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Unformatted text preview: 1210. A particle travels in a straight line such that for a
short time 2 s S t S 65 its motion is described by
v = (4/a) ft/s, where a is in ft/sz. If v = 6ft/s when
t = 2 5, determine the particles acceleration when t = 3 5. 1211. The acceleration of a particle as it moves along
a straight line is given by a = (2! — l) m/sz, where t is in
seconds. lfs = 1 m and v = 2 m/s when! = 0, determine
the particle's velocity and position when t = 65. Also,
determine the total distance the particle travels during
this time period. Ivdv= L44! 6 1v1—18=4:s
2 v1 81+20
Au= 3s, choosing lhcposiu'vcroot v = 6.63 11/: a = J'— = 0.603 M1
6.63 rdy=r(21—l)dl
1 o 3:671]: Slnocv at Omen 4:67—1:66!!! l
l
l
‘l
 *1212. When a train is traveling along a straight tracl<
at 2 m/s, it begins to accelerate at a = (60 If") m/s’,
where v is in m/s. Determine its velocity v and the pOSlthn 3 s after the acceleration. " d
Lsmgfz 6011))" l S
3— 300(1) 32) Prul). ll— 12 v = 3.925 m/s= 3.93 m/s aha vdv ¢=Ld°=105dv a 60 x 1 L925 5
d:=——I vdv
Jo 6° 1 1 vs 1,15
Halall = 9.98 in 1213. The position of a particle along a straight line is
given by s = (1.5!3 ~ 13.512 + 22.51) ft. where l is in
seconds. Determine the position of the particle when l =
6 s and the total distance it travels during the 6s time
interval. Him: Plot the path to determine the total distance traveled. Position :Ihepositionofmepmicle whent= 6: is slug, = 1.5(6’)  13.5(61) +22.5(6) —27.on Ann M (*0 0 Total Diuanu hauled : The velocity of the particle can be mm \ by Ipplyins F4 12 1. W4
0 t = § = 4.50:1 —27.o:+ 22.5 .5: 1315 ft .5 210}: =0 5%”:
1'55 1" 65 fag 1:15
Thetimes whmﬂieparﬁdestopsue 4.50:127.o:+22.s=0
(=18 Ind i=5: Thepositionofmepmiclentt=05. lsmdSaue M. =1.5(o’)13.s(o’)+22.5(0)o
,'_l_ = 1.5(1’) —13.s(1’) +22.5(1) = 10.5 n
,[hh =1.5(5’)13.5(5’)+22.5(5)= 37.5 n : Fun thepam‘de‘s puh. tbsmmldimnceis :m = 10.5+48.0+ 10.5 = 69.0 ft *1224. At 1 = 0 bullet A is fired vertically with an
initial (muzzle) velocity of 450 m/s. When I = 3 s, bullet
B is fired upward with a muzzle velocity of 600 m/s.
Determine the time 1, after A is ﬁred, as to when bullet 1 1
B passes bullet A. At what altitude does this occur? *1 ‘4 = ("h + m)” r + 5", 1
5‘ = o + 450 1 + 56981) t1 1
+1 5, 0.). + (mow 5a.? 1
x, o + 6000 — 3) + 2(—9.81)(t— 3)’ Require.“ = x. 4501 4.905 :1 = 500: — 1800 — 4.905 :1 + 29.43 t — 44.145 r = 10.3 5 An ;A = 5, = 4.11 km Ans 11225. A particle moves along a straight line with an
acceleration of a = 5/83”3 + 55/2) m/sz, where s is in
meters. Determine the particle’s velocity when 5 = 2 m, if it starts from rest when 5 = 1 m. Use Simpson’s rule to evaluate the integral. 5* fond» '(3sl+s;) 1
mush—.1I
2 v: 1.29 m/s Ans 1226. Ball A is released from rest at a height of 40 ft
at the same time that a second ball B is thrown upward
5 ft from the ground. If the balls pass one another at a
height of 20 ft, determine the speed at which ball B was
thrown upward. Forballﬂ 1:
(+i)s=xo+uot+;qtz
m=o+o+§(32.2)rz
r=1.11465 For ballﬂZ:
(+ T).r=.ro +vot+ {412
15 = 0+ u.(1.1146)+;(—32.2)(1.1146)1 u, =31.4 ms *1244. A motorcycle starts from rest at s = O and
travels along a straight road with the speed shown by the
v—r graph. Determine the motorcycle‘s acceleration and
position when! = 8s and I = 12 5. Ana _ l _ _1 l 3 3
30 — 2(‘5)(5) + (10 4)(5) + 2(1510)(5v)  i(3)(5)(3)(5) .r=48m Au 1245. An airplane lands on the straight runway, originally
traveling at 110 ft/s when r = O. [f it is subjected to the
decelerations shown, determine the time I’ needed to stop
the plane and construct the 3—: graph for the motion. v0 =110ﬂ/s
Av = Ind: 0 — 110 = —3(15—5)  8(2015)  3(r'— 20) r’ = 33.3 s 1253. Two cars start from rest side by side and travel
along a straight road. Car A accelerates at 4 m/s2 for 10 s
and then maintains a constant speed. Car 8 accelerates
at 5 m/s2 until reaching a constant speed of 25 m/s and
then maintains this speed. Construct the (1—1, 12—1, and 5—1
graphs for each car until I = 15 5. What is the distance
between the two cars when t = 15 5? 9105’s“ CarA: v=vo+acl vA =0+4t m:=lOs. VA=4omls :.=0+0+;(5)t’=2.5tz 1 2
=1 +vol+ar a
a 2‘ Nl=55. s.=62.5m 10 5A=0+0+%(4)11=2!2 [)55' d5=Vd1 I. '
Au=lOs. 5,4 =200m £1,434,254; l>105. ds=vdt ,3 62.5=251— 125 j" d: ' 40d: 3. =25:—62.5
100 10 When 1:15 5, J.=312.5 D'suince between the cars is A: :34 s. #003125 = 87.5 m Cu'AisaheadofcarB. Velocity : The velocity express in Cartesian vector form can be obtained by 1266. A particle, originally at rest and located at point
applying 134.12 9. (3 ft, 2 ft, 5 ft), is subjected to an acceleration of a = [6ti + 12 rzk} ft/sz. Determine the particle’s position (x, y, z) att=ls. dv=ndr J'dv=J'(6xi+12:1k)d:
0 0
V = {3:1l+4:’k) ﬁ/s Position : The position express in Canesian vector form can be obtained by
applying 1:41.12 — 7. dr=vdr
J" dr=r(3:2i+4:’k)dr
r. 0 r—(3i+2j+5k)=z’i+z‘k
r={(:’+3)i+21+(r‘+5)k} ft When (=15, r=(1°+3)i+2j+(1‘+5)k={4i+2j+6k}it
The coordinates of the pam‘cleare (4, 2. 6) ft Acclllraliau : The acceleration express in Cartesian vector form can be obtained 1267. The velocity of a particle is given by v =
bypplyinsBiJZ9. {1612i +4I3j + (5! +2)k} m/s, where t is in seconds. If the
particle is at the origin when t = 0, determine the
magnitude of the particle’s acceleration when t = 2 s a=ﬂ=ﬂu+mzj+su ms;
Also, what is the x. y, z coordinate position of the particle d! at this instant? When (=25. a= 32(2)l+12(2’)j+5k = (64l+48j+5k} m/s‘. The magnitude
oftheaccelemionis ’ a=‘/a}+a}+a}=1/641+482+51=80.2m/sz Ans Position : The position express in Cartesian vector form can be obtained by
applying Eq.12—7. dr=vdr
rdr=Jol(1611i+4:3j+(51+2)k) d:
I! r = [gr’i + 1‘) + (2:2 + 2t)k] m When t=25. r= 1—36(2’)i+ (2‘)j+[§(22) +2(2)]k = (4171+ 16.0j+14.0k} m. 'nius. the coordinale of the particle is (42.7. 16.0. 14.0) r: m“ ‘“ “2.4.1.4 «misinmimrm‘» m“*"‘mmir_m;yi mmmmmﬂ “New. i
z
i 1270. The car travels from A to B, and then from B to
C, as shown in the ﬁgure. Determine the magnitude of
the displacement of the car and the distance traveled. Displacement: Ar=(2—31} km Ar=./21+3z =3.“ km Am
Distance uqued: d=2+3=5km Ans 1271. A particle travels along the curve from A to B in
2 s. It takes 4 s for it to go from B to C and then 3 s to
go from C to D. Determine its average speed when it goes
from A to D. y (21000)) + 15 + gum» = 335‘ / C v s —r = 3856 = 4.7.8 111/: Ans
” I, 2+4+3 *1272. A car travels east 2 km for 5 minutes, then north
3 km for 8 minutes. and then west 4 km for 10 minutes.
Determine the total distance traveled and the magnitude 3 = 2+3+4 =9 km A"
of displacement of the car. Also, what is the magnitude
of the average velocity and the average speed? Total Distance Traveled and Displacement : The total distance traveled is and the magnitude of medisplscement is Ar=¢21+31=3.606km=3.6lkm Average Velaeily and Spud : The total time is At = 5+ 8+ 10
= 23 min = 1380 s. The magnitude of average velocity is Ar_ 3.606(10‘) UV =—_———_—=
.. A, 1380 2.6lm/s Ans and the average speed is U —————_
(.,)"l A, 1380 6.52mi: Ans 1273. A car traveling along the straight portions of the road has the velocities indicated in the figure when it v‘ = 20:
arrives at points A, B, and C. If it takes 3 s to go from A
. v. = 2121 l + 2121]
to B, and then 5 s to go from B to C, determine the
average acceleration between points A and B and vc =40: between points A and C. A v 21.21: + 2121; — 2m
"' = F? = ——3——
.4, = (0.4041 + 7.07 J} mls’ 5... Av 40: — 2m
l‘c = E = _—__8 (2.50:)m/s’ Velocity : Taking the ﬁrst derivative of the path y = 22’. we have 1274. A particle moves along the curve y = e?" such
that its velocity has a constant magnitude of v = 4 ft/s.
Determine the x and y components of velocity when the y'= ZeZ’X‘ particle is at y = 5 ft. [1] However. i = v, and Y = v, » “HIS. 1=Jfl{113‘=°"“‘cs v = 221‘ v1 Hem, u=4ﬁls.Then Solving Eqs.[2] and [3] yields 1 8 :41
01:4 1+4e‘x and u’— 1+4e‘x Aty=5ft, 5=e“,x=o.8047 ft. Thus. 1
muon) u] = 4 = 0.398 NS Ans [+49 “(01947)
8 .———— = 3.98 ftls
1+4¢4(0.l047) 3‘75 The path of a particle is deﬁned by y2 = 4k):, and
e co""POnent of velocity along the y axis is vy = cl. ,1 =4Jcr z’he’e both k and c are constants. Determine the x and y
Omponents of acceleration. 2,4,, = MN, 2(cr)‘ + 2y: = 41a:x a, =ECl;(y+cr2) Ans — 1285. From a videotape, it was observed that a pro _ ‘ ¢ .
football player kicked a football 126 ft during a measured u time of 3.6 seconds. Determine the initial speed of the
ball and the angle 6 at which it was kicked. 5=so+vot 126=0+(V0)x(36) } (v0)z = 35 fth l
A 1
o =o+(v,,), (3.6)+2('322)(36)2 nurt (vu), = 57.96st w=./(35)1+(57.96)2=67.7st An. sink/5 Jo — “ 3—1—93 = 89" A Gun 35 ) 5. n: A
“445/; 1286. During a race the dirt bike was observed to leap
up off the small hill at A at an angle of 60° with the
horizontal. If the point of landing is 20 ft away, determine (—3) 3 = ’0 + vs I
the approximate speed at which the bike was traveling
just before it left the ground. Neglect the size of the bike
for the calculation. 20=0+v‘cos60°1 l
(+DS=So+vo+Eagx7 0 = 0 + v‘ sin60° r 4» $9321) ,1 Solving
“\\ x = 1.4668 s
\
\
/ \
\ v, = 27.3 ft]: .4... AR 1291. It is observed that the skier leaves the ramp A at
an angle 0,4 = 25° with the horizontal. If he strikes the
ground at 5, determine his initial speed 11‘ and the time
of ﬂight I“). (—3) mm:
4 0
100(3) = VA c0325 In
1 1
(+T) :=:o*‘b‘+5a=‘
3  o 1 931‘
_4_100(3)=0+v‘ stS In +59  )‘4' Solving. VA = 19.4111}: All! ,u =4.54s Anl l
E
l; *12108. Starting from rest. the motorboat travels /+ _;
around the circular path, p = 50 m, at a speed 1) = p=50m ="
(0.2?) m/s, where I is in seconds. Determine the «/ magnitudes of the boats velocity and acceleration at the
instant I = 3 s. Velocity : When 1: 3 s, the boat travels at a speed of u=ot2(32) :lcSOm/s Ans Acceleration : The tangential acceleration is a, = 1') = (0.4r) m/sl. When
I: 3 s. a, = 0.4(3)=1.20 m/sl
To dciemu'ne the normal acceleration, apply Eq. 12 — 20. "2 1'80: 00648 Isl
=—=——= ‘ m
”" 50 p Thus, the magnitude of acceleration is a=‘/u,2+a,, = J1.202+0.060482 = 1.20 m/s2 Ans 12109. A car moves along a circular track of radius i2550 Bind itszspeed for a short. period of time 0 s (S 2 s V = 30+ [1)
I) ~_ (I + t ) ft/s, where (IS in seconds. Determine the
ma ' ‘
. gmtude .of its acceleration when t = 2 5. How far has ‘1”
it traveled int=25? “I =F=3+6t
I
When I = 2 s a, =3+6(2)= ism/s2 v2 304.22) 2
“v =3 =[ﬁ—l = 1.296(1/51 250 a = ¢(15>2+(i.296)z = 15.1 ﬁ/s2 Ans dr=vdr Iir=I°13(r+rz)d: 61 *12116. The jet plane is traveling with a constant speed
of 110 m/s along the curved path. Determine the magnitude of the acceleration of the plane at the instant
it reaches point A (y = 0). _ 1+(o.1375)1]”2
‘ [0.002344! 01 (110)1 =—=———=26.9 ’
a. p 4494 m/s Since the plane travels with a constant spew. a. = 0. Hence a = a. = 26.9 m/s2 Am 12117. A train is traveling with a constant speed of
14 m/s along the curved path. Determine the magnitude
of the acceleration of the front of the train, B, at the
instant it reaches point A (y = 0). Au mm.....,‘..mm«.m.~mm—mwyv ‘. l «anvilw? w A *12120. The car B turns such that its speed is increased
by 135 = (0.5e’) m/sz, where ris in seconds. lfthe car starts
from rest when 6 = 0°, determine the magnitudes of its
velocity and acceleration when t = 2 s. Neglect the size
of the car. Also, through what angle 6 has it traveled? 12121. The box of negligible size is sliding down along
a curved path defined by the parabola y = 0.4x2. When it
is at A (x,. = 2 m,y,1 = 1.6 m), the speed is v” = 8 m/s
and the increase in speed is de/dr = 4 m/sz. Determine
the magnitude of the acceleration of the box at this
instant. “= M +a.% = N4)2 + (7.622)2 = 8.61 m/s2 1' Jv = I' Oje'dl
0 0 v = 015 e’li, = 0.5(e' ~ 1) t = 2 s, v = 05M  1) = 3.1945 = 3.19 m/s
0, = 0.5 e’ = 3.6945 m/sz vz (3.1945)’ 1
= _. _—_ _.._— = . 1
a. P 5 204 m/s a = ((169491 + (2.041): = 4.22 m/s’ 12145. A truck is traveling along the horizontal circular
curve of radius r = 60 m with a speed of 20 m/s which is
increasing at 3 m/sz. Determine the truck’s radial and transverse components of acceleration. 12146. A particle is moving along a circular path having
a radius of6 in. such that its position as a function of time
is given by 0 = sin 3!, where 0 is in radians, the argument
for the sine is in degrees, and t is in seconds. Determine
the acceleration of the particle at 0 = 30°. The particle
starts from rest at 0 = 0°. r=6in,. , 'r=1.20m/a V.= 83 r9 = 0.4189(3) = 1.26 m/s .. .2
ﬂy=r—ra =o_ r=60 u,=3nls’
2 1 A = L = _._("’”) = 6.67111152
r 60 4.. 4. a 4.57 mls’ u,a,3mls2 i=Q i=0 9 = sin3r nm' 9 = 3 cow 9 = 2.5559 red]; 9: —9 sin}: a = ‘4'"24 I‘d/"z At 9 = 30°,  " z “r  r  r9 = 0— 60.5559)’ = 49,196 30° ,  . Wﬂ=sm31 a.=r0+2r0=6(‘4.7124)+0=—28 274
r: ID. 525 s a = (49196): + (43.274): = 48.3 in Isz A ' ll! 12147. The slotted link is pinned at 0, and as a result of the constant angular velocity 0 = 3 rad/s it drives the peg P for a short distance along the spiral guide 7 :1 (0.4 0) m, where 9 is in radians. Determine the radial '9 = 3 my, _ an transverse components of the velocity and r  04 a
acceleration of P at the instant 0 = 7r/3 rad. 'r — 04 '0 'r' = 0.4 '6
7?
A16 = 5. r = 0.4189 'r = 04(3) = 1.20 'r'= 04(0)  o All:
All:
0.4189(3)’ = —3.77 m/s2 "9 = ’9’" 2'”? = 0 + 20.2mm = 72.0 m/s2 All All Ans Ans —————_g *12152. At the instant shown, the watersprinkler is rotating with an angular speed 9 = Zrad/s and an ’= 02 angular acceleration 6 = 3 rad/52. If the nozzle lies in r s 3 9 =2
the vertical plane and water is flowing through it at a constant rate of 3 m/s, determine the magnitudes of ,r, =0 '9 =3
the velocity and acceleration of a water particle as it exits the open end, r = 0.2 m. V, = 3 v. =o.2(2) = 0.4 V = V (3)1 + (0.4)2 = 3.03 11113 1
a, = o — (02x2)1 = —o.so m/s Ans 1
a. s 0.20) + mm) = 12.6 mls 1 A
a = /(—o.80)2 + (12.6)1 = 12.6 m/s ns 12153. "Hie automobile is traveling from a parking deck
down along a cylindrical spiral ramp at a constant speed
of v = 1.5 m/s. If the ramp desecnds a distance of 12 m
for every full revolution, 8 = 217 rad, determine the
magnitude of the car's acceleration as it moves along the
ramp. r = 10 in. Him: For part of the solution, note that
the tangent to the ramp at any point is at an angle of d: =
tan‘ (12/[271(10)])=10.81° from the horizontal. Use
this to determine the velocity components 2),, and 2):,
which in turn are used to determine band 2. ’2
11:15"),S ¢
277 ('O’rw Vr:0 va:l.5c0310.81°:l.473 mjs since 9:0 V: = ‘15 sin10.81° = 0V2814 m/s (1, = r—er =0—10(0.l473)2 :0.217
Since (19 = n9+ Zré? =10(0)+2(O)(O.1473)= 0 10 a: (‘0.217)2+(0)2+(0)1:O’217m/52 Ans 12158. For a short time the arm of the robot is extending
at a constant rate such that f = 1.5 ft/s when r = 3 ft,
3 = (4t2) ft, and 0 = 0.5t rad, where t is in seconds. Determine the magnitudes of the velocity and acceleration
of the grip A when t = 3 s. 9 = 0.51rad r = 3 n z 4:211
(9 = 0.5 md/s f = l.5 ftls 2 = 8! ft/s r“ :0 i=8 11/52 v = 24.1 ft/s Ans
a, = 0— 3(05)2 = —o.75
0,. = o + 20.5)(05) = [.5
a2 = 8 a = ‘/(—0.75)2 + (1.5)2 + (8)2 = 8.17 ms? Ans ( 12159. The partial surface of the cam is that of a loga "'°50) mm, where 9 is in radians.
If the cam is rotating at a constant angular rate of 9 =
4 rad/s, determine the magnitudes of the velocity and ac
celeration of the follower rod at the instant 0 = 30°. If _ M5 —
= 4080.051; r : 40,; l (6) = 41.0610 . MW, . . 0.05 — 6 = 4 rad/S
2e 9 ’ = 2" (6 )(4) = 8.21 mm/s Ans 0. 160.0% J 2 + 280.050 0. 1:005 ‘ V...”“mm—.uyﬂwmmw'r‘ﬁ‘ A i < (4)2 + 0 = [.64 ﬁlm/$2 *7:
>6 “— *12176. Determine the displacement of the log if the
truck at C pulls the cable 4 ft to the right. 250+(SB~5c)=l 35; ~56 =I
3A5] ‘AIC = 0
*—1
S“ “v =4! mm 93:19:39“?
3555 = 4 A:5=—l.33fl=l.33fl+ Ans 12177. The crate is being lifted up the inclined plane
using the motor M and the rope and pulley arrangement
shown. Determine the speed at which the cable must be
taken up by the motor in order to move the crate up the
plane with a constant speed of 4 ft/s. Position  Coordinate Equation :Dnturn is established atﬁxed pulley B.
The position of point P and emeA with respect to datum are x, and :A. reSpwcively.
27A +(.rA ~43.) =1
35‘ .r, = 0 Time Derivative .‘ Taking the time derivuivc ol' the above equalion yields 31)‘ — v, =0 [1]
Since 1),. =4ftls, from Eq.[l]
(+ ) 3(4)— 1),, =0
v,=12ﬁ/s Ans 100 H *12180. The pulley arrangement shown is designed for
hoisting materials. If BC remains ﬁxed while the plunger P
is pushed downward with a speed of 4 ft/s, determine the
speed of the load at A. 5.x, +(.:,l ~5A) =1 6:, —5A = 6v, — VA = 6(4) = VA VA = 24 ids Ans 12481. If block A is moving downward with a speed of
4 ft/s while C is moving up at 2 ft/s, determine the speed
of block 3. 54 +lr5 +5C =[
VA+2V5 +vc =0 4+2VB~'2=0 v» =—lft/s=lfl/sT Ans 102 12187. The motion of the collar at A is controlled by a
motor at B such that when the collar is at 5A = 3 ft it is moving upwards at 2 ft/s and slowing down at 1 ft/sz.
Determine the velocity and acceleration of the cable as it is drawn into the motor B at this instant. «3} +42 +5, :1 l 7 .1
502 +16) ‘(Zsim +55 :0 5B = “5A5.4(Ii +16)_i 5" z ii“) (‘5 * “3V; (xi + Mini“ (39: + myth, a )J — ___(‘”*)Z J“ — ’
(51“16)’ (xi +16)i Evaluating these equations : 3, = —3(—2)((3)Z+16)_i= 1.20 ﬁ/s l Ans mm (2)2+3(i)
)—'——=— Z_
((3)z+16)1 ((3)z+16)§ I'Hﬁ/S .llﬁ./32T An, * v$24.88. The roller at A is moving upward with a 4 f1:;it2y of vA = 3 ft/s and has an acceleration of a,‘ =
15 when s,‘ =4 ft. Determine the velocity and acce eration of block B at this instant. “WW/ma . I I
“'9 +E[ln )2 +32]'1(2:A “A = O 105 ~— ‘12196. Two planes, A and B, are ﬂying at the same
altitude. If their velocities are vA = 600 km/h and v, = 500 km/h such that the angle between their
straightline courses is 9 = 75°. determine the velocity of
V: = VA + Vnu plane B with respect to plane A.
[500 4—} = 1600‘s,] + Vm
(2.) 500 = —600 cos75° + (Vm).
(vm), = 655.29 +
(4. T) o = — 600sin75° + (v,,,. ), (vm), = 579.56 T (Vm) = (65529)2 + (57955): v“, = 375 Ian/h Am
57956 9 = tan" —— = o A
(655.29) “5 Al“ 1219‘7. At the instant shown, cars A and B are traveling
tit speeds of 30 mi/h and 20 mi /h, respectively. If B is
increasmg its speed by 1200 mi/hz, while A maintains a
constant speed, determine the velocity and acceleration
of B with respect to A. v, = VA 4' VIM 30°
20 \l = 30 4' (V314); 4' (VIM), 30°
.9 T B <—
\ 0.3 mi
v, = 20 mi/h (.3) 40mm = 30 + (vm). A
I = O '
[A 3 mi/h4— (+13 2000530" = (VIM), Solving
(vm), = 20—9 (vm), = 17.32 T v,“  [my + (17.32)z = 255 mifh Ans a = tan—'(1—72—‘g—2) 409° A0 Am 1
(‘1’).I = 52—02— =1333.3 '5 = ‘A + “HM 1200 30° 1333.3
V + a = O + (aim): + (45M),
30° + T (—3) 4200mm + 1331mm" = (am). (+1) 1200m30° + 1333.3sin30“ = (am), Solving
(am). a 5547 y ; (am), = 1705.9T
am 3 (554.7): + 1705.9)z = 179(10’) mim’ Ans 4(192) = 72.0° A” Ans ° = m“ 554.7 12201. At the instant shown, the bicyclist at A is traveling
at 7 m /s around the curve on the race track while increasing
his speed at 0.5 m/sz. The bicyclist at B is traveling at 8.5 m/s along the straightaway and increasing his speed at VA = v. + v
0.7 m/sz. Determine the relative velocity and relative m
acceleration of A with respect to B at this instant. [ 7 [\1 = [3.5 _,l + “me _,1 + W”), l]
40’
U, = 8.5 III/S +
_ 7 W N _ (—1) 7mm" = 8.5+(vm), (+1) 7mm" = (vm), Thus.
(VA/11); = 4.“) 111/3 4— (vm), = 5.36 111/3 1 (vm) = «4.00): + (5.315)1 v.41: = 669 Ill/l Ans
_ 5.36
9 = an ‘(20—0) = 53.3%? An: 1 7
a = — = _ 1
(,4) so 0980mls .A = I: + .All [0.980] 3 + [0.5] A“. = [0.7 —vl + “ax/n): vl + Ham), J'1
(:1) —o.9so «mm + 0.5 311140" = 0.7 + (am), (am). = 1.129 Inls‘q— (+ .L) 0.980 511140" + 0.5 oos40° = (11",), (a.,,), = 1.013 ml:1 .L (ax/g) = 1/(1.129)z + (1.013)1 a“, = 1:2 Ill/81 A...
1.013
9 un—l — = °
= (mg) 41.9 t? A” 113 *12204. The two cyclists A and B travel at the same
constant speed 1). Determine the speed ofA with respect
to B if A travels along the circular track, while 8 travels
along the diameter of the circle. VA = vsin0i+ucos€j v9 ui V‘,B=VAVB (a = (unnBHveosBD— vi = (vs:n9—v)i+ ucusej um = WusinH—u)z+(uc050)2
= «21)Z 2ulstn9
=wztl—sin6) Ans 12205. A man can row a boat at 5 m/s in still water.
He wishes to cross a 50mwide river to point B. 50 m
downstream. If the river ﬂows with a velocity of 2 m/s,
determine the speed of the boat and the time needed to
make the crossing. VI: = Vr+Vbn
1),, sin 45°i — chos 45°j = —2j + 5cos 9i — Ssin €j ywm«wwwr *‘ “ “fingi Endj component, we have vbsin 45°=SCos 8 [1]
—U,,cos 45°=—2—Ssin8 [2]
854511] and [2) yields
0: 28.576
D, = 6.210 m/s = 6.21m/s Ans m‘mnimd by the boat I0 travel from pointA to B is ‘/502+502 _ 5,45
1= — = ——__ 11.4
12,, 6.210 S Ans 114 ...
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This note was uploaded on 07/09/2011 for the course EGN 3311 taught by Professor Hudyma during the Fall '08 term at UNF.
 Fall '08
 Hudyma
 Dynamics

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