Dynamics_CH12_HW

Dynamics_CH12_HW - 12-10. A particle travels in a straight...

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Unformatted text preview: 12-10. 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. 12-11. 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 | *12-12. 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 Hal-all = 9.98 in 12-13. 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 6-s 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 4.50:1-27.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 *12-24. 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 + 569-81) 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 1-12-25. 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 12-26. 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 *12-44. 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 3-0 — 2(‘5)(5) + (10 4)(5) + 2(15-10)(5v) - i(3)(5)(3)(5) .r=48m Au 12-45. 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(20-15) - 3(r'— 20) r’ = 33.3 s 12-53. 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 12-66. 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 12-67. The velocity of a particle is given by v = by-pplyinsBiJZ-9. {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 12-70. 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 12-71. 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 = 38-56 = 4.7.8 111/: Ans ” I, 2+4+3 *12-72. 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 12-73. 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 12-74. 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-{11|3‘=°"“‘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, =EC-l;(y+cr2) Ans — 12-85. 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(3-6) } (v0)z = 35 fth l A 1 o =o+(v,,), (3.6)+-2-('-32-2)(3-6)2 nurt- (vu), = 57.96st w=./(35)1+(57.96)2=67.7st An. sink/5 J-o — “ 3—1—93 = 89" A Gun 35 ) 5. n: A “445/; 12-86. 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 12-91. 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; *12-108. 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 12-109. 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 *12-116. 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 12-117. 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 «anvil-w? w A *12-120. 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? 12-121. 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’ 12-145. 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. 12-146. 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,=3|nls’ 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! 12-147. 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 - 0-4 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 *12-152. At the instant shown, the watersprinkler is rotating with an angular speed 9 = Zrad/s and an ’= 0-2 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 12-153. "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 12-158. 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 ( 12-159. 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 “— *12-176. 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 12-177. 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 *12-180. 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 12-187. 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 ~— ‘12-196. 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 straight-line 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“ 12-19‘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 = 5-2—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 12-201. 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 straight-a-way 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: = 6-69 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 *12-204. 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=VA-VB (a = (unnBHveosBD— vi = (vs:n9—v)i+ ucusej um = WusinH—u)z+(uc050)2 = «21)Z -2ulstn9 =wztl—sin6) Ans 12-205. A man can row a boat at 5 m/s in still water. He wishes to cross a 50-m-wide 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.

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Dynamics_CH12_HW - 12-10. A particle travels in a straight...

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