= ^ (- s in ? + sin? + ?cos?)i + A(cost - cost + ts in t)j
- ^ (rc o s t)i + A (tsin t)j
a=
- Accost - ?sin?)i + /l(sin? + ?cos?) j
,
(a) When r and a are perpendicular, r • a = 0
vf[(cos? + ?sin?)i + (sin? - ?cos?) j] • A [(cos? - ?sin?)i + (sin? + ?cos?) j = 0
(cos? + ?sin?)(cos? - ?sin?) + (sin? - ?cos?)(sin? + ?cos?) = 0
(cos2 ? - ?2 sin2 ?j + (sin2 ? - ?2 cos2 ? = 0
1 - ?2 = 0
? = 1s A
2
(b) When r and a are parallel, r x a = 0
. (cos? + ?sin?)i + (sin? - ?cos?)j j x ..'"(cos? - ?sin?)i + (sin? + ?cos?)j = °
((cos? + ?sin?)(sin? + ?cos?) - (sin? - ?cos?)(cos? - ?sin?)J = 0
. in?cos? + ?sin2? + ?cos2? 4- ?2sin?cos?j - (sin?cos? - ?cos2? - ?sinz ? + ?2sin?cos?| = 0
2? = 0
?= 0 <
PROBLEM 11.94
The damped motion of a vibrating particle is defined by the position
vector r = x , [ l - l / ( / + l ) ] i + (y\e~m ' 2 cos 2n tj j, where t is expressed
in seconds. For xx - 30 in. and y x = 20 in., determine the position, the
velocity, and the acceleration of the particle when (a) / = 0,
(b) t = 1.5 s.
SOLUTION
r = 30 1 -
Given:
t +l
i + 20: e ntl2 cos 2^/ j
Differentiating to obtain v and a,
* = in
+ 20 ~ ~ e ntil cos2nt - 2ne r ! sin 2nt |j
v =—
30
dt
(t + i y
i - 20n
2
0/2 | 1 cos 2nt + 2 sin 2nt
(r +1)
ft ,-KtH —cos 2nt + 2 sin 2nt
— = -3 0 — - - Ti - 20n
dt
(/ + !)
2
2
I0n2e^,/2
~ 6 0 _ ii _- t O r 1 --* * 12 (4 sin2nt
+ e *r/2(-7Tsin 2nt + 4n c o s2 n t)j
- 7.5cos2/r/)j
(f + 1)3
f
iN
r = 30 1 i + 20(1) j
v
1/
(ia) At / = 0,
v = 30
11i - 207T 0)-
+0
60.
- j- i - 10^r2 (l)(0 - 7.5) j
a =
(b) At t = 1.5 s,
2
r = 30 1 -
2.5
r = 20 in. ( 4
v = 43.4 in./s ^
46.3° 4
a = 743 in./s2 7 ^ 85.4° 4
i + 20e °'25/r cos 3ttj
= (18 in.)i + (-1.8956 in.)j
r = 18.10 in. ^
6.0° 4
= —— y i - 20ne~OJ5* f —cos3n + 0 j j
( . )
U
r
2
5
= (4.80 in./s)i + (2.9778 in./s) j
a =
60
v = 5.65 in./s ^
31.8° 4
a = 70.3 in./s2 ^
86.9° 4
.
i + 10;r2e~° ’5;r(0 - 7.5cos3/r)j
(2-5)
= (-3.84 in./s2)i + ( 70.1582 in./s2) j
PROBLEM 11.95
The three-dimensional motion of a particle is defined by the position
vector
r = (Rt cos cont)i + ct j + (Rt sin cont)k.
Determine
the
magnitudes of the velocity and acceleration of the particle. (The space
curve described by the particle is a conic helix).
SOLUTION
Given:
r
- (Rt coscont)i + ct\ +
sin<y„t)k
Differentiating to obtain v and a.
= R (cos cont - a>ntsma>nt)i + cj + i?(sine;„t + contcoscont) k
v =
a = — = R(-a>n sincont - consincont - a>n2tcoscont)i + R (a>„coscont + concoscont - co„tsmo)nt \ k
dt
'
’
v
'
-
-2consincont - ton2tcoscont ji + (2concoscont - a>n2tsma>nt )
k
Magnitudes of v and a.
2
V
2
P V
2
2
+ vy + V
= j R(cosco„t - coytsin (OJ~\ + ( c f + [/?(sin<y„t + contcoscont) ]2
=■ R 2 [ cos2 mnt - 2mnt svo (Dnt cos mnt + (o2nt 2 sin'
+ c2
+ R 21sin2 cont + 2a)nt sin a>„t cos <ont + co2t2 cos2 cont J
= R 2(l + a>lt21 + c2
v = -Jr 2( i + co2nt2^ + c2 A
a2 = a] + a 2 + a2
= R 2 j -2 a>n sincont - co2t cos contj + {2a>„coscont - ©2t sin n y j
= R 2\ 4co2 sin2 cont + 4co2nt sin cont cos <ont + co^t2 cos2 cont + 4 co2 cos2 cont
- 4co;t sin o>nt cos oont + co*t2 sin2 a>nt \
= R 2{4co2 + co4nt2)
a = Rcon^ 4 + colt2 A
PROBLEM 11.96
The three-dimensional motion of a particle is defined by the position
vector r = ( /l/c o s /) i + ^A^jr +1 j j + ( # /s in /) k , where r and / are
- ——-r ~ ~~z ■ 1
expressed in feet and seconds, respectively. Show that the curve
described
by
the
particle
lies
on
the
hyperboloid
( y / X f - ( x l A f - ( z / B ) ~ = 1.
A =3
For
B = 1,
and
determine
(a) the magnitudes of the velocity and acceleration when / = 0, (b) the
smallest nonzero value of / for which the position vector and the velocity
vector are perpendicular to each other.
SOLUTION
r = (zl/cos/)i + ^ A \lr + 1jj + (5 /s in /)k
Given:
x = Atcost,
y = A\lt2 + 1,
x
cos / = —
At
.
z
sin / = —
Bt
z = 5 /sin /
7
/* =
2
/
For A = 3
Z
-1
KA
.2
cos2 / + sin2 / = != > ! — I +
AXt))
Then,
from which
I X_
U
r *
x2
( A
B
and
]
kb)
I
2 J
A,
I
= (3/cos/)i + 13V/2 + 1jj + (/s in /)k
B = 1,
Differentiating to obtain v and a.
dr
v = — = 3(cos/ - /sin /)i + 3 , 1
dt
j + (sin/ + /co s/)k
yjt2 + 1
a = — = 3 (-2 sin / - /co s/)i + 3----- -— - j + (2cos/ - /s in /)k
dt
At / = 0,
And
Then,
(/2 + l) 2
v = 3(1 - 0)i + (0) j + (0)k
a = —3 (0 )i + 3(1 )j + (2 - 0)h
a2 = (3 )' + (2 )' = 13
(b) If r and v are perpendicular, r • v = 0
(3/cos/)[3(cos/ - /sin /)] + llylt2 + 1j
V
or
= 3 ft/s <
^ — + ( /s in /) ( s in /+ /co s/) = 0
/ d t2 + 1
^9/cos2/ - 9/2 sin/cos/) + (9/) + (/sin2/ + r s in /c o s /) = 0
= 3.61 ft/s2 «
PROBLEM 11.96 CONTINUED
With t * 0,
9cos2t - 8f sin t cost + 9 + sin21 = 0
10 - 8tsintcost + 8cos2t = 0
or
The smallest root is
The next root is
7 + 2cos2t - 2tsin2t = 0
I t - 7.631 s
t = 4.38 s
t = 3.82 s <
PROBLEM 11.97
TIJim
*
i
A baseball pitching machine “throws” baseballs with a horizontal
velocity v0 . Knowing that height h varies between 788 mm and
1068 mm, determine (a) the range of values o f v0, (b) the values o f a
corresponding to h = 788 mm and h = 1068 mm.
SOLUTION
y 0 = 1.5 m, (vy )o = 0
(a) Vertical motion:
or
y = y , + (vy \ t - ^ g t
At point B,
y =h
2 (yo - h)
tB = J —---- — -
or
(2)(1.5 - 0.788) . 0 3810.
When h = 788 mm = 0.788 m,
9.81
When h = 1068 mm = 1.068 m,
3
Horizontal motion:
V
*0 = °> ( vx)o = V0>
x = Vr.t
X
or
and
Xf,
vn = — = —
t
tD
we get
With xB - 12.2 m,
v0 =
v0 =
12.2
- 32.02 m/s
0.3810
12.2
= 41.11 m/s
0.2968
or
32.02 m/s < v0 < 41.11 m/s
(b) Vertical motion:
115.3 km/h < vn £ 148.0 km/h A
vy = ( vy \ - S t = ~gt
Horizontal motion:
Vx = V0
_ _<fy =
ta n a =
dx
(vy)B = gts_
i vx)B
(£ £ 1 ^ 0 3 8 1 0 ) =
For h = 0.788 m,
For h = 1.068 m,
0.2968,
9.81
32.02
^
. (9.81)(0.2968) , Q
41.11
vo
a = 6.66° <
a = 4.05° <
PROBLEM 11.98
While delivering newspapers, a girl throws a paper with a horizontal
velocity v0. Determine the range of values of v0 if the newspaper is to
land between points B and C.
----- 11a -----J
SOLUTION
Sketch the limiting trajectories.
Point A:
x = x0 = 0
y = y0 = 0
Point B:
xB = 2.2 m
y B = -1.0 m
Point C:
xc = 4.0 m
y c = -0.6 m
Vertical motion with
=0:
y = y0 + (vy)of -
g t2
or
-2 y
/ =
(1 )
g
Horizontal motion with (yt )0 = v0:
x = vtf
or
v0 = —
Substituting equation (1) into equation (2) gives
(2 )
Vn =
x
g
-2y
(3)
Applying equation (3):
at point B,
v0 = 2.2
at point C,
v0 = 4.0
Range of v0:
9.81
= 4.87 m/s
(—2) (—I -0)
I
9 l\
(- 2) (0-6)
= 11.44 m/s
4.87 m/s < v0 < 11.44 m/s ^
PROBLEM 11.99
A ski jumper starts with a horizontal take-off velocity of 80 ft/s and lands
on a straight landing hill inclined at 30°. Determine (a) the time between
take-off and landing, (b) the length d of the jump, (c) the maximum
vertical distance between the jumper and the landing hill.
SOLUTION
y - - x tan 30°
(a) At the landing point,
Horizontal motion:
* = *0 + (vx)</ = V
Vertical motion:
y = y 0 + (vy )0t - \ g t 2 ~ ~ \ g t 2
2y _ 2xtan30° _ 2v0ttan30°
_
from which
g
g
g
2v0 tan 30°
Rejecting the / = 0 solution gives
g
' =
(b) Landing distance:
cos30°
(2)(80)tan30°
‘
t = 2.87 s A
32.2
y
(80)(2.87)
cos30°
cos30°
d = 265 ft A
h = x tan 30° + y
(c) Vertical distance:
1 9
h - v0ttan30° - —gt
or
Differentiating and setting equal to zero,
dt
Then,
g
,
_ (v0)(v0 tan30o)tan30°
“ max ~
g
1
tT g
2
Vp tan2 30°
(80)2(tan30°)2
2-g
(2)(32.2)
v0 tan 30° 2
g
/ W = 33.1 ft A
PROBLEM 11.100
A golfer aims his shot to clear the top of a tree by a distance h at the peak
of the trajectory and to miss the pond on the opposite side. Knowing that
the magnitude of v0 is 85 ft/s, determine the range o f values o f h which
must be avoided.
SOLUTION
Horizontal motion:
Vertical motion:
At ground level, y = 0,
A tx = 150 ft,
X
*0 + ( v*)</ = V -
y = To + ( v y ) ^ - ^ g t 2 = T o- \ & 2
so that
y0 -
t =—
or
or
y=yo
gx
2v02
gx
2vl
_ (32.2)(150)2 _
2'
= 5 0 ' 14ft
(2)(85)2
h = y 0 - 40 = 10.14 ft
A tx = 160 ft,
(32.2)(160)2
/ ,2 = 57.05 ft
(2 )(8 5 r
h = y 0 - 40 = 17.05 ft
Range to avoid:
10.14 ft < h < 17.05 ft A
PROBLEM 11.101
A golfer hits a golf ball with an initial velocity of 48 m/s at an angle of
25° with the horizontal. Knowing that the fairway slopes downward at an
average angle of 5°, determine the distance d between the golfer and
point B where the ball first lands.
SOLUTION
x = xA + v 0 cos 25°/ - v0 cos 25°/
Horizontal motion:
At point B,
xB = d cos5° = v0cos25°/s
1 2
1 2
y = y 0 + v0 sin 25°/ - —gt = v0 sm 25°r - —gt
Vertical motion:
(
y B = -c/sin5° = v0sm 2 5 °
V
Ratio:
or
Then,
v„
<?in5° (vnsin25° - ^ g to ] tR
—
— = ------------ =
xB
cos 5°
v0 cos 25%
1 A
gtB
2
or
\
-gtB
2
- v 0cos25°ta n 5° = v0 sin25° -
i^/L(cos25°tan5° + sin25°) = — —— - ( cos25°tan5° + sin25°) = 4.912 s
? V
1
9.81 V
'
j _
_ Vqcos25°rfi _ (48)(cos25°)(4.y]zi
cos 5°
cos 5°
cos 5°
xb
J = 215m ^
PROBLEM 11.102
Water flows from a drain spout with an initial velocity of 0.76 m/s at an
| angle of 15° with the horizontal. Determine the range of values of the
distance d for which the water will enter the trough BC.
3m
«/ ------- |
I— o A S m
SOLUTION
t \
1 2
y = yo + (vo ) / - - g r
Vertical motion:
- & 2 - { v 0)y ‘ -{y<>- y ) = o
With y 0 = 3 m
and
y -= 0.36 m at level BC,
y 0 - y = 2.64 m
(v0) t = -0.76sinl5° = -0.19670 m/s
4.905r + 0.19670r - 2.64 = 0
Then,
Solving the quadratic equation,
-0.19670 + J(0.19670)2 - (4)(4.905)(-2.64)
(2)(4.905)
Horizontal motion:
For water impact at C,
For water impact at B,
So,
x = (v0) I = (0.76cos 15°)(0.7139) = 0.524 m
xc = 0.524 ft
with
d = xc - 0.62 = -0.096 m
xB = d = 0.524 m
-0.096 m < d < 0.524 m
Owing to the foot of the wall, d cannot be negative; hence, the allowable range of d is
0 < d < 0.524 m A
PROBLEM 11.103
In slow pitch softball the underhand pitch must reach a maximum height
of between 6 ft and 12 ft above the ground. A pitch is made with an initial
velocity v0 of magnitude 43 fit/s at an angle of 33° with the horizontal.
Determine (a) if the pitch meets the maximum height requirement, (b) the
height of the ball as it reaches the batter.
aFPT
SOLUTION
v0 - 43 ft/s, a = 33°, x0 = 0, y0 = 2 ft
vy = v0s in a - gt
Vertical motion:
y = yo + (v 0s i n a ) r - | g t 2
vy = 0
At maximum height,
or
v0s in a
g
f= “
(«)
5131 = 07273 ,
32.2
ymax= 10.51 ft <
^max - 2 + (43sin33°)(0.7273) - i(32.2)(0.7273)2
yes A
6 ft < 10.51 ft < 12 ft,
Horizontal motion:
At x = 50 ft,
(b) Corresponding value of y:
(
\
x = x0 + (v0cos a ) t
t x
or
x - xn
t = ------- —
vncosa
50 -9.- = 1.386 s
43cos33°
y = 2 + (43sin33°)(l.386) - |(3 2 .2 )(l.3 8 6 )2
y te 3.52 ft <
PROBLEM 11.104
-'p-'.
A tennis player serves the ball at a height h with an initial velocity of
120 ft/s at an angle of 4° with the horizontal. Knowing that the ball
clears the 3-fit net by 6 in., determine (a) the height h, (b) the distance d
<-| from the net to where the ball will land.
SOLUTION
/ \
* = * 0 + (W
Horizontal motion:
X - Xc
t = (v \
\ x)q
or
(vx)Q = 120cos4° = 119.71 ft/s,
x0 = 0
t = 40 " ° = 0.3341 s
119.71
At x = 40 ft,
Vertical motion:
y = y 0 + (vy \ t - \ g t 2
(vy) - -1 2 0 sin4° = -8.371 ft/s
y = 3.5 ft
(a) At the net,
g = 32.2 ft/s2
and
and
t - 0.3341 s
Jo = y + (vy)ot + ± g t 2 = 3.5 + (-8.37l)(0.334l) + l(3 2 .2 )(0 .3 3 4 l)2
y 0 = 8.09 ft 4
(b) At y = 0,
j g r 2 - ( y y)ot - y 0 = °
j( 3 2 .2 )t2 - (-8.371)/ - 8.094 = 0
-8 .3 7 1 + ^/(8.37l)2 -(4 )(l6 .l)(-8 .0 9 4 )
32.2
At t = 0.4952 s,
Then,
jc =
(119.71)(0.4952) = 59.29 ft
d = x - 40
d - 19.29 ft 4
PROBLEM 11.105
A homeowner uses a snow blower to clear his driveway. Knowing that
“f the snow is discharged at an average angle of 40° with the horizontal,
’I" determine the initial speed v0 of the snow.
SOLUTION
Using a = 40° the horizontal and vertical motions are
x = (v0 cosar)f,
or
t =
y = y 0 + (v0 s in a ) f - - g t 2,
and
vn cos a
andj
^ gt'7
y = y 0 + x tan a - —
2 _ 2(y0 + x ta n a - y )
From which
g
Using
y 0 = 0.6 m, xB = 4.2 m,
and
y B = 1.1m
,
2 (0 .6 + 4 .2 ta n 4 0 ° - 1.1)
, _ ,
tl =
= 0.6166 s2
"
9.81
tB = 0.7852 s
From the horizontal motion,
Vn =
tco sa
4.2
(0.7852cos40°)
vn = 6.98 m/s A
PROBLEM 11.106
A basketball player shoots when she is 5 m from the backboard. Knowing
that the ball has an initial velocity v0 at an angle of 30° with the
horizontal, determine the value of v0 when d is equal to (a) 228 mm,
(b) 430 mm.
SOLUTION
x = (vgcosa)/
Horizontal motion:
or
t
v0cos«
1 2
y = yo + V0s m a t--g
t
Vertical motion:
y = y 0 + xtan a - j g t 2
2 = 2 (y0 + x tan or - y )
g
(a) When d = 228 mm = 0.228 m,
x = 5 - 0.228 = 4.772 m
2 2 (2 + 4.772 tan 30° - 3.048)
t2 = -A
1 = 0.3480 s2
9.81
/ = 0.590 s
Vn =
4.772
0.590 cos 30°
vn = 9.34 m/s 4
(b) When d = 430 mm = 0.430 m,
x = 5 - 0.430 = 4.57 m
2 2(2 + 4 .5 7 ta n 3 0 ° -3.048)
,
t2 =
= 0.3243 s2
9.81
t = 0.569 s
Vn =
4.57
0.569 cos 30°
v0 = 9.27 m/s 4
PROBLEM 11.107
The nozzle at A discharges cooling water with an initial velocity v0 at an
angle of 6° with the horizontal onto a grinding wheel 13.8 in. in
diameter. Determine the range of values of the initial velocity for which
the water will land on the grinding wheel between points B and C.
7 S in
SOLUTION
Choose the origin at the center of the grinding wheel, so that the horizontal and vertical motions are:
x - x0 = v0cosat,
And
from which
v0 = -------- ,
tc o sa
or
y ~ y 0 = v0 s i n a / - j g r = ( x - x 0)ta n a - ^ g t 2
r 2
2 [(^o - y ) + ( * - *o)tan« ]
g
Data:
a = -6 °
jr0 = -0.75 in.
r = ^ d = ^-(13.8) = 6.9 in.
(a) Stream lands at B.
y 0 = 8 in.
g = 386.4 in./s2
x = r sin 10° = 1.198 in.
y = r cos 10° = 6.795 in.
,
2 T 8 - 6.795 + (1 .1 9 8 + 0 .75)tan(-6°)l
,
r = L
V
'
V /J - 0.005177 s2
386.4
t = 0.07195 s
1.198 - ( - 0 .7 5 )
,
vn = --------------- ----- r = 27.22 in./s
0 0.07195 cos(-6°)
(b) Stream lands at C.
v0 = 2.27 ft/s <
x - r cos 30° = 5.976 in.
y = r sin 30° = 3.45 in.
,
2 T 8 - 3.45 + (5.976 + 0.75)tan (-6 °)!
r = L
V
386.4
,
0.019892 s2
t = 0.14104 s
5.976 + 0.75
„ . .
Vn = --------------- ----- - = 47.95 in./s
0 0.14104 cos (-6°)
v0 = 4.00 ft/s <
PROBLEM 11.108
A ball is dropped onto a step at point A and rebounds with a velocity v0
at an angle of 15° with the vertical. Determine the value of v0 knowing
that just before the ball bounces at point B its velocity v B forms an angle
of 12° with the vertical.
SOLUTION
The horizontal and vertical components of velocity are
vx = v0 sin 15°
v = v0cosl5° - gt
At point B,
v
or
v0sinl5°
_
v0c o s l5 ° - g r
^
v0sinl5° + v0 cos 15° tan 12° = g /ta n l2 °
0.46413v0 = g t ta n U 0
t = 2.1836—
g
Vertical motion:
1 ,
y - y 0 = v0 cos 15°t - —gt
2
2.1836cos 15°— - ^ g (2 .1 8 3 6 )2
ygy
= -0.27486 —
g
v02 = -3.638g(y - y 0) = - ( 3 .6 3 8 ) ( 3 2 .2 ) ^ - - - 0
= 78.10 ft2/ s2
vn = 8.84 ft/s M
PROBLEM 11.109
The conveyor belt, which forms an angle of 20° with the horizontal, is
used to load an airplane. Knowing that a worker tosses a package with an
initial velocity v0 at an angle of 45° so that its velocity is parallel to the
belt as it lands 1 m above the release point, determine (a) the magnitude
of v0, (b) the horizontal distance d.
SOLUTION
The horizontal and vertical components of velocity and position are
vr = v0cos45°
v = v0 sin 45° - gt
x = v0cos45°t
y = v0s i n 4 5 ° / - - g / 2
At landing on belt,
gt
= tan 4 5 ° --------s ------ = tan 20°
vv
v0cos45°
v
l = (ta n 45° - tan 20°) cos 45°-^- = 0.44974-^g
g
y = v0sin45° 0.44974— 1 - ~ g ( 0.44974— 1
g)
2 *{
g }
y = 0.21688—
g
Vq = 4.6108gy = (4.6108)(9.81)(1.000) = 45.232 m: /s2
v0 = 6.73 m/s A
(a)
(0.44974)(6.73)
9.81
(b)
x = (6.73cos45°)(0.3083)
x = 1.466 m A
PROBLEM 11.110
s f , H Fy
(_ » ._ !
n,_
V jjp ,
| |
'
A golfer hits a ball with an initial velocity of magnitude v0 at an angle a
3___| with the horizontal. Knowing that the ball must clear the tops o f two trees
and land as close as possible to the flag, determine v0 and the distance d
when the golfer uses (a) a six-iron with a = 31°, (b) a five-iron with
a = 21°.
SOLUTION
The horizontal and vertical motions are
x = (v0c o sa )t
X
v0 = -------tco sa
or
(1)
y = (v0s in a ) t - i gt2 = r t a n a - ~ g t2
2 (x ta n « - y)
2
or
(2)
8
y, = 0,
At the landing point C:
/ = 2”«Sin‘2
g
,
s,
2vn»inacosa
xc = (v0c o s a jt - —---------------
And
(3)
(a) a = 31°
x A = 30 m, y A = 12 m
To clear tree A:
From (2),
From (1),
To clear tree B :
From (2),
From (1),
t\
= 2 ( 3 0 ta n 3 1 " -!2 ) = 1 2 2 g s l
9.81
30
(v0) . = ------------------31.58 m/s
v ° ' A 1.1084cos31°
xB - 100 m,
y B = 14 m
= 9357s2,
(t5 )2 _ 2(100tan31° - 1 2 )
9.81
,,
3.0652 .
B
(v0) = -------— ------ = 38.06 m/s
v 0,B
3.0652cos31°
v0 = 38.06 m/s
The larger value governs,
From (3),
^
v0 = 38.1 m/s 4
( 2 ) ( 3 W * J W .» 3 .9.81
O
1
T-*H
II
d = 20.4 m <
PROBLEM 11.110 CONTINUED
(,b) « = 27°
tA = 0.81846 s,
By a similar calculation,
ig = 2.7447 s,
(v0)^ = 41.138 m/s,
(v0)s = 40.890 m/s,
v0 - 41.138 m/s
= 41.1 m/s 4
xc = 139.56 m,
f = 29.6 m 4
PROBLEM 11.111
A model rocket is launched from point A with an initial velocity v of
m/s. If the rocket’s descent parachute does not deploy and the rocket
lands 104 m from A, determine (a) the angle a that v forms with the
vertical, (b) the maximum height h reached by the rocket, (c) the duration
of the flight.
0
8 6
0
V
/
• 1 0 4 I II ■
SOLUTION
Given:
v =
0
8 6
m/s,
xA = 0 ,
y A = 0,
x = (v s in a )/
Horizontal motion:
xR = 104 ft,
t = — ——
v sin a
or
0
yH = 0
( )
1
0
Vy = v0cos a - gt
Vertical motion:
y ~
( v 0
.
cosar)/
1
2
gx2
,
(v cosar)x
gr = v
.
vo sin a
0
a
(v c o sa )x fl
— \J —
v0sin a
(a)
2
s in a c o s a =
2
0
2
gx2
0
At point B,
.
(v sin a )
2
2
*
2
v0 sin a
= sin a
2
vo
sin2o = (9'8 I)( 104) _
( )-
0 . 1 3 7 9 4
l a = 7.93°
8 6
v,, = v()co sa - gt =
At maximum height,
t =
( ) Then,
6
(c) From equation (1),
a = 3.96°
A
Tmax = 375 m
A
tH = 17.51 s
A
0
vn c o sa
cos 13.46°
------- = ------------------= 8.746 s
g
9.81
8 6
Tmax = (86cosl3.46°)(8.746) - i(9 .8 l)(8 .7 4 6 )2
B
xB
v0sin a
104
(86)(sin3.96°)
PROBLEM 11.112
~~7_ The initial velocity v0 of a hockey puck is 170 km/h. Determine (a) the
largest value (less than 45°) of the angle a for which the puck will enter
j. \
the net, (b) the corresponding time required for the puck to reach the net.
SOLUTION
Horizontal motion:
= (v0cosa ) t
t -
or
v0cosa
■ a \) t - - 1g t -2
y = t(v0sin
Vertical motion:
= x tan a -
2vq cos2 a
2
= x tana - - ^ - ( l + tan2 a j
2v0
tan" a
Data:
2v2
— ta n a +
g*
v0 = 170 km/h = 47.222 m/s,
=0
)
x = 4.8 m at point C,
y = 1.22 m at point C.
2vo _ (2)(47-222)
gx '
(9.81)(4.8)
2 ^ = (94.712)(l.22) =
gx2
4.8
tan2 a - 9 4 . 7 1 2 a + 25.073 = 0
(«)
tan a = 0.26547
a = 14.869°
(b)
t =
c
v0co sa
or
and
94.45
89.4°
4.8
(47.222)cos 14.869°
a = 14.9° A
t = 0.1052 s*^
PROBLEM 11.113
*■ t
il •
Iz r - f ^
v*
^
The pitcher in a softball game throws a ball with an initial velocity v0 of
40 mi/h at an angle a with the horizontal. If the height of the ball at
point B is 2.2 ft, determine (a) the angle a , (b) the angle 0 that the
velocity of the ball at point B forms with the horizontal.
J
SOLUTION
x = (v0c o s a )t
Horizontal motion:
Vertical motion:
or
t =
v0cosa
y = To + (v0s in a )t
gx'
= y0 + x t a n a - — -----—
2Vo cos a
= y 0 + x ta n a - -^-j-fl + tan2 a)
2Vn v
'
2
tan" a
from which
Data:
2Vq
-ta n a gx
2yo ( t ~ To)
+1
=0
gx
v0 = 40 mi/h = 58.667 ft/s, y 0 = 2.0 ft, y = 2.2 ft,
x = 50 ft
2vj = (2)(58 667)2 = 42755
gx
(32.2)(50)
2vo ( t ~ To)
(4.2755)(2.2 - 2.0)
gx2
50
tan2a - 4.2755tana + 1.01710 = 0
tan a = 4.0226
(a) Solving the quadratic equation,
a = 76.04°
(b)
and
and
0.25285
a = 14.19° <
14.19°
v
vfts i n a - g /
tan 6* = — - = — 2---------- — = - t a n a
v0cosa
gx
(v0c o s a )‘
= -0.25285 + ------(32'2)(5°)---- j = 0.24483
(58.667 cos 14.19°)‘
0 = 13.76° <
PROBLEM 11.114
A worker uses high-pressure water to clean the inside of a long drainpipe.
If the water is discharged with an initial velocity v0 of 35 ft/s, determine
(a) the distance d to the farthest point B on the top of the pipe that the
water can wash from his position at A, (b) the corresponding angle a.
SOLUTION
ay = - g
Vertical motion:
or
vy - M l = 2 a ( y - yo)
or
o ~
sin a = (vVly;°
v0
at point B.
(vl o = 2 g ( y e ~ y °)
( v / r = (2) (32.2) (3.6) = 231.84 ft2/s2
Vy = ( v j 0 - g t =
vy = 0
with
or
(vy) = 15.226 ft/s
(v )
tB = ^ 20. = 0.47287 s
g
35
= 0.43504
a = 25.79°
Horizontal motion:
(a)
(b) From above,
x = (v0c o s a )/
xB = (35 cos25.79)(0.47287)
xB = 14.90 ft 4
a - 25.8° 4
PROBLEM 11.115
A nozzle at A discharges water with an initial velocity of 12 m/s at an
j angle a with the horizontal. Determine (a) the distance d to the farthest
point B on the roof that the water can reach, (b) the corresponding
angle a. Check that the stream will clear the edge of the roof.
SOLUTION
1 = — ——
v0cosa
Horizontal motion:
x = (v0c o sa )/
or
Vertical motion:
1
1
2
y = y0 + (v0s i n a ) f ---- gt2 = y 0 + x tan a - —-----—---- j
2
2 ( V()COSa)‘
2
y = y 0 + x tan a -
Let
u = x ta n a
so that
y = y 0 + u ------j l x 2 + «2)
2v0‘ v
’
2y2
x 2 = — - l u + y 0 - y ) - u2
g
Solving for x 2:
The maximum value of x~2 is required:
d (x2)
—1
— - - 0.
du
d l x 2)
jv2
—^— - = —- - 2« = 0
dU
g
Data:
1 + tan2ar)
2v„ v
'
v0 = 12 m/s2,
y 0 = 1.2 m/s,
or
yB = 6 m
vz
u = —
g
(12)2
u = -— — = 14.6789 m
9.81
(*max)2 = ^ ^ - ( 1 4 - 6 7 8 9 + 1.2 - 6 )-(1 4 .6 7 8 9 )2 = 74.553 m2
(a) Maximum distance:
d = xmax - 4.5
xmax = 8.6344 m
d = 4.13 m A
PROBLEM 11.115 CONTINUED
si*\
1
(b)
Angle
o r.
a
u - 14.6789 = 1.7000
ta n a = *maxtana! = ------*
*max
*max
8.6344
a = 59.535°
Check the edge.
y = y0 + x tan a +
a = 59.5° A
ox2
2(v0co sa)
2
= 1.2 + (4.5)(1.7000)------ (9-81)(4-5) ~2
2[l2cos59.535°]
Since y > 6 m, the stream clears the edge.
y = 6.17 m A
PROBLEM 11.116
A projectile is launched from point A with an initial velocity v0 of
40 m/s at an angle a with the vertical. Determine (a) the distance d to
the farthest point B on the hill that the projectile can reach, (b) the
corresponding angle a, (c) the maximum height of the projectile above
the surface.
SOLUTION
Let P = 9 0 ° - a
x = (v0cos p ) t
Horizontal motion:
t = — -—
v0 cos P
y = (v0 sinp ) t - —t2 = xtan p - —gx2 sec2 p
§
2
Vertical motion:
y B = xB tan 30° = xB tan P - -j gx\ sec2 plvl
At point B,
Solving for xB,
or
fy 2
2
xB = —^-(tan /? - tan30°)cos2/? = ^ -( s in /9 c o s /? - tan30°cos2/?!
g
g v
7
Differentiating with respect to f3 and setting equal to zero,
i
—JL
dP
j 2
“ 'i-icos2 p _ sjn2 p + 2 tan 30° cos P sin /■ i
g v
7 2
= —~ (c o s2 /? + tan 30° sin i p ) = 0
8
tan2/? = - c o t 30° - -a/3
or
p = -30° and 60°.
2 p = -60°
and
120°
Use p = 60°.
Maximum values of xB and d.
(x*)
(a)
(b)
d
max
. 2
2
= ^ - ( t a n 6 0 ° - tan30°)cos2 60 = 0.57735-^8
8
_ (Xs ) max _ 2 v 02 _ 2 (4 0 )2
^ aq
« a
cos 30°
3 g
3 9.81
0 1
a = 90° - P - 90° - 60°
max = 108.7 m <
a = 30° A
PROBLEM 11.116 CONTINUED
At the point of maximum height, the projectile path is parallel to the surface so that
or
= tan 30°
t =^
or
y = (v0 sin 0 ) 0.57735g J
1 Vn
3 g
1 (<
3 9.81
^
g
v0sinyff-g? _
= tan 30°
v0 COS P
(ta n P - tan30°) = .
0
1
r
2
v
5 7 7 3 5
^.
g
\2
0.57735 —
gj
54.366 m
/
\
x = (vqCos/7) 0.57735 — = 0.288675-2g
v
S'y
(0.288675)(40)
9.81
(c)
_
= 47.083 m
hmax= y - x tan 30° = 54.366 - 47.083 tan30°
Iax - 27.2 m <
PROBLEM 11.117
- n
■
A
<
At an intersection car /I is traveling south with a velocity of 40 km/h
when it is struck by car B traveling 30° north of east with a velocity of
50 km/h. Determine the relative velocity of car B with respect to car A.
n
&
'
50
1
0> km/li
SOLUTION
VS
=
V BM
=
^
V «
+
\ B/A
+
y A
= v f l + ( - v /f)
.
Sketch vector addition on a diagram as shown.
Law of cosines:
J/„
/
o
*
- ^
V
v bia
.
vb
=
502 + 402 - (2)(50)(40)cosl20°
=
6100(km/h)2
*
VB/A =
+
^8.1
va
- 2 vb va
cos 120°
=
*C m ^ 1
sin a
—r N
Law of sines:
-j
sin a
=
a
=
=
sin 120°
i
— --------- 1—
| VB m |
40 sin 120°
0.44353
78.1
26.3°,
a + 30° = 56.3°
------------------------- =
y B,A
=
78.1 km/h ^ 5 6 .3 °
<
PROBLEM 11.118
Small wheels attached to the ends of rod AB roll along two surfaces.
Knowing that at the instant shown the velocity v ,( o f wheel A is 1.5 m/s
to the right and the relative velocity \ WA of wheel B with respect to
wheel A is perpendicular to rod AB, determine (a) the relative velocity
\ WA, (b) the velocity \ H of wheel B.
SOLUTION
First solve geometry triangle ABC.
^
. .
Law of sines:
^OO mw,
.<*■
sin B
sin 120°
— — = ---------500
800
\2s> \
\
\
sin B = — sin 120° = 0.54127
800
c
P = 32.77°
V
a + P + 120° = 180°
g Law of angles:
a = 60° - p = 27.23°
\ B = \ A + y B/A
Now
Va
\
QO°
W -cL /
Sketch the vector addition triangle.
90° - a = 62.77°
9 + 60° + 90° - a = 180°
..
\
6
/
\
\
\
A /
^ /
/
e = 180° - 60° - 62.77° = 57.23°
V B/^
»
.
Law o f sines:
sin60°
sin#
---------= ------Vria
Va
1
. .
(«)
vbia =
v,sin60°
1.5sin60°
- „
= .
= 1-545 m/s
sin#
sin 57.23
\ B/A = 1.545 m/s
sin(90° - a )
sin#
---- 1--------- - -------VB
VA
Law of sines:
(b)
62.77° <
v/<sin(90o - a )
1.5sin62.77°
, co, ,
Vr> = —---- ---------- - = --------------- = 1.586 m/s
sin#
sin 57.23°
vs = 1.586 m/s 'N 60° M
PROBLEM 11.119
As slider block A moves downward at a speed of 1.6 ft/s, the velocity
with respect to A of the portion of belt B between the idler pulleys C and
D is \ CD/A = 6.4 ft/s
9. Determine the velocity of portion CD of the
belt when (a) 6 = 45°, (b) 6 = 60°.
SOLUTION
V CD
~
V /( +
V COM
Sketch the vector addition triangle. It need not be to scale.
a = 65° - 9
By law of cosines:
VCD = VA + VCD/A - 2 v AVCD!A C O S «
sin p _ sin o r
By law of sines:
VA
VCD/A
. „ v .s in a
sm p = —------VCD/A
Finally, r - 9 - p
(a)
9 = 45°,
vcd
a = 65° - 45° = 20°
= O-6)2 + (6-4 )2 - (2)(l.6)(6.4)cos20° = 24.275 ft2/s2
vCD = 4.927 ft/s
sin P = 1-6sin20° = 0.1H07,
4.927
y — 45° —4.9° = 38.6°,
(b)
9 = 60°,
vcd
P = 6.4°
\ B = 4.93 ft/s ^
38.6° <
a = 65° - 60° = 5°
= 0 -6 )2 + (6.4)2 - (2)(l.6)(6.4)cos5° = 23.118 ft2/s2
vcd
= 4.808 ft/s
sin p = 1,6sm5° = 0.02900, P = 1.7°
4.808
y = 6 0 °-1 .7 ° = 58.3°,
vd = 4.81 ft/s ^
58.3° <
PROBLEM 11.120
Shore-based radar indicates that a ferry leaves its slip with a velocity
v = 10 knots 7 ^ 65°, while instruments aboard the ferry indicate a speed
of 10.4 knots and a heading of 35° west of south relative to the river.
Determine the velocity of the river.
M
r
SOLUTION
The given velocity vectors are:
y F = 10 knots T’’’ 65°
1 F IR
= 10.4 knots
A
35°
\ F - y R + \ F/R - y F/R + y R
Construct the velocity addition triangle as shown. By law of cosines:
V R
=
VF
+
V F IR
~
^ V F V R /F
COS 1 0 °
= 102 + 1 0 .4 2 - ( 2 ) ( 1 0 ) ( 1 0 . 4 ) c o s 10°
= 3.320 knots2
vR = 1.822 knots
By law of sines:
sin a
sin 10°
F IR
or
sin a = vF 8 in iy = 10stnl0° = 0.95302,
1.822
VF IR
a = 72.37°
7 2 .3 7 °-5 5 ° = 17.37°
y R = 1.822 knots ^
or 1.822 knots at 17.37° south of east
17.37° <
PROBLEM 11.121
The velocities of commuter trains A and B are as shown. Knowing that
the speed of each train is constant and that B reaches the crossing 10 min
after A passed through the same crossing, determine (a) the relative
velocity of B with respect to A, (b) the distance between the fronts of the
engines 3 min after A passed through the crossing.
SOLUTION
B
~
y A
+
y B IA
Sketch the vector addition as shown in the velocity diagram.
By law of cosines:
v b/a
=
va
+
vb
-2 v ^ v flcosl55°
= 802 + 602 -(2 )(8 0 )(6 0 )c o sl5 5 °
= 18.7005 x 103(km/h)2
vB/A = 136.7 km/h
sin a
o Law of sines:
sin 155°
B
V B IA
60 sin 155° ntoCA^
s m a = ---------------= 0.18543
136.7
a = 10.69°
(a)
y B /A
= 136.7 km/h
10.69° <
Determine positions relative to the crossing.
r., = v ,/ = 80— = 4 km
A
60
r B
=M o
r B
~
r A
+
+
y B t
r B /A
= 60( ^ ] / +60f ^ ] / = 7 km P7 25°
Sketch the vector addition as shown.
By law of cosines :
rL
= rA + rl
- 2rArR cos 25°
= 42 + 72 - (2)(4)(7)cos25° = 14.25 km2
(b)
d = r,B / A
d = 3.77 km A
PROBLEM 11.122
Airplanes A and B are flying at the same altitude and are tracking the eye
of hurricane C. The relative velocity of C with respect to A is
rCIA = 470 km/h T7 75°, and the relative velocity o f C with respect to B
is vc/B = 520 km/h ^ 40°. Determine (a) the relative velocity of B with
respect to A, (b) the velocity of A if ground-based radar indicates that the
hurricane is moving at a speed of 48 km/h due north, (c) the change in
position of C with respect to B during a 15-min interval.
SOLUTION
vc =
Begin with
Subtracting
+ \ aA - \ B - \ C/B =
0=
y B/A
o r
=
+ \ OA and vc = vfl + vc/B
~ y AJB
=
y CIA +
( _ V C /fl) =
+ \ CIA - \ aB
~ y C/B
+
V C/A
Sketch the vector addition as shown. By law of cosines:
VB/A =
VC/A
+ VB/A -
2 vC/Av B/A COS 65°
= 4702 + 5202 - (2)(470)(520)cos65°
Va/A
= 284.72 x l O 3 (km/h)2
vB/A = 533.6 km/h
Law of sines:
sin or
sin 65°
C/B
B/A
520sin65°
sin a = --------------= 0.88322
533.6
a = 62.03°,
7 5 ° - a = 12.97°
\ BIA = 534 km/h T7 12.97° A
(a)
Vr = V . + V,CIA
or
v , = y r + ( VC/jf)
Sketch the vector addition as shown.
Law of cosines:
v 2a
vA= v£ + v^/A - 2vc vc/A cos(l80° - 15°)
= 482 + 4702 - (2)(48)(470)cosl65° = 266.79 x 103(km/h)2
vA = 516.5 km/h
PROBLEM 11.122 CONTINUED
Law of sines:
sin p
sin 165°
VC ! A
y A
. _ 470 sin 165°
sin P =
= 0.23551
516.5
or
p = 13.62°
90° - 13.62° - 76.4°
\ A = 517 km/h ^
(b)
(c)
rc/B = W
= (520)
15
60
rc/B = 130 km
76.4° <
40° A
PROBLEM 11.123
Slider block B starts from rest and moves to the right with a constant
acceleration of 300 mm/s2. Determine (a) the relative acceleration of
portion C of the cable with respect to slider block A, (b) the velocity of
portion C of the cable after 2 s.
SOLUTION
Let d be the distance between the left and right supports.
Constraint of entire cable: xB + [xB - x A) + 2 (d - x A) = constant
2vs - 3vA = 0
9
and
9
a 4 = —Og = —(300) = 200 mm/s2
2aB - 3 aA = 0
aA = 200 mm/s2 —-
or
Constraint of point C:2(d - x A) + y aA = constant
~ 2 va +
(a)
v c /a
= °
30(1
~ 2ai
+
a ctA
=
0
ac/A = ~2a A = -2 (-200) = 400mm/s
a(V/) = 400 mm/s2 | A
Velocity vectors after 2s: v A = (200)(2) = 400 mm/s —y c/A = (400)(2) = 800 mm/s }
v c
(b)
=
y A
+
y CIA
vc = 894 mm/s
63.4° A
PROBLEM 11.124
Pin P moves at a constant speed of 8 in./s in a counterclockwise sense
along a circular slot which has been milled in the block A as shown.
Knowing that the block moves up the incline at a constant speed of
4.8 in./s, determine the magnitude and the direction relative to the xy axes
of the velocity of pin P when (a) 9 = 30°, (b) 9 = 135°.
SOLUTION
Vector
1PIA
Vector
x A = 4.8 in./s
a
= 8 in./s
a = 30° + 9 + 90° = 120° + 9
30°
v /> -
y A
v
P/A
(a) 9 = 30°, a = 150°
Method 1: Sketch the vector addition.
By law of cosines:
vp
=
va
+
v p /a
~ 2 va vp i a c o s 6 0 °
= 4.82 + 82 -(2 )(4 .8 )(8 )c o s6 0
vP = 6.97 in./s
By law of sines:
sin/7
sin 60°
P = 36.6°,
(b) 9 = 130°,
or
4.8sin60°
sin B = -------------- = 0.59604
6.97
P + 30° = 66.6°,
\ P = 6.97 in./s ^
66.6° <
a = 120° + 135° = 255°
Method 2: Use unit vectors i and j.
Vector v P/A = 8 c o sa i + 8 sin a j = -2.0706i - 7.7274j
Vector
= 4.8cos30°i + 4.8sin30°j = 4.1569i + 2.4j
xp =
vp
+ x P/A = 2.0863i - 5.3274j
= J(vp )1 + M l = V2.08632 + 5.32742 = 5.72 in./s
_ ( vp)y _ -5.3274 _
= 2.5535
tan <p =
2.0863
or
<p = -68.6°
v„ = 5.72 in./s ^
68.6° <
PROBLEM 11.125
As the truck shown begins to back up with a constant acceleration of
4 ft/s2, the outer section B of its boom starts to retract with a constant
acceleration of 1.6 ft/s2 relative to the truck. Determine (a) the
acceleration of section B, (b) the velocity of section B when t = 2 s.
SOLUTION
For the truck,
aA = 4 ft/s2
For the boom,
aB/A = 1.6 ft/s2
50°
ia ) a fl = a ,4 + a B/A
Sketch the vector addition.
By law of cosines:
a2B = a2A + a%A - 2aAaBIA cos 50°
6 /.A
42 + 1.62 - 2 (4) (1.6) cos 50°
aB = 3.214 ft/s2
Law of sines:
sin a =
a = 22.4°,
(b)
aB!A. sin 50° _ 1.6 sin 50°
an
~ 3.214
0.38131
aB = 3.21 ft/s2 ^
22.4° <
vb = (vb)0 + aB* = 0 + (3-214)(2)
v d - 6.43 fit/s2 ^
22.4° 4
PROBLEM 11.126
A boat is moving to the right with a constant deceleration of l ft/s: when
a boy standing on the deck D throws a ball with an initial velocity relative
to the deck which is vertical. The ball rises to a maximum height of 25 ft
above the release point and the boy must step forward a distance d to
catch it at the same height as the release point. Determine (a) the distance
d , ( b ) the relative velocity of the ball with respect to the deck when the
ball is caught.
&
SOLUTION
Horizontal motion of the ball:
v* = K ) 0 ’
Vertical motion of the ball:
*baii = ( v < V
V.v=Mo ~S'
y» =Mo' - \ g'2' M ‘ ~Mo =-2&
At maximum height.
vr = 0
y = y mM
and
(vy f = 2gynua = (2)(32.2)(25) = 1610 ft2/s2
(vv)o = 40.12 ft/s
At time of catch.
or
Motion of the deck:
y = 0 = 40.12/ + ^ ( 3 2 .2 ) r
/calch = 2.492 s
= 40.12 ft/s j
and
vx = ( vx)0 + a</’ xdcck =(vx)or--aof2
Motion of the ball relative to the deck:
( vw d ) x
* b /o
= (vx)0 - [(v,)„ + a , / ] =
=
K
)
0t -
Mo), =M 0~
(a) At time of catch,
(b)
-a tf
( v x )Qt + - a r f 2
= — a,J~
2
1
yniD=yB
d = xaB = -_ (- l) ( 2 .4 9 2 )( vs/d )* = -(-l)(2 -4 9 2 ) = +2.492 ft/s
d
= 3.11 ft <
or 2.49 ft/s —►
(v#/o)v = 40.12 ft/s |
\ WD = 40.2 ft/s ^
86.4° <
PROBLEM 11.127
Coal discharged from a dump truck with an initial velocity
(v c )Q = 1.8 m/s 7^50° falls onto conveyor belt B. Determine the
required velocity \ B of the belt if the relative velocity with which the
coal hits the belt is to be (a) vertical, (b) as small as possible.
SOLUTION
First determine the velocity vc of the coal at the point where the coal
impacts on the belt.
Horizontal motion:
(vc )x
=
[Mx\
= - L 8 cos50°
= -1.1570 m/s
(vc)2 = [(vc)>,]o-2s(>'-.Vo)
Vertical motion:
= (l.8sin50°)2 - (2 )(9 .8 l)(—1.5)
= 31.331 m2/ s 2
(vc )r = -5.5974 m/s
tan P =
vc =
-5 5974
-1.1570
= 4.8379,
p = 78.32°
Ml + Ml = 32669 m2/s2
vc = 5.7156 m/s,
vc = 5.7156 m/s T7 78.32°
vc = (-1.1570 m /s )i+ (-5.5974 m/s)j
or
Velocity o f the belt:
vs = vfl(-c o sl0 °i + sinl0°j)
• ob = V,. - V„ = V, + (-v»)
Relative velocity:
(a) \ aB is vertical.
(vc/B)^ = 0
(vc/B) = -1.1570 - vfl(-c o sl0 °) = 0,
vfl = 1.175 m/s
vB = 1.175 m/s ^
10°
(b) vc/B is minimum. Sketch the vector addition as shown.
v |/c = vB +
v£
- 2vfivc cos 88.32°
Set the derivative with respect to vB equal to zero.
2vg - 2vc cos 88.32° = 0
vB = vc cos 88.32° = 0.1676 m/s
v B = 0.1676 m/s ^
10° A
PROBLEM 11.128
Conveyor belt A, which forms a 20° angle with the horizontal, moves at
a constant speed of 1.2 m/s and is used to load an airplane. Knowing that
a worker tosses duffel bag B with an initial velocity of 0.76 m/s at an
angle of 30° with the horizontal, determine the velocity o f the bag
relative to the belt as it lands on the belt.
SOLUTION
v* = v0cos30°,
Horizontal motion of bag:
vy = v0 sin 30° - gt
Vertical motion of bag:
t
•
1
y = To + (v0 sin30'“) / - ~ g t ~
2
2
or
t
2
2vosin30°
2 ( y - y 0)
S— ------1 + - 4 — — =
g
g
y = xtan20° = (v0 cos 30° tan 20°)/
At the landing point on the belt:
Then,
x = (v0cos30°)/
t 2 - ^ - ( s in 3 0 ° - cos30°tan20°)f - = & = 0
g
g
9.81
’
v
9.81
t 2 -0.028632/ -0.101937 = 0
/ = 0.33391 s, - 0.30528 s
vx = 0.76 cos 30° = 0.65818 m/s
v,. = 0.76sin30° - (9.8l)(0.3339l) = -2.8957 m/s
vB = 0.658181 - 2.8957j m/s
Motion of the belt:
—1.2cos20°i + 1.2 sin 20° j
= 1.12763i + 0.41042j m/s
Relative motion:
= v« "
= - 0 .4 6 9 4 5 i- 3.306 lj m/s
vb / a
= (-0.46945 )2 + (—3.3061 )2
y BiA ~ 3.34 m/s
81.9° A
PROBLEM 11.129
As the driver of an automobile travels north at 20 km/h in a parking lot,
he observes a truck approaching from the northwest. After he reduces his
speed to 12 km/h and turns so that he is traveling in a northwest direction,
the truck appears to be approaching from the west. Assuming that the
velocity of the truck is constant during that period of observation,
determine the magnitude and the direction of the velocity o f the truck.
SOLUTION
Let the x-axis be directed east and the y-axis directed north.
According to the first observation, the velocity may be represented as
vr =
+ (v™ ), “ 20 t +{vm ) t ^
45° = 20j + (v7-M)i (cos45°i - sin45°j)
From the second observation,
= ( v^ )2 + ( v™ )2 = 12 ^
45° +
( vt/a ) 2
=
-12cos45°i + 12sin45°j +
( v T/A)2 1
Subtracting the two equations gives
0 = 20j + ( v ^ ) (cos45°i - sin45°j) + 12cos45°i - 12sin45°j - (vrM) I
x-components:
(cos45°)(vrMX -
y-components:
-(sin45°)^vr//(
Solving (1) and (2),
Then,
or
{ v T/A^
= -12cos45°
(1)
= -2 0 + 12 sin 45°
(2)
(v™ \ = 46.284 km/h, (% ^ )2 = 20 km/h
\ T = 20j + 16.284(cos45°i - sin45°j) = 11.515i + 8.485j km/h
vr = -12cos45°i + 12sin45°j + 20i = 11.515i + 8.485j km/h
vr = 14.30 km/h ^
36.4° <
PROBLEM 11.130
The paths o f raindrops during a storm appear to form an angle of 75°
with the vertical and to be directed to the left when observed through a
left-side window of an automobile traveling north at a speed of 64 km/h.
When observed through the right-side window of an automobile traveling
south at a speed of 48 km/h, the raindrops appear to form an angle of
60° with the vertical. If the driver of the automobile traveling north were
to stop, at what angle and with what speed would she observe the drops to
fall?
SOLUTION
Let the x-direction be north, and the ^-direction be vertically upward.
From the north bound observer, the velocity vector of the rain is
Vr =
+ v m = 64
+ vWN
A
75° = 64i + vaN (-sin75°i - cos75°j)
According to the south bound observer, it is
N 60° = —48i + vR/s (sin 60°i - cos 60°j)
Subtracting,
0 = 112i + VfuN (-sin75°i - cos75°j) - vR/s (sin60°i - cos60°j)
x-components:
(sin75°)vJj/Ar + (sin60°)vws =112
( 1)
y-components:
-(cos75°)vww + (cos60°)vjj/s = 0
(2)
Solving (1) and (2),
vWN = 79.196 km/h, vR/s = 40.995 km/h
Then,
\ R = 64i + 79.196(-sin 7 5 °i - cos75°j) = -12.4971 - 20.497j km/h
v R = -48i + 40.995 (sin 60°i - cos60°j) = -12.4971 - 20.498j km/h
\ R = 24.0 km/h T 7 58.6° A
PROBLEM 11.131
When a small boat travels north at 3 mi/h, a flag mounted on its stem
forms an angle 9 = 50° with the centerline of the boat as shown. A short
time later, when the boat travels east at 12 mi/h, angle 9 is again 50°.
Determine the speed and the direction o f the wind.
SOLUTION
Let the x-axis be directed east, and the y-axis be directed north.
From data obtained as the boat travels north, the wind velocity is
vh' = (v«), + (vhvb), = 3 \ + (% /B)| N 50° = 3j + ( v ^ J ^ s in S O 0! - cos50°j)
From data obtained as the boat travels east, it is
v »' = ( v « )2 + ( 'V /s ) , = 12 ~ ~ + ( vw/b) ^
50° = 121 + ( v ^ ) 2(-c o s5 0 °i - sin50°j)
Subtracting,
0 = 3j +
(sin50°i - cos50°j) - 12i - (% /B^ (-c o s 5 0 ° i - sin50°j)
x-components:
(sin50°)(vw,/fl)| + (cos50°)(tV/«)2 = 12
(1)
y-components:
( - cos50°)( v^ /b )i + (sin50°)(vH,/B)2 = -3
(2)
Solving (1) and (2),
Then,
or
= 11121 mi/h, (vM
,B)2 = 5.415 mi/h
x w = 3j + 11.12l(sin50°i - cos50°j) = 8.519i - 4.148j mi/h
vH- = 12i + 5.415(-cos50°i - sin50°j) = 8.519i - 4.148j mi/h
v w = 9.48 mi/h ^
26.0° <
PROBLEM 11.132
Instruments in airplane A indicate that with respect to the air the plane is
"I
headed 30° north of east with an airspeed of 300 mi/h. At the same time
radar on ship B indicates that the relative velocity of the plane with
respect to the ship is 280 mi/h in the direction 33° north of east. Knowing
that the ship is steaming due south at 12 mi/h, determine (a) the velocity
of the airplane, (b) the wind speed and direction.
A
11 mi'li
SOLUTION
Let the x-axis be directed east, and the v-axis be directed north.
Airspeed:
= 300 mi/h ^
30° = 300(cos30°i + sin 30°j) mi/h
Plane relative to ship:
v ^g = 280(cos33°l + sin 33°j) mi/h
Ship:
v„ = 12 mi/h 1 = -12j
(a) Velocity of airplane.
\ A = \ B + y ^ g = —12j + 280(cos33°i + sin33°j)
= 234.831 + 140.50j
\ A = 274 mi/h ^
30.9° <
v^. = 26.7 mi/h ^
20.8° <
(b) Wind velocity.
VH- =
+ y W!A = '/ A - ' / AIW
= 234.831 + 140.50j - 300(cos30°i + sin30°j)
= -24.98i - 9.50j
PROBLEM 11.133
The diameter o f the eye of a stationary hurricane is 20 mi and the
maximum wind speed is 100 mi/h at the eye wall, r = 10 mi. Assuming
that the wind speed is constant for constant r and decreases uniformly
with increasing r to 40 mi/h at r = 110 mi, determine the magnitude of the
acceleration of the air at (a) r = 10 mi, (b) r = 60 mi, (c) r = 110 mi.
SOLUTION
(a)
(b)
(c)
r = 10 mi = 52800 ft,
r = 60 mi = 316800 ft,
v = 100 mi/h = 146.67 ft/s
v2
(146.67)2
r
52800
v = 100
4 0 _ 1 0 °(6 0
110 —10 v
v2
(l 02.67 )2
r
316800
r = 110 mi = 580800 ft,
a = 0.407 ft/s2 <
10) = 70 mi/h = 102.67 ft/s
’
a = 0.0333 ft/s2 <
v = 40 mi/h = 58.67 ft/s
v2
(58.67)2
r
580800
a = 0.00593 ft/s2 <
PROBLEM 11.134
At the instant shown, race car A is passing race car B with a relative
velocity of 3 fl/s. Knowing that the speeds of both cars are constant and
that the relative acceleration of car A with respect to car B is
0.9 ft/s2 directed toward the center of curvature, determine (a) the speed
of car A, (b) the speed of car B.
SOLUTION
Velocities:
v .</a = y A ~
= 3 ft/s |
Va
Accelerations:
a,
a A!B ~ &A ~ a l) ~ 0 . 9 ft/S -
(a)
VA
VA
aa =
—
Pa
300
Pb
( v, - 3 ) 2
290
aH =
a AlB
- A _ b L Z~ 2n n) l - o <u y;
300
290
•
8700
Vj
+
145 A
29
VA - lSOv^ +8100 = 0
(b)
vB = 90 - 3
vA = 90 ft/s <
vH = 87 ft/s <
PROBLEM 11.135
Determine the maximum speed that the cars of the roller-coaster can
reach along the circular portion AB of the track if the normal component
of their acceleration cannot exceed 3g.
SOLUTION
2
—
P
*
—
- Pan
v2ax = (25)(3g) = (25)(3)(9.81) = 735.35 m2/s2
vmax = 27.125 m/s
vmax = 97 6 km/h <
PROBLEM 11.136
As cam A rotates, follower wheel B rolls without slipping on the face of
the cam. Knowing that the normal components of the acceleration o f the
points of contact at C of the cam A and the wheel B are 0.66 m/s2 and
6.8 m/s2, respectively, determine the diameter of the follower wheel.
SOLUTION
[w .L *
P
a
vc2 = P a [(*c \ ] a = P b [{»c )h\
PjL
Pa
_
A -
= 0.09706
6.8
p B = 0.09706/?^ = (0.09706)(60) = 5.8235 mm
dB = 2p H = 11.65 mm A
PROBLEM 11.137
Pin A, which is attached to link AB, is constrained to move in the circular
slot CD. Knowing that at / = 0 the pin starts from rest and moves so that
its speed increases at a constant rate of 0.8 in./s2, determine the
magnitude of its total acceleration when (a) / = 0, ( b ) t = 2 s.
SOLUTION
Given:
(VA = 0’
Then,
V A
(a) 1 = 0,
M , = ^ r = 08s
= ( VA + K f ) , ' =
=^
v ,= 0 ,
0 8
'
=0
= 0.800 in./s2 A
°A = M ,
vA = 0 + (0.8)(2) = 1.6 in./s
(b) 1 = 2 s.
(a )
V
=
A’n
V a
p
-
=
=
3.5
[ ( « .) ; + M
0.731 in./s2
lf
= [{ 0 .8 /
+
(0-731 )2] ' 2
a A
= 1.084 in./s2 A
PROBLEM 11.138
The peripheral speed of the tooth of a 10-in.-diameter circular saw blade
is 150 ft/s when the power to the saw is turned off. The speed of the
tooth decreases at a constant rate, and the blade comes to rest in 9 s.
Determine the time at which the total acceleration of the tooth
is 130 ft/s2.
SOLUTION
v = v0 + a,t
For uniformly decelerated motion:
At t = 9 s,
0 = 150 - at (9),
or
at = -16.667 ft/s2
a1 = a 2 + a 2
Total acceleration:
an = [ a 2 - a 2] ^ = [(130)2 - (-16.667)2 ^ = 128.93 ft/s2
Normal acceleration:
v2
an = — ,
p
where
1
5
p = — diameter = — ft
2
12
( c\
v2 = pan = — (128.93) = 53.72 ft2/s2,
1,12/
Time:
v —v0
7.329 - 150
t = ------ 2- = --------------a,
-16.667
v = 7.329 ft/s
t = 8.56 s
PROBLEM 11.139
An outdoor track is 130 m in diameter. A runner increases her speed at a
constant rate from 4 to 7 m/s over a distance o f 30 m. Determine the total
acceleration of the runner 2 s after she begins to increase her speed.
SOLUTION
Tangential component o f acceleration:
yf = Vq +
2at - s0)
a = ^ ~ v° = M — (4) = 0.550 m/s2
2{sx - s 0)
(2)(30)
At %*= 2 s,
v = v0 + att - 4 + (0.550)(2) = 5.10 m/s
Normal component of acceleration:
a,
—
P
p = ~(130) = 65 m, an - —
2
65
where
p = — diameter
2
= 0.400 m/s2
1/2
Total acceleration:
a = [ a 2 + a 2] V2 = (0.550)2 + (0.400)5
a = 0.680 m/s2 M
PROBLEM 11.140
A motorist starts from rest at point A on a circular entrance ramp when
I = 0, increases the speed of her automobile at a constant rate and enters
the highway at point B. Knowing that her speed continues to increase at
the same rate until it reaches 100 km/h at point C, determine (a) the speed
at point B, (b) the magnitude of the total acceleration when t = 20 s.
j— KM) in -»J
SOLUTION
v0 = 0
Speeds:
V, = 100 km/h = 27.78 m/s
5 = ^ (1 5 0 )+ 100 = 335.6
Distance:
Tangential component of acceleration:
v ^ . ( 2 7 78) ^ 0 = , 1495m/s,
2s
(2)(335.6)
'
At point B,
v,' = Vq + 2a,s
v\ = Vq + la ,sB
sB = —( l 50) = 235.6 m
where
v | = 0 + (2)(1.1495)(235.6) = 541.69 m2/s2
vB = 83.8 km/h 4
vB = 23.27 m/s
At / = 20 s,
v = v0 + a,t = 0 + (1.1495)(20) = 22.99 m/s
p = 150 m
Since v < vB, the car is still on the curve.
Normal component of acceleration:
Magnitude of total acceleration:
p
150
|a| = -Jaf + o2 = y(T1495)~ + (3.524)"
U = 3.71 m/s2 <
PROBLEM 11.141
-3 0 0 m -
At a given instant in an airplane race, airplane A is flying horizontally in a
straight line, and its speed is being increased at a rate of 6 m/s2. Airplane
B is flying at the same altitude as airplane A and, as it rounds a pylon, is
following a circular path of 200-m radius. Knowing that at the given
420 km/h
y
2(K) m
pi^
instant the speed of B is being decreased at the rate o f 2 m/s2, determine,
f°r
positions shown, (a) the velocity of B relative to A, (b) the
acceleration of B relative to A .
SOLUTION
420 km/h
(«)
V8 =
+ y B/A
v s = 520 km/h "'v 60°
or
V' BIA
mi =
- V
yB» - V i = V,
Sketch the vector addition as shown.
-
Va
VB/A = VA + V B - 2 v AVb
COS60°
= (420)2 + (520)2 - (2)(420)(520)cos60°
or
v bia
sin a
sin 60°
520 " 477.9
= 477.9 km/h
a = 70.4°
or
v b/a = 478 km/h P
(b)
ctA = 6 m/s2 —~
(##), = 2 m/s2
70.4° n*
60°
vB = 520 km/h = 144.44 m/s
(a g ) =-AL = S.14,44) = 104.32 m/s2 P
v ti,n
p
200
30°
a BIA = a B ~ a A = ( a f i) , + ( a f l) „ ~ &A
= [2 ^
60°] + [104.32 P
30°] - [ 6 — ]
= 2 (-cos60°i + sin60°j) + 104.32(-cos30°i - sin30°j) - 6i
= -97.34i - 50.43j m/s2
aB/A = 109.6 m/s2 P
21 A 0 <
PROBLEM 11.142
Racing cars A and B are traveling on circular portions of a race track. At
the instant shown, the speed of A is decreasing at the rate of 8 m/s2, and
k l lSO
' J ki i ill
500 in
/M
\
the speed of B is increasing at the rate of 3 m/s2. For the positions
shown, determine (a) the velocity of B relative to A, (b) the acceleration
Ini
- ---------------------------------8 0 0 I II
■»
SOLUTION
(m)
\ H = 162 km/h = 45 m/s ^
= 180 km/h = 50 m/s N 30°,
v bia = vfl “
V .4
45°
= 45(cos45°i - sin45°j) - 50(cosl20°i + sinl20°j)
= 56.82i - 75.12j = 94.2 m/s ^
52.9°
vB/A = 339 km/h X 52.9° <
(b)
(a ,<), = 8 m/s2 ^
(
v A,n
60°,
a
pA
400
( a .) =
v R,n
pB
300
=
a « “
(a B)f = 3 m/s2 ^
= 6.25 m/s2
30°
= 6.75 m/s2 ^
45°
* a = ( a «), + (a «)„ " ( a ^), -
45°
( a .r)„
= 3(cos45°i - sin45°j) + 6.75 (cos 45°i + sin45°j)
- 8 (cos 60°i - sin 60°j) - 6.25 ( - cos 30°i - sin30°j)
= 8.3li + 12.70j m/s2,
or
aBIA = 15.18 m/s2 ^
56.8° <
PROBLEM 11.143
A projectile is launched from point A with an initial velocity v0 of
40 m/s at an angle of 30° with the vertical. Determine the radius of
curvature of the trajectory described by the projectile (a) at point A,
(b) at the point on the trajectory where the velocity is parallel to the
incline.
PROBLEM 11.144
The motion of particle P on the elliptical path shown is defined by the
equations x = (2 cos n t - l)/(2 - cos nt) and y = 1.5sin/rr/(2 - cos^t),
where x and y are expressed in meters and t is expressed in seconds.
Determine the radius o f curvature o f the elliptical path when (a) t = 0,
(b) t = 1/3 s, (c) / = 1 s.
SOLUTION
Compute x- and y-components of velocity and acceleration.
Ic o s n t - 1
x =
.
x =
,
2 - cos^-r
- ^ n 1 cos n t
X
-3 ;rsin nt
(2 - cos n t)
6^ sin n t{n sin n t)
2
(2 -
T
(2 -
C O S /T f)
1.5sin;z?
COS7Tf)
. _ 1.5^(2cos^r - l)
2 -co s^ -r’
-3;r2sin;rt
y
II
-1,
=
y
^
=
0,
P
a"
= x =T ’
~
x■ = n0,
n
-y = _ ,
4 n 24r>
v2
’
P ~
v2
a „
s h
i
2a 2
p = 0.75 m -4
ii
=
~ y
x = -3 n 2,
o'
ll
•?\
i
ii
x
=
y = 1.5n,
1
II
■X
1
II
>
a" =
cosjt?)3
v2
(l.5;r)2
p = — = ----- j —
an
3n
y =T
..
l,
v
x - 0,
y = 0,
* = 0,
- j -
(2 -
A.
ii
U)
5*
II
o
x = 1,
1
=
3^-(2cos^rr - l)(^ s in ^ r)
(2 - c o s n tf
v = y = 1.5n,
t
(2 -c o s ^ -r)2
a n
y -
—
n
2
,
*2 3
------J
4
p
3.2n2
~
n
x
-
1.155 m -4
* 2
3
’
p
= 0.75 m 4
PROBLEM 11.145
From a photograph of a homeowner using a snowblower, it is determined
that the radius of curvature of the trajectory of the snow was 30 ft as the
snow left the discharge chute at A. Determine (a) the discharge velocity
v ., of the snow, (b) the radius of curvature of the trajectory at its
maximum height.
SOLUTION
(a) The acceleration vector is 32.2 ft/s {.
At point A. Tangential and normal components of a are as shown
in the sketch.
an = a cos 40° = 32.2 cos 40° = 24.67 ft/s2
CL
v2a=
P
a
M
„ = (
30)(24.67) = 740.0 ft2/s2
\ A = 27.2 ft/s ^
40° <
vx = 27.20 cos 40° = 20.84 ft/s
(b) At maximum height,
v = vv = 20.84 ft/s
an = g = 32.2 ft/s2,
v2
(20.84)“
P =— =—
a„
32.2
p = 13.48 ft <
PROBLEM 11.146
A child throws a ball from point A with an initial velocity v A of 60 ft/s at
an angle of 25° with the horizontal. Determine the velocity of the ball at
the points of the trajectory described by the ball where the radius of
curvature is equal to three-quarters of its value at A.
SOLUTION
g = 32.2 ft/s2,
a = 25°
(a) At point A. Sketch the tangential and normal components of aA as
shown.
(° a )„ = gco sa
v = vA = 60 ft/s
p, = L l = _ J k _ = —
-------= 123.36 ft
an
geos or 32.2 cos 25°
vx ~
va c o s
(b) At point B.
Vfi _
a -
60 cos 25°
=
54.378 ft/s
/% = f A = f (123.36) = 92.52 ft
v*
cos 9
= geos 9 = 32.2 cos 9 ft/s2
and
or
a„ -
(54.37S)2
32.2 cos6
(cos2 9^(92.52)
Pb
cos39 = 0.99257,
vR
B
cos 9 = 0.99752,
9 = ±4.0°
54.378
------------= 54.5 ft/s
0.99752
v g = 54.5 ft/s ^
4.0° <4
and 54.5 ft/s ^
4.0° <
PROBLEM 11.147
Coal is discharged from the tailgate of a dump truck with an initial
velocity v A = 2 m/s 50°. Determine the radius of curvature of the
trajectory described by the coal (a) at point A, (b) at the point o f the
, trajectory 1 m below point A.
SOLUTION
aA = g j = 9.81 m/s21
(a) At point A.
Sketch tangential and normal components of acceleration at A.
(® h )t / + >
(aA)n = geos 50°
a,
(2 )2
P
a =
(aA)n
p A = 0.634 m A
9.81 cos 50°
(6) At point B, 1 meter below point A.
Horizontal motion: (vfl)^ = (v*)^ = 2cos50° = 1.286 m/s —
(v*)J = (vAfy + 2ay {yB - y A)
Vertical motion:
= (2 cos 40° )2 + (2 )(—9.81)(—l)
= 21.97 m2/s2
(vB) = 4.687 m /si
_ (v»)y
tan 9 =
(V»L
4.687
------1.286
or
9= 74.6°
aH = geos 74.6°
-
Pb =
vb
-
_ (1.286)' + 21.97
9.81 cos 74.6°
p B = 9.07 m
A
PROBLEM 11.148
A horizontal pipe discharges at point A a stream of water into a reservoir.
Express the radius of curvature of the stream at point B in terms o f the
magnitudes of the velocities \ A and v R.
PROBLEM 11.149
A projectile is fired from point A with an initial velocity v0. (a) Show
that the radius of curvature of the trajectory of the projectile reaches its
minimum value at the highest point B of the trajectory. (b) Denoting by
6 the angle formed by the trajectory and the horizontal at a given point
C, show that the radius of curvature of the trajectory at C is
P = />min/c <>s30.
PROBLEM 11.150
A projectile is fired from point A with an initial velocity v0 which forms
an angle a with the horizontal. Express the radius of curvature o f the
trajectory of the projectile at point C in terms of x, v0, a , and g.
t
SOLUTION
Let 9 be the slope angle of the trajectory at an arbitrary point C.
V,
(ac ) = g c o s9 = - £ ”
pc
Then,
or
pc = — <
—
geos 9
But the horizontal component of velocity is constant, (vc )v = ( vA)
where
I =
or
* = (vo)x' = {v0co sa )l
(v-*)x = ( vc ) ,
v0 cos a
(l)
and (vc )x = v0cos9
( v^)x = v0cosa
v0co sa = vc-cos#
Then,
pc -
so that
( 2)
gv0cosa
The vertical motion is uniformly accelerated
(vc)„ = (vo)v, ~ g f = v0 sin a -
&
v0cosa
(3)
/
But
vc = (v0); + (v0fv = (v0c o s a )2 + v0s in a - gv0cosa
= v„
or
j _ 2gx tan a +
g 2x 2 ^
4
vn cos 2 a
j _ 2gx tan a +
Vc = vo
0
g 2x 2 ^
Vq
co s2
3/2
(4)
a
Finally, substituting (4) into (2) gives
2 /
v0
geos a{
2gxtanar
7
j 2 \^
gx
v0 cos"2 a,cJ
PROBLEM 11.151
Determine the radius of curvature of the path described by the particle of
Prob. 11.95 when t = 0.
SOLUTION
r = (/ft cos <o„i) i + ctj + (/ft sin «„i)k
Given:
Differentiating to obtain v and a,
v = — = R (cos cont - o)„tsina>nt)i + cj + R(sincont + o)ntcosa>nt) k
a=
dt
dv
~dt
= R{-con sma>nt -
oj
„sin cup - a 21 cos copy + R{con cos a>nt + con cos cop - ou^t sin copy
= /?[(-2tf>n sincont - fi>2/cos< „r)Ji + (lcun coso)nt - <y2/sin&>„/jl
0
Magnitudes of v and a.
2
2
2
2
v = v; + vy + v;
S)
II
= [/?(cos<a„i - <y„/sinty„i)]2 + ( c f + [/?(sino>,/ + <y„icoso„t)]
[c o s2
a>nt ■- 2cup sin cop cos cop + ou2p 2 s i n 2 cup
+ /? 2 [ s i n 2
cont +
2 < y „ /sin
= R 2(l + co2t2j + c2
<3
II
= R
K
cop cos a>nt - co2t 2 c o s
or
v —
^ R 2( l
+
< #)
+ c!
+ a2 + a2
{~2con sin cont - a%tcos cop'j + {2con cos cup - cop sin cup'j
- R 2 [4<a2 sin2 co„t + 4<y2/sin cont cos a>nt + coA
p 2 cos2 cont
+ 4&>2 cos2 cop - Aco\t sin cup cos cop + cuAt2 sin2 cup^
= R 2(A(ol + coAt 2j
or
Tangential component of acceleration:
a = Rcon\j4 + a>2t2
a, =
dv
m
dt
At t = 0,
a = 2Rcun,
v2 = R 2 + c2,
[/?2(l + «>2/ 2) + c2]
at - 0
a„ = ija2 - a2 = 2Rcon
Normal component of acceleration:
But
R 2cun2t
a„ =
v2_
p
.2
v
or
P =—
P =
R +c
2R(o„
PROBLEM 11.152
Determine the radius of curvature of the path described by the particle of
Prob. 11.96 when t = 0, A = 3, and 5 = 1.
SOLUTION
With A = 3
r =
(3tcosf)i
and 5 = 1,
+
13y j t 2
+
ljj
the position vector is
+
(fsin t)k
Differentiating to obtain v and a,
v =
'
= 3(cost - f sinf)i +
3T
^
j + (sint + tco sf)k
\y jt2 + 1 j
+ \-t
I
a=
\y jt~ + 1 J
- 3 (-sin r - sint - tc o st)i + 3
t2 + 1
+ (cost + cost - tsin /)k
3
jjj j + (2cos t - tsin t)k
= -3 (2 sin t + tco st)i +
( ,= + i)
Magnitude ofv2.
2
91
v2 = v2 + v2 + v2 = 9(cost - tsin t) + —
+ (sint + tcost)
r +1
Differentiating,
2v— = 18(cost - tsin r)(-2 sin t - tcost) + — — 2
dt
(l + t2)
+ 2(sint + rcost)(2cost - tsinr)
When t = 0,
a = 3j + 2k,
dv
v = 9,
2v— = 0
dt
a1
= 32 + 22 = 13
dv
a, = — = 0
'
dt
Tangential acceleration:
Normal acceleration:
But
a 2
an = —
p
or
-
a 2 - a ,2
= 13
P =— =
an
Vl3
or
an =
Vl3
p
= 2.50 ft <
PROBLEM 11.153
A satellite will travel indefinitely in a circular orbit around a planet if the
normal component of the acceleration of the satellite is equal to g ( R / r ) ,
where g is the acceleration of gravity at the surface of the planet, R is the
radius of the planet, and r is the distance from the center of the planet to
die satellite. Knowing that the diameter of the sun is 1.39 Gm and that the
acceleration of gravity at its surface is 274 m/s2, determine the radius of
the orbit of the indicated planet around the sun assuming that the orbit is
circular.
Earth: (wmean)orbit = 107 Mm/h.
SOLUTION
For the sun,
g = 274 m/s2
i r r
R = - D = f ~ ) (l.39 x 109) = 0.695 x 10' m
and
Given that an =
gR2
r
v2
and that for a circular orbit an = —
r
gR2
Eliminating an and solving for r,
For the planet Earth,
Then,
r =~ v
v = 107 x 106m/h = 29.72 x 10’ m/s
(274)(0.695 x 109f
= 149.8 x 109 m
r =-------^
(29.72)
r = 149.8 Gm <
PROBLEM 11.154
A satellite will travel indefinitely in a circular orbit around a planet if the
normal component of the acceleration of the satellite is equal to g (/? / r ) ,
where g is the acceleration of gravity at the surface o f the planet, R is the
radius o f the planet, and r is the distance from the center o f the planet to
the satellite. Knowing that the diameter o f the sun is 1.39 Gm and that the
acceleration of gravity at its surface is 274 m/s2, determine the radius of
the orbit of the indicated planet around the sun assuming that the orbit is
circular.
Saturn: (wmean)orbit = 34.7 Mm/h.
SOLUTION
g = 274 m/s2
For the sun,
R - -^D = |^ j ( l . 3 9 x 109j = 0.695 x 109 m
and
Given that a =
gR~
v2
.and that for a circular orbit: an = —
r
r
Eliminating an and solving for r,
v = 34.7 x 106m/h = 9.639 x 103m/s
For the planet Saturn,
Then,
gR2
V
r = —=-
r =
(274)(0.695 x 109 !
/v
’ = 1.425 x 1012 m
(9.639)
r = 1425 Gm <
PROBLEM 11.155
Determine the speed of a satellite relative to the indicated planet if the
satellite is to travel indefinitely in a circular orbit 100 mi above the
surface of the planet. (See information given in Probs. 11.153-11.154).
Venus: g = 29.20 ft/s2, R = 3761 mi.
SOLUTION
gR2
a„ =
”
r2
From Problems 11.153 and 11.154,
v2
a„ - —
r
For a circular orbit,
Eliminating an and solving for v,
For Venus,
v = R J—
g = 29.20 ft/s2
R = 3761 mi = 19.858 x 106 ft.
r
Then,
= 3761 + 100 = 3861 mi = 20.386 x 106 ft
6JV-----29-20
- = 23.766 x 103 ft/s
20.386 xlO6
v = 19.858 x 10
v = 16200 mi/h
-4
PROBLEM 11.156
Determine the speed of a satellite relative to the indicated planet if the
satellite is to travel indefinitely in a circular orbit 100 mi above the
surface of the planet. (See information given in Probs. 11.153-11.154).
Mars: g = 12.24 ft/s2, R = 2070 mi.
SOLUTION
gR2
an = ——
r
From Problems 11.153 and 11.154,
,2
V
For a circular orbit,
an =
Eliminating an and solving for v,
v= R g
For Mars,
r
g = 12.24 ft/s2
R = 2070 mi = 10.930 x 106 ft
r = 2070 + 100 = 2170 mi = 11.458 x 103 ft
Then,
6JV -----12-24
,
11.458 x 10
v = 10.930 x 10
= 11.297 x 103 ft/s
v = 7700 mi/h 4
PROBLEM 11.157
Determine the speed of a satellite relative to the indicated planet if the
satellite is to travel indefinitely in a circular orbit 100 mi above the
surface of the planet. (See information given in Probs. 11.153-11.154).
Jupiter: g = 75.35 ft/s2, R = 44,432 mi.
SOLUTION
ann =
From Problems 11.153 and 11.154,
zR2
2
r
v2
an = —
r
For a circular orbit,
Eliminating an and solving for v,
v= R
g = 75.35 ft/s2
For Jupiter,
R = 44432 mi = 234.60 x 106 ft
r = 44432 + 100 = 44532 mi = 235.13 x 106 ft
Then,
v
=
6)J-—
-35 ,
235.13 x 10
(234.60 x 10
^
= 132.8 x 103 ft/s
v = 90600 mi/h
A
PROBLEM 11.158
Knowing that the radius of the earth is 3960 mi, determine the time of
one orbit of the Hubble Space Telescope if the telescope travels in a
circular orbit 370 mi above the surface of the earth. (See information
given in Probs. 11.153-11.154).
SOLUTION
sR2
an = A-__
r
From Problems 11.153 through 11.157,
2
an = —
r
For a circular orbit,
Eliminating an and solving for v,
v = /?,/—
2nr
t = ----v
For one orbit the distance traveled is 2nr\ hence, the time is
or
For the earth,
Also,
Then,
t =
2xr' 1
rprRgj
g = 32.2 ft/s2
and
R = 3960 mi - 20.908 x 106 ft.
r = R + h = 3960 + 370 = 4330 mi = 22.862 x 106 ft
2^(22.862 x l 0 6f /2
t = 7---- i---------- ------^ 1/2
77 = 5.789 x 103s
(20,908 x 106)(32.2)
t = 1.608 h <
PROBLEM 11.159
A satellite is traveling in a circular orbit around Mars at an altitude of
290 km. After the altitude of the satellite is adjusted, it is found that the
time o f one orbit has increased by 10 percent. Knowing that the radius of
Mars is 3333 km, determine the new altitude of the satellite. (See
information given in Probs. 11.153-11.154).
SOLUTION
gR2
From Problems 11.153 through 11.157,
an - — = -
r
v2
For a circular orbit,
an = —
r
Eliminating a n and solving for v,
v
=
2n r
t = ------
For one orbit the distance traveled is 2n r , hence, the time is
V
2 n r 2! 2
t = ------TTT-
or
Rg1'2
Data:
rx = R
h2
JO
=
3333
+
290 - 3623 km
o
II
r2 =
+ \
K?
II
o►1
ii
J* |- r
N)
u>
Applying to orbits 1 and 2 and taking the ratio, noting that g and R are constant,
(3623)(1.10)^3
=
3861km
= r2 - R = 3861 - 3333
hz = 528 km M
PROBLEM 11.160
Satellites A and B are traveling is the same plane in circular orbits around
the earth at altitudes of 190 and 320 km, respectively. If at / = 0 the
satellites are aligned as shown and knowing that the radius of the earth is
R = 6370 km, determine when the satellites will next be radially
aligned. (See information given in Probs. 11.153-11.154).
SOLUTION
From Problems
1 1 .1 5 3
through
gR 2
an = —=r
1 1 .1 5 7 ,
2
V
an = —
r
For a circular orbit,
v = R .j—
Eliminatinga„ and solving for v,
For one orbit the distance traveled is 2nr\ hence, the time is
2nr
v
2*r3' 2
t =
or
/ =
Rg'rFor satellites A and B,
t .
=
2 nr?*1
----7
7 7
Rg'n-
-
2 m-B^2
and
t R
= -----7 7 7 -
Rg'r-
Let n = number of orbits of B. For the next alignment,
/
.\
(n + l)lA = n tB
or
n
tR I( r„ \ 3' 2
—— = - » = - £
+
1
n
1
fr .V 2
1
fA
,
' A>
Data:
R = 6370 km = 6.370 x 10 m
3
rA = 6370 + 190 = 6560 km = 6.560 x 10 m
3
rB = 6370 + 320 = 6690 km = 6.690 x 10 m
3
Then,
- = f 6-69° X
]
n
1,6.560 x 1 0 v
1 0
- 1 = 0.02987
or
n = 33.475
PROBLEM 11.160 CONTINUED
Time for orbit of satellite B is
3 /2
2/r|(6.690 x 106]
= 1.5137 h
(6.370
x
106)(9.8l)1/2
Time for next alignment is
ntB = (33.475)(1.5137)
ntB = 50.7 h A
PROBLEM 11.161
The rotation o f rod OA about O is defined by the relation
9 = 0.5e~° 8' sin 3nt, where 9 and / are expressed in radians and
seconds, respectively. Collar B slides along the rod so that its distance
from O is r = 0.2 + 1.92/ - 6.72/2 + 6.4/3, where r and / are expressed
in meters and seconds, respectively. When / = 0.5 s, determine (a) the
velocity of the collar, (b) the acceleration of the collar, (c) the
acceleration of the collar relative to the rod.
SOLUTION
Differentiate the expressions for r and 9 with respect to time.
r = 0.2 + 1 .9 2 /- 6.72/2 + 6 .4 /3
r = 1 .9 2 - 13.44/ + 19.2/2
r = -13.44 + 38.4/
9 = 0.5e~°8' sin 3n t
9 = -0.4e~"8' sin 3/r/ + 1.5/re-0 8' cos 3n t
9 = 0.32e"
sin 3nt - 1.2/re“
cos 3nt
-1.2/re ° 8'cos 3/r/ - 4.5/r2e 0 8'sin 3/r/
r = 0.28 m,
At / = 0.5 s,
e"08' = 0.67032,
9 = -0.33516 rad,
(a)
r = 0,
sin 3/zr = -1,
r = 5.76 m/s2,
cos 3/zz = 0
9 = 0.26812 rad/s,
9 = 29.56 rad/s2
Velocity of the collar.
v = rer + r9e0
v = (0.075 m/s)e<; A
vr = 0 ,
(b)
v0 = 0.0751 m/s A
Acceleration of the collar.
a = (r - r 9 2je r + ( rO + 2 r9 ^e0 = arer + a ^te
(c)
ar = 5.76 - (0.28)(0.26812)2
ar = 5.74 m/s2 A
a0 = (0.28)(29.56) + (2)(0)(0.26812)
a0 = 8.28 m/s2 A
a = ^5.74 m/s2)er + ^8.28 m/s2 je„
A
rer = ^5.76 m/s2)er
A
Acceleration of the collar relative to the rod.
PROBLEM 11.162
The oscillation of rod OA about O is defined by the relation
9 = (4 /;r)(sin n t), where 6 and t are expressed in radians and seconds,
respectively. Collar B slides along the rod so that its distance from O is
r = 10/(f + 6), where r and t are expressed in mm and seconds,
respectively. When / = 1 s, determine (a) the velocity of the collar,
(b) the total acceleration of the collar, (c) the acceleration o f the collar
relative to the rod.
SOLUTION
Differentiate the expressions for r and 9 with respect to time.
10
r
.
r =
mm,
t+6
4
9 = —sin rt rad,
n
10
r - — mm;
7
At t = 1s,
10
•
9 - 4 cos nt rad/s
r =
9= 0,
(a)
. . .
^ mm/s,
(/ + 6)2
10
mm/s,
49
9 = -4 rad/s,
20
r =
.2
r mm/s'
(/ + 6)
9 = 4;rsin nt rad/s
i
..
20
.2
r =
mm/s
343
9 =0
Velocity of the collar.
vr = r = 0.204 mm/s,
ve = r9 = -5.71 mm/s
vB = (0.204 mm/s)er - (5.71 mm/s)efl A
(b)
Acceleration of the collar.
ar = r - r& =
- ^ ^ j ( - 4 ) 2 = -22.8 mm/s2
a0 = r 9 + 2 r 9 = ( y j (0) +
(-4) = 1-633 mm/s2
a fl = -^22.8 mm/s2je r + ^1.633 mm/s2
(c)
A
Acceleration of the collar relative to the rod.
20
»bioa = r*r = —
*bioa = (0.0583 mm/s)er A
PROBLEM 11.163
The two-dimensional motion o f a particle is defined by the relations
r = 6(4 - 2e~'j and 9 = 2^2t + 4e~2tJ, where r is expressed in feet, t
in seconds, and 9 in radians. Determine the velocity and the acceleration
o f the particle (a) when t = 0, (b) as t approaches infinity. What
conclusions can you draw regarding the final path of the particle?
SOLUTION
Differentiate the expressions for r and 9 with respect to time.
r = 6(4 - 2<T') ft,
9= 2
6>=2(2r + 4e~2' ) rad,
(a) At t = 0 s,
r = \2e~l ft/s,
r = 12 ft,
9 = 8 rad,
( 2
-
8
r = -12e~'ft/s2
r = 12 ft/s,
9 = -12 rad/s,
vr = r = 12 ft/s,
9 = 32e_2'rad/s2
e~2' ) rad/s
r = -12 ft/s
9 = 32 rad/s
2
2
ve = r 9 = -144 ft/s
v = (12 ft/s)er - (144 ft/s)ee A
ar = r - r92 = -12 - (l2 )(l2
= -1740 ft/s
) 2
2
ae = r9 + 2r9 = (12)(32) + (2 )(l2 )(—12) = 96 ft/s
2
a = -(l7 4 0 ft/s )er + (96 ft/s )e„ A
2
{b) A
e~‘—- 0
t —00,
r « 24 ft,
9 w 4/ rad,
vr = r » 0,
r « 0,
2
e ~2t—►O
and
r « 0
( 9 - 4 rad/s,
^~ 0
« 96 rad/s
v = (96 ft/s)e
or » r —
= -(2 4 )(4 )2 = -384 ft/s2,
0
* 0
a = -(3 8 4 ft/s )er A
2
The particle is moving on a circular path of radius of 24 ft and with a speed of 96 ft/s. The acceleration is the
normal acceleration v2/r = ^ '
24
= 384 ft/s directed toward the center of the circle.
2
PROBLEM 11.164
/
p
/
j/
/\ g
\
The path of a particle P is a limacon. The motion o f the particle is defined
by die relations r - b{2 + cos n t) and 9 = nt, where t and 9 are
expressed in seconds and radians, respectively. Determine (a) the velocity
and the acceleration of the particle when t - 2 s, (b) the value o f 9 for
which the magnitude of the velocity is maximum.
\
'
V
SOLUTION
Differentiate the expressions for r and 9 with respect to time.
r = b { l + cos^t),
r = -;rf>sin;tf,
r = - j?b cos nt
9 - nt,
9
n,
-
'9= 0
{a) At t - 2 s,
sin^f = 0,
r - 3b,
cos^t = 1
r = 0,
vr = r = 0 ,
r - -t? b ,
9 = 2;rrad,
9 = n rad/s
v0 - r9 - 3nb,
v - 3 nbe0 <4
ar = r - r& = -t? b - (3Z>) 7? = -Aj?b
ae = r9 + 2r9= 0,
a - - 4 n 2ber -4
(b) Values of 9 for which v is maximum.
vr = r = -/zAsin nt
Vg = v 9 — - b ( l + cos n t)n
V2
= v2 +
V g
= i?b2 j^sin2 nt + (2 + cos nt)
2]
= 7?b2 ^sin2 nt + 4 + 4 cos nt + cos2 nt~^
= 7?b2(5 + 4cos;ff)
v2 is maximum when
But
cos nt = l
9 = nt,
or
hence
nt = 0,
2n,
4n,
bn, etc
9 = 2Nn, N = 0 ,1 ,2 ,... <4
PROBLEM 11.165
The motion of particle P on the parabolic path shown is defined by the
equations r = 2t\j\ + At2 and 0 = tan_12t, where r is expressed in
meters, 6 in radians, and t in seconds. Determine the velocity and
acceleration of the particle when (a) t = 0, (b) t = 0.5 s.
SOLUTION
Differentiate the expressions for r and 0 with respect to time.
r = 2tVl + 4t2,
r = 2-Jl + 4/2 + 8t2 (l + 412)”'/2
r = 24r(l + 4r2)-1/2 - 32r3(l + 4r2)”3/2,
0 = arctan2t
0 = 2^1 + 4t2j ,
9 = -1 6 r(l + 4t2)"2
(a) At t = 0,
r = 0,
r - 2 m/s,
0 = 0,
0 = 2 rad/s,
vr = r = 2 m/s,
ar = r - r& = 0,
(b) At t = 0.5 s,
0 =0
ve = r0 = 0,
v = (2 m/s)er
ae = r0 + 2r0 = 8 m/s2,
r = y/2 m,
0 = — rad,
4
r =0
r = 3>/2 m/s,
0 = 1 rad/s,
vr = r = 4.24 m/s,
a = ^8 m/s2^
4
-4
r = 5>/2 m/s2
0 = -2 rad/s2
vg = r0 = 1.414 m/s
v = (4.24 m/s)er + (1.414 m/s)e^ 4
ar = r - r & = 5 y f2 - y[2(l)2 = 5.66m/s2
ae = r0 + 2 r'0 = V 2(-2) + (2)(3>/2)(l) = 5.66 m/s2
a=
(5.66 m/s2
+ ^5.66 m/s2
4
PROBLEM 11.166
The two-dimensional motion of a particle is defined by the relations
r = ------and tan# = 1 + 4", where r and # are expressed in
sin# - cos#
t
meters and radians, respectively, and t is expressed in seconds. Determine
(a) the magnitudes of the velocity and acceleration at any instant, (b) the
radius of curvature of the path. What conclusion can you draw regarding
the path of the particle?
SOLUTION
x
cos # = — and
r
Change to rectangular coordinates.
Equation of the path:
Differentiating,
r
y ~x
y = x + 1.
. y
x +1 ,
1 ,
1
tan # = — = ------ = 1 + — = 1 + -=X X
x
t
x = t2
vx =
and
y - t2 + 1
= 21,
Vy = y — 2t
ax = x = 2,
ay = y = 2
(a) Magnitudes:
X
NJ
from which
or
+
Also,
1
y_ _ *
r
r
y - x =1
n.
II
from which
1
sin # - cos #
y
sin # = —
r
/ 2
2
a = ,Jax + ay
V
= 2yf2t 4
a = 2-J2 4
▲
ll
Hence,
8
(b) y = x + 1 is the equation of a straight line.
PROBLEM 11.167
To study the performance of a race car, a high-speed motion-picture
camera is positioned at point A. The camera is mounted on a mechanism
which permits it to record the motion of the car as the car travels on
straightaway BC. Determine the speed of the car in terms of b, 9, and 9.
SOLUTION
Sketch the directions of the vectors v and
vg = v • t e = -vcos 9
ve = r9
But
Hence,
r0 = -v c o s0
r=
But from geometry,
b'9
= -vcos 9
cos 9
or
cos 9
v=-
b9
cos2 9
Speed is the absolute value of v.
v=
b9
cos2 9
PROBLEM 11.168
Determine the magnitude of the acceleration of the race car of
Prob. 11.167 in terms of b, 6, 9, and 9.
SOLUTION
r =
From geometry,
cos 9
r =
Differentiating with respect to time,
6sin 99
cos2 9
Transverse component of acceleration
a0 = r9 + 2 r 9 =
b9
cos 9
2bs\n9fr
cos' 9
( 1)
Sketch the directions of the vectors a and
a() = a • e0 = - a c o s 9
(2)
Matching from (1) and (2) and solving for a,
b'9
2b sin 9&
a = ------ ^-------------j---cos' 9
cos 9
= -----^ —[9+ 2 tan##2)
cos'
'
Since magnitude of a is sought,
|a| =
cos2 9
9 + 2 tan 9&
PROBLEM 11.169
After taking off, a helicopter climbs in a straight line at a constant angle
jj. its flight is tracted by radar from point A. Determine the speed of the
helicopter in terms of d, jj, 6, and 9.
PROBLEM 11.170
Pin C is attached to rod BC and slides freely in the slot of rod OA which
rotates at the constant rate co. At the instant when /? = 60°, determine
(a) r and 6, (b) r and 0. Express your answers in terms o f d and co.
SOLUTION
Looking at d and /? as polar coordinates with d = 0,
Vp = d p = da),
vd = d = 0
ap = d p + 2d p = 0 ,
r = d j 3 for angles shown.
Geometry analysis:
(a) Velocity analysis:
ad = d - d f t = - d a r
Sketch the directions of v, er and e#.
vr = r = v • e = do>cosl20°
,? e
r = — da) A
2
v0 = rO = v • eg = da)cos30°
i _ <ftycos30° _ da)^-
7
Q = —a) A
"W T
2
Sketch the directions of a, er and
a. - a • er = a cos 150° =
73
2
da?
r -r & = - J L d a ,2
2
r = - ^ - d c o 1 + rd2 =
a
-
d a r + d\[3 —a)
— dar A
4
d u ?
ae = 3 ' eg = d<y2 cos 120° = -T-r/a;2
ag = r d + 2r 0
PROBLEM 11.171
For the race car of Prob. 11.167, it was found that it took 0.4 s for the car
to travel from the position 6 = 60° to the position 6 = 35°. Knowing
that b - 80 ft, determine the average speed of the car during the 0.4-s
interval.
PROBLEM 11.172
'
1
For the helicopter of Prob. 11.169, it was found that when the helicopter
was at B, the distance and the angle of elevation of the helicopter were
r = 900 m and 0 = 20°, respectively. Five seconds later, the radar
station sighted the helicopter at r = 996 m and 9 = 23.1°. Determine
the average speed and the angle of climb /? of the helicopter during the
4-s interval.
SOLUTION
Sketch the initial and final position vectors.
Using the law of cosines
(Ar)2 = 9002 + 9962 - (2)(900)(996)cos(23.1° - 20.0°)
= 11839 m2
Ar = 108.8 m
Ar
108.8
vnv„
avg = — = -------- = 27.2 m/s
At
Average velocity:
Vavg = 97.9 km/h A
Using the law of sines,
Ar
sin(A£?)
900
s i n ( / ? - 23.1°)
• in
900sin3.1°
sin(y9- 23.1°) = — — ^— = 0.44731
108.8
p - 23.1° = 26.57°
or p = 49.7°
A
PROBLEM 11.173
A particle moves along the spiral shown. Determine the magnitude o f the
velocity o f the particle in terms of b, 6, and 0.
SOLUTION
r = beU2(?,
r = bem ^ 0 6
vd =
r9= bem * 6
+
KJ
N
>
II
+
<Si
II
vr = r = be1'2* 60,
v
= b e ^ 2(6 2 + l f 26 <
PROBLEM 11.174
A particle moves along the spiral shown. Determine the magnitude of the
velocity of the particle in terms of b, 9, and 6.
SOLUTION
PROBLEM 11.175
A particle moves along the spiral shown. Knowing that 9 is constant and
denoting this constant by co, determine the magnitude of the acceleration
of the particle in terms of b, 9, and co.
SOLUTION
r = bem e\
1/20"
r = bem
e' 99,
ar = r - r9 2 = beme
r = bem °21(9 9 )' + 9 Z + 99
(99)2 + 9 2 + 9 9 - 9 2j = bem$21(##)2 + 99
J/202
a0 = r9 + 2r9 = be1
12"*9 + 2bem °2992 = bem<>1 [ # + 2##2]
But
9= co
ar = bey2^ (9cof
and
and
9 -0
ae = beV2^ [29at2^
a2 = a2 + a 2 = [bem ^ { ( ? + 4(t)a>A
a = bem *
\l/2 ,
+ 4j at 4
PROBLEM 11.176
A particle moves along the spiral shown. Knowing that 9 is constant and
denoting this constant by co, determine the magnitude o f the acceleration
of the particle in terms of b, 9, and co.
SOLUTION
b
r =— ,
92
.
2b x
r = — T0,
9
..
2b v 6b x2
r =—
+ —I 9 Z
93
94
a = r - r92 = ~ Qi 9 + %
e A9 2 - \0 2d 2 - \ { - 2 9 e9 \ +\ 69 2 - 0 20 2))
a0
But
' 2b'
\ 9
92
v 9\
9= co
ar = -K-lo - 9 2)a>2
r 94 '
’
9 2 = \ ( 9 9 - 402)
9
1
and
and
9 -0
ad =
\co2
93
„4\ 7 16b2
b2 i .............. ,
a2 = aj2 + aj = — (36 - 129 2 + 9 4)co2 +
-co
9b
= -^-(36 + 49 2 + 9 4)co2
a = 4 -(3 6 + 49 2 + 0 4f 2 co2 <
&
PROBLEM 11.177
Show that r = h/fis'inO knowing that at the instant shown, step AB of the
step exerciser is rotating counterclockwise at a constant rate <f>.
SOLUTION
Sketch the geometry.
Law of cosines:
r 2 = d 2 + h2 - 2dhcos<p
Differentiating with respect to time and noting that d and h are constant,
2rr = 2dhs\n (pip
r =
Law of sines:
so that
dhs'm<p.
<p
r
sin (p
r
sin 0
d
r = hs\nd<p
Q.E.D 4
PROBLEM 11.178
The three-dimensional motion o f a particle is defined by the cylindrical
coordinates (see Fig. 11.26) R = A /( t + 1), 6 = Bt, and z = C tl(t + l).
Determine the magnitudes o f the velocity and acceleration when
(a) t = 0, (b) t = oo.
SOLUTION
R = —— ,
t+1
Given:
6 = Bt,
z = ^
t+ l
Differentiating with respect to time,
t . —
* .
t>=B,
(,+ i
0 = 0,
R=—
(t + i r
(a) t = 0.
R = A,
R = -A ,
0 = B,
z =C
R = 2A,
0 = 0,
z = -2 C
v0 = R 0 = AB,
R - R 0 2 = 2 A - AB2a \ = 4 A2
ae = R0 + 2R0 = 0 - 2AB
t
=
oo.
R
=
vz = z = C
0,
a 2e
0
a2
4 A 2B 2 +
A 2B 4
=
4 C2
+ a2 = 4A2 + A 2B 4 + 4C2
z
= co,
=
R = 0,
vr
v = ylA2 + A 2B 2 + C 2 <
a2e = 4A 2B
= z = -2 c
a2 = a \ +
(b )
z = 0
= v \ + v2g + v] = A2 + A2B 2 + C 2
aR =
a z
(i
i =- 0 = 0,
vR = R = -A ,
V2
f(, + i
= R = 0,
ar = R - RO2 = 0,
C,
R
0 =
v0 = R0 = 0,
=
0,
= \l4A 2 + A 2B 4 + 4C2 <
a
0,
0
=
iB ,
z
=
0,
z =0
vz = z = 0,
ae = R0 - R 02 = 0 ,
v = 0 -4
az = z = 0,
a =0A
PROBLEM 11.179
The motion of a particle on the surface of a right circular cylinder is
defined by the relations R = A, 9 - 2 K t , and z = A t2 / 4, where A is a
constant. Determine the magnitudes of the velocity and acceleration of
the particle at any time t.
SOLUTION
R = A,
In cylindrical coordinates.
A t2
z ------4
9 = 2 M,
Differentiating with respect to time,
Velocity vector:
R = 0,
9=
R = 0,
9= 0,
2 k,
vr = R = 0 ,
z = —
2
z =—
2
v. = z =
ve = R 9= 2 k A ,
v2 = v2r + V0 + v2 =
0
At
T
+ 4 7?A 2 + ^ 4 2t 2
v=
Acceleration vector:
— A \ J i 6 tc2
2
+ t2 <
ar = R - R & = 0 - 4 tT A
ag = R9 + 2R9 =
0,
a2 = a2R + a2g + a2 =
a, = z = A12
16 k 4 A2
+
0
+
i
A2
a = —A \ I 6 4 k 4 + 1 A
2
PROBLEM 11.180
For the conic helix of Prob. 11.95, determine the angle that the oscillating
plane forms with the y axis.
SOLUTION
r = (flrcos<y„f)i + ct\ + (fltsin cont)k.
From problem 11.95, the position vector is
Differentiating to obtain v and a,
dr
dt
d\
a = — = R[-co„s\ncont dt
'
v = — = R (cosa>nt - co„tsincont ) i + cj + /?(sin*y„f + cont coscont)k
sincont - co\tcoscont ) i + R (concoscont + concoscont - co2:t sincont )k
'
'
'
= R ^-2 co ns'm cortt - co2t cos
vx a=
I
j
k
V
Yx
Vvy
V
vz
+ (2&»„cos<w„/ - c^/sin <u„/jkj
= (VA “ vzay ) I + (v;ax - vxaz ) j + [yxay - vv.ax)l
ax Oy o2
=
r
[2con cos cont - <y^rsin<yn/ j j i + j^/?2( s i n + <y„f cos<y„f)(-2<yns in u y - copcoscop^
- R 2(coscop - cop sin cop )^2co„ cos cop + ^-cR^-lcO" sinft>„/ -
6
sin<y„/)J j
), cos<y„/)jk
7
= cRcon( 2 coscop - o)nts\ncont)\ - R 2co„\2 + co2t 2^jj + cRco„(2s\ncop + cont coscont ) k
| v x a |=
\c W n(4 +a ,y ) + RWn(2 + coy)2
vx a
|vx a|
The binormal unit vector eh is given by
eh =-
Let a be the angle between the y-axis and the binormal.
(v x a ) ■j
cos a = eA • j = ------—- --- ----------i-x° i
Lei
A = R 2(on(2 + o ) l r \
P
B
B = cRo]„ 14 +
R 2con(2 + cont)
+* y )+
j
+
« > y )]
C = -Ja 2 + B2 sothatcoser =
as
shown in the sketch. The angle that the osculating plane makes with the
y-axis is the angle p.
A
R (2 + a> y)
2\'/2
:(4 * my )
P = tan"
R (2 + *> y)
c(4 +a ,y f
PROBLEM 11.181
Determine the direction of the binormal of the path described by the
particle of Prob. 11.96 when (a) t = 0, (b) t = n i l s.
SOLUTION
For A
= 3 and B = 1, r
= (3fcos/)i + |3\ l r
+1j j + (/sin /)k
Differentiating to obtain v and a.
v = — = 3(cosr - fsin /)i + 3 r ~ — j + (sinf + rco s/)k
dt
dv
a = — = 3 (-2 sin f - rco s/)i + 3—
*
3/2
j + (2cot/ - tsin /)k
(t2 + l)
(a) At t = 0,
v = 3(1 - 0)i + (0) j + (0)k = 3i
a = -3(0)i + 3(l)j + (2 - 0)k = 3j + 2k
and
i J k
v x a = 3 0 0 = - 6 j + 9k
0 3 2
| v x a | = yj62 + 92 = 10.817
eh = . v. x-a.. = -0.55470j + 0.83205k
|vxa |
cos6>v = 0,
cos0v = -0.55470,
cos ft = 0.83205
9X = 90°,
n
(b) At t = — s,
2
v = -4.71239i + 2.53069j + k
a = -6i + 0.46464j - 1.5708k
i
j
v x a = -4.71239 2.53069
-6
k
1
0.46464 1.5708
= -4.43985 i - 13.4022 j + 12.9946 k
ft. = 123.7°,
f t = 33.7° <
PROBLEM 11.181 CONTINUED
|v
X
a I= [j4.43985)2 + (13.4022)2 + (12.9946)2J 2 = 19.1883
e. -
V — = -0.23138i - 0.69846j + 0.67721k
|v xa |
cos 0X = -0.23138,
cos Gy = -0.69846,
cos 9Z = 0.67721
Gx = 103.4°,
0y = 134.3°,
)z = 47.4° A
PROBLEM 11.182
The acceleration of a particle is defined by the relation a = A - 6t2,
where A is a constant. At t = 0, the particle starts at x = 8 m with
v = 0. Knowing that at t = 1 s, v = 30 m/s, determine (a) the times at
which the velocity is zero, (b) the total distance traveled by the particle
when t = 5 s.
SOLUTION
— = a = A -6 t2
dt
JId v -
- 6t 2^dt
or
30 = 4 1 ) - (2)(1)3
When t = 1 s, v = 30 m/s.
v = A t - 213
A = 32 m/s2
or
v = 3 2 1 —2t2 — 2 f ( l 6 - t 2)
Then,
t = 0,
(a) When v = 0,
and
t = ±4 s,
t -
Rejecting the negative time,
(b) For 0 < t < 4 s,
v is positive,
x is increasing.
For t > 4 s,
v is negative,
x is decreasing.
dx = vdt
or
0
,
r = 4s
dx = Jj(32f - 2t3^dt
x - 16f2 - V
2
At t = 0,
At t - 4 s,
At t = 5 s,
x = 0
^ = (16)(4)2 - ^ ( 4 ) 4 = 128m
* = (16)(5)2 - } ( 5 ) 4 = 87.5 m
0 < t < 4 s,
distance traveled = 128 - 0 = 128 m
4 s <, t < 5 s,
distance traveled = 128 - 87.5 = 40.5 m
Total distance traveled = 128 + 40.5
d = 168.5 m <
PROBLEM 11.183
At t = 0, a particle starts from x = 0 with a velocity v 0 and an
acceleration defined by the relation a = -5 /(2v0 - v ) , where a and v are
expressed in ft/s and ft/s, respectively. Knowing that v = 0.5v0 at
t = 2 s, determine (a) the initial velocity of the particle, (b) the time
required for the particle to come to rest, (c) the position where the
velocity is 1 ft/s.
SOLUTION
- —
a - —
2v0 - v
dt
Given:
dt =
or
Integrating, using appropriate limits
(2v0 - v)<afv
£<* = ~C0j (2vo - v ) ^
1 2
2
t = —-V0V + - 1- V2 - — v
5
10 vo 10
2
3 2
v0v + — vn
5 0
10 0
V = —Vn
(a) At t - 2 s,
2 = I —— —+ — ] Vq = —Vq
40 5 10/
8
Then,
or
Vq = 16 ft2/s2,
0
Vn
= ±4 ft/s -4
v= 0
(b) Time to come to rest.
t = 0 - 0 + — vl = — (4)2
t = 4.8 s <
10V !
10 0
(c) Position where velocity is 1 ft/s.
vdv - adx
or
dx =
= _ “ (2v0 - vjvdv
Integrating, using appropriate limits,
£ ^ = “ - r 0(2vov - v 2) ^ = -
1
VnV2 - - V
3
v0
X
2
=
1
v0v - - V
3
2
3
~ -v 0
±4v2 - | v 3 - |( ± 6 4 )
With v = 1 ft/s,
x =—
5
x = 7.80 ft and -7.67 ft <
PROBLEM 11.184
10uii/h
wC m la O .
Assuming a uniform acceleration of 11 ft/s2 and knowing that the speed
„ of a car as it passes A is 30 mi/h, determine (a) the time required for the
car reach B, (b) the speed of the car as it passes B.
160fl ------ -j
SOLUTION
xA = 0,
Let the origin lie at point A.
vA = 30 mi/h = 44 ft/s,
(a)
XB = X A
xH = 160 ft
a = 11 ft/s2
1 2
+V + -o r
5 . 5 r + 4 4 / - 160 = 0
/ = -10.71 s, 2.71 s
/ = 2.71s <4
(b)
vB = vA + at = 44 + (l l)(2.7l) = 73.86 ft/s
vB = 50.4 mi/h -4
PROBLEM 11.185
Block B starts from rest and moves downward with a constant
acceleration. Knowing that after slider block A has moved 400 mm its
velocity is 4 m/s, determine (a) the accelerations of A and B, (b) the
velocity and the change in position of B after 2 s.
SOLUTION
Constraint of cable with x A increasing to the left and y B increasing downward.
xa
+ 3AyB = 0
= constant,
+
VA
+ 3vs = °.
aA
+ 3as =
0
Block B moves downward so that vB and aB are positive.
Since, vA = -3vB,
is negative, i.e. to the right.
For block A,
va
(a)
(
b)
"
va
- 21*A
aB
=
a
At t
=
~ ^ °
a
M )2
(2)(-0.4) -
_
-
-
~
6.67 m/s"
= 2aAAxM
, 0 m/~2
20 ^
{
n
r
U 1
a
a
Oft
—
—
^
a #
£m\J II Yl l /l d/ c ______
•
6.67 m/s
=
^
^
1 4
2 s,
vB
A
=
yB =
aBt
=
\ ° B '2
(6.67)(2)
=
{(6-67)(2)2
vB
=
Ay B
13.33 m/s2
=
13.33 m
[ <
PROBLEM 11.186
a(m/**)
A particle moves in a straight line with the acceleration shown in the
figure. Knowing that the particle starts from the origin with v0 = -2 m/s,
(a) construct the v - t and x - t curves for 0 < / < 18 s, (b) determine
the position and the velocity of the particle and the total distance traveled
when t = 18 s.
8
-075
#<*:
12
SOLUTION
Compute areas under a - t curve.
a <V>/s*'
A, = (-0.75)(8) = -6 m/s
'
A2 = (2)(4) = 8 m/s
■ /K
A.>
-o.is
2K
#
A3 = (6)(6) = 36 m/s
•'
m zzSi
t (
s)
v0 = -2 m/s
v8 = vo + 4 =
m/s
v,2 = vg + A2 = 0
(tyi/s")
vlg = 36 m/s A
VI8 = v !2 + A i
Sketch v —t curve using straight line portions over the constant
acceleration periods.
(5/
Compute areas under the v - 1 curve.
1
A> = j ( - 2 - « ) ( 8 ) = -40 m
X (* )
Si
4 = { (-» )(« ) * - ‘6 m
1
A6 = -(3 6 )(6 ) = 108 m
t(s')
xg = Xq + A4 = -40 m
xl2 =
*18 = * 1 2
Total distance traveled = 56 + 108
+ Ai = “ 56 m
+
4
,
x|g = 52 m
A
d = 164 m
A
PROBLEM 11.187
A temperature sensor is attached to slider/IS which moves back and forth
through 60 in. The maximum velocities of the slider are 12 in./s to the
right and 30 in./s to the left. When the slider is moving to the right, it
accelerates and decelerates at a constant rate of 6 in./s2; when moving to
the left, the slider accelerates and decelerates at a constant rate of
20 in./s2. Determine the time required for the slider to complete a full
cycle, and construct the v - / and x - t curves o f its motion.
SOLUTION
Moving to the right
V ( >v1 / & )
vm ax= 12in-/s
(v > 0),
l«max 1= 6in./s2
12 .
t, = vm
—axx = —
= 2s
l«max I
t3 - t 2
6
= — = 2s
A = ^ ( 12)( 2) = 12 in- A = |( 1 2 ) ( 2) = 12“ •
Total distance = At + A2 + A3 = 60 in.
4 2 = 60 - 12 - 12 = 36 in.
A
t2 - t \ = ------vmax
36
„
— = 3s
12
tj = 2 s, t2 = 5 s, t3 = 7 s
x2 = 12 + A2 = 48 in.
Moving to the left (v < 0),
30
m. ni = ~
x, = Ax = 12 in.
x3 = 48 + A3 = 60 in.
vmin = -30 in./s,
= l . 5s>
20
t
- f
|amax | = 20 in./s
vm;„ 1
30
^max I
20
= 1W
JB & l = —
= 1.5 s
A4 = |(-3 0 )(1 .5 ) - -22.5 in.,
A6 = |(3 0 )(l.5 ) = -22.5
A4 + A5 + A^ = -60
As = -60 - A6 = -15 in.
so that
Ac
-15
t5 - t4 = -22- =
= 0.5 s;
vmin
-3 0
t4 = 8.5 s, t5 = 9 s,
x4 = 60 - 22.5 = 37.5 in.,
t6 = 10.5 s 4
x5 = 37.5 - 15 = 22.5 in.,
x6 = 0
PROBLEM 11.188
A group of children are throwing balls through a 0.72-m-inner-diameter
tire hanging from a tree. A child throws a ball with an initial velocity v0
at an angle of 3° with the horizontal. Determine the range of values of
v0 for which the ball will go through the tire.
ii t '
u
SOLUTION
x = v0 cos at,
Horizontal motion:
t =
Vn =
v0 cos a
tc o s a
1 •>
1 ,
y = y 0 + v0sin a / - - g f = y 0 + x ta n a - - g f
Vertical motion:
(2 _ 2(y0 + x t a n a - y)
g
To strike point C:
y = 0.25 m
>2. 2Q.5 t 6 ,a „ 3 ° - 0.25) .
9.81
t = 0.56476 s
vn = -------------0 0.56476cos3°
=10.64m/s
y = 0.25 + 0.72 = 0.97 m
To strike point B:
2 (1 .5 + 6 - 3 - - M 7 )
.,
9.81
t = 0.41492 s
6
Vn =
0.41492 cos 3°
= 14.48 m/s
Range of values for v0 :
10.64 m/s < v0 < 14.48 m/s 4
PROBLEM 11.189
$
4
i
l
^
A
J
I* * * - - , *
An oscillating garden sprinkler which discharges water with an initial
velocity v0 of 8 m/s is used to water a vegetable garden. Determine the
distance d to the farthest point B that will be watered and the
corresponding angle a when (a) the vegetables are just beginning to
grow, (b) the height h of the com is 1.8 m.
SOLUTION
Horizontal motion:
x = (v0) t - v0cosa t
Vertical motion:
t \ t - - g1 t ~’ = v0sina t - - 1g r 2
y = {v0)y
or
y = x ta n a
Set y = 0.
x tan a =
x =0
or
t -
v0cosa
gx~
— T~
2v0 cos“ a
gx~
2vn2 cos 2 a
2 va .
•'n . _
x = —-s in a cosa = — sin 2a
g
g
or
or a = 45° A
sin 2a = 1
But, for maximum x,
2
(a) Then,
X
=
i - I S
g
9.81
6 5 2 m
<
a = 58.2°
A
x = 5.84 m
A
*m ax
=
(b) The height of com is 1.8 m.
Set x = 1.5 m and y = 1.8 m
for stream to clear the com.
h = x ta n a
„2
4 , = x t a n a - -^-H l + tan2 a)
2v” V
2Vo cos2 a
”
tan" a
or
2
tan" a -
2v2/»
2v;
- ta n a + 1 + —
= 0
gx
gx
\2 i
(2)(8)2
(2)(8)J (1.8)
■ — ■-. tan a+ 1 + '
—2 = 0
(9.81)(1.5)
(9.8l)(l.5)
tan2 a - 8.6986 tan a+ 11.4383 = 0
ta n a = 7.0839 and 1.6147,
Using
a = 82.96°, 58.23°
a = 58.23° and y = 0
x = — sin 2 a = ^ - s i n ( l 16.46°)
g
9.81
V
'
PROBLEM 11.190
Knowing that at the instant shown block A has a velocity of 8 in./s and
an acceleration of 6 in./s2 both directed down the incline, determine
(a) the velocity of block B, (b) the acceleration of block B.
SOLUTION
Constraint:
2xa
+
x bi a
Then,
=
constant,
2 v A + v B/A
v b/a = ~ 2 v a = " ( 2 ) ( 8 ) =
a B/A
=
~2aA
=
0,
2 a A + a BIA
=
0
-16in./s
= ~ (2)(6) = -12 in./s2
(a) The velocity vectors are as follows:
= 8 in./s '^ 2 5 ° , \ WA = 16in./s i^ 4 0 °
v« =
Sketch the vector addition as shown.
Law of cosines:
vl =
va
+ vb/a ~ 2 vavb cos 15°
= 82 + 162 -(2 )(8 )(l6 )co sl5 °
vfi = 8.53 in./s
. .
Law of sines:
vB
8.53
= 0.24280,
sin B sin 15°
— — = ---------
P = 14.1°
P + 40° = 54.1°,
v„ = 8.53 in./s ^
54.1° A
0b) The acceleration vectors are as follows:
aA = 6in./s2 ^
25°,
aB/A = 12 in./s2
40°
Using a vector addition diagram like that used for velocity.
o| =
aA
+ a\iA - l a AaWAcos\S° = 6: + 122 - (2)(6)(l2)cosl5
aB = 6.40 in./s2
. _ a .s in l5 °
6sinl5°
sin P =
= ------------ = 0.24280,
aB
6.40
P+ 40° = 54.1°
/? = 14.1°
a fl = 6.40in./s2 i* . 54.1° 4
PROBLEM 11.191
To test its performance, an automobile is driven around a circular test
track of diameter d. Determine (a) the value o f d if when the speed of the
automobile is 72 km/h, the normal component of the acceleration
is3.2 m/s2, (b) the speed of the automobile if d = 180 m and the normal
component of the acceleration is measured to be 0.6g.
SOLUTION
v = 72 km/h = 20 m/s,
(a)
a„=
v2
P
,
P=
v2
an
d - 180 m,
(b)
-
an = 3.2 m/s2
(20)2
' = 125 m,
d = 2 p = 250 m 4
3.2
p = —d = 90 m,
2
an = 0.6g = (0.6)(9.8l) = 5.886 m/s2
2
a„ = —
P
or
v2 = pan = (90) (5.886) = 529.74 m2/s2
v = 23.016 m/s
v = 82.9 km/h 4
PROBLEM 11.192
A motorist traveling along a straight portion of a highway is decreasing
the speed of his automobile at a constant rate before exiting from the
highway onto a circular exit ramp with a radius of 560 ft. He continues to
decelerate at the same constant rate so that 10 s after entering the ramp,
his speed has decreased to 20 mi/h, a speed which he then maintains.
Knowing that at this constant speed the total acceleration of the
automobile is equal to one-quarter of its value prior to entering the ramp,
determine the maximum value of the total acceleration of the automobile.
SOLUTION
On the ramp at constant speed:
v = 20 mi/h = 29.333 ft/s
± . ! » * £ . ts x s r t
p
560
a, = 0
a = 1.5365 ft/s2
a = (4)(l .5365) = 6.146 ft/s2
Before entering ramp :
an = 0
for straight path
a; = a2 - a2n = 37.774 ft2/s4
a, = -6 .1 4 6 ft/s2
(negative for decelration)
Speed v() at entering the ramp:
v = v0 + a,t
29.333 = v0 - (6.146)(10),
v0 = 90.79 ft/s
At entering the ramp:
"
Vo_ = (90.79) = 14.7205 ft/s2
p
560
a, = -6.146 ft/s2
a =
+ a~ = 15.95 ft/s2
a = 15.95 ft/s2 A
PROBLEM 11.193
A golfer hits a golfball from point A with an initial velocity of 50 m/s at
an angle of 25° with the horizontal. Determine the radius of curvature of
the trajectory described by the ball (a) at point A, (b) at the highest point
of the trajectory.