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¡Descarga Solucionario dinamica de hibbeler capitulo 12 y más Ejercicios en PDF de Dinámica solo en Docsity!
12-1. A bicyclist starts from rest and after traveling along a straight path a distance of 20 m reaches a speed of 30 km/h. Determine his acceleration if it is constant. Also, how long does it take to reach the speed 0£30 km/h? 122 A car starts from rest and reaches a speed of 80 ft/s alter traveling 500 ft along a straight road. Determine its constant acceleration and the time of travel. 12-3, A baseball is thrown downward from a 50-ft tower with an initial speed of 18 ft/s. Determine the speed at which it hits the ground and the time of travel. » = 30 kan/b = 833 m/s vs vie 2als 5) (8397 =0+24,(20-0) % = 14m? Ans YM=+gr 333 =0+ 1.749 1= 4305 Ans 3 velas 1) (80) = 0 + 24,($00-0) q = 640 ns Ans YY +ar 30 =0+ 640 1= 12535 Ane 4 + 24981) Y = (18)? + 2(32.21(30--0) Ya » 59.532 = $95 (Us Ans Y=y+ar 59.532 = 18 + 32.240 11296 Ans +12-4, A particle travels along a straight line such that in 2 s it moves from an initial position s4 = +0.5m1tó a position sg = —1.5 m. Then in another 4 s it moves from sg to sc = +2.5m. Determine the particle's average velocity and average speed during the 6-s time interval. what distance does the car travel during this time? at this time. Prob. 124 126, A freight train travels at v = 60(1 — e”!) ft/s, where ¿ is the elapsed time in seconds. Determine the i distance traveled in three seconds, and the acceleration yu SIE Tosal displacemoit(d, — 51) » 2m “Total distance traveled (05 + 1.5+ 15+ 25) =6m Total time traveled (2 + 4) = 62 2 Yog = q 0333 més Aus ve im Aus $ r=n+ar 125, Traveling with an initial speed of 70 knvh, a car accelerates at 6000 km/h? along a straight road. How long 1205 20 + 6000(9) will it take to reach a speed of 120 km/h? Also, through Pit zas) 1 8.3X10%) he = 303 HE dns (20 = 2 + 2 6000)(4 0) s = 0.792 km < 79% 1 LJ vas [es $= 123 i POTES d= 606” = 299 np? 5= ó0(t+ e) 2 >» A) An Ana 12-10. A particle travels in a straight line such that for a short time 25 <+f=6s its motion is described by v = (4/a) lt/s, where a is in ft/s?. Tf y = 6 fi/s when £ = 2, determine the particle's acceleration when £ = 3 s, 1-4. The acceleration of a particle as it moves along a straight line is given by a = (24 — 1) m/s?, where ¿is in seconds, lfs = 1 mand y = 2 m/s when! = 0, determine the particle's velocity and position when t = 6s. Also, determine the total distance the particle travels during this time period. 2. da Fra $ sde Par Y 31+20 A£í= 3 s, chocsing the positive root Y = 6.63 TU8 4 ES - = 0.603 fs Ans , an 05 fo. £ ar-ye v=P-1+2 . . fs - Jen. 1,1 =P 12 201 s=3P ze Wiear=6s, *12-12. When a train is traveling along a straight track at 2 mís, it begins to accelerate at a = (60 47*) m/8?, where vis in m/s. Determine its velocity + and the position 3 s after the acceleration. hs — - Prob, 12-12 12-13, The position of a particle along a straight tine is given by s = (1.5é -- 13.5% + 22.51) ft, Where £ is in seconds, Determine the position of the particie when f = 6 s and the total distance it travels during the 6-5 time interval. Hint Plot the path to determine .the total distance traveled. Position : The position of tho particie when £= 6 is =15(6) -13.5(6) +22.5(6) =-270 Ans pas Total Distance Treveled : Tho velocity of the particle can be determined by applying Eg. 121, 0 40? 2.009 22,5 de The times when the particle stops aro 4.504 --27.07422.5=0 tels ad treSs “The position of the particlea£1=05, 15 a0d $3 are Mino =1.5(0)-13.5(0) +22.5(0) 8 Ah, 2 1.5(8) -13.5(P) +225(1)=10.58 Slias, =1:5(5) -19,5(5) +22,5(5) 237.58 From the paricle's path, the total distanos is Se: =10,5+48.0+ 10,5 = 69.0 ft Ara a E > (do f.8- $, 09 35= (0-3) 300 1 = 3,925 mís= 3.93 m/s o LOs Re 10858 M E. A SER 53220 50 Subte É+8s tu ós dao talg 12-17. Two particles A and B start from rest at the origin s = O and move along a straight hine such that a, = (6t — 3) fus? and ag = (121 — 8) fus?, where 1 is in seconds. Determine the distance between them when / =4s and the total distance each has traveled in 1 =4 s. Velocity: The velocity of particles A und B can be determino using Eg. 12-2. des =usdr 0. . f dea =( (60 — 3)dt o o va =30 3 deus =agdt f dos =/ (120 — 8d lo lo va = 40 — 81 Tho times when particle A stops are %-31=0 1=0sad=ls The times when particle B stops are di—8=0 1=0samdi=Y28 Position: The position of particles A and 8 can be determine using Eg 121. dez = vadr . . Í dsa = [cama o . dsg == vgdr ñ dsg = Í (a — 8nde a 48 The positions of particlo A at 1 =1 5 and 4 are 3 SA limas =P (17) = 0.500 Ml Sa hnos =P — (6) =3400 8 Particle A has traveled da=205)+400=41.0f Ans The positions of particlo £ att =V2 5 and 4 s are = (VÍ 442) =-4 Nh 58 lyas = (4 497 = 192 61 Particle B has traveled de = 249) + 192 = 200 fí Ans ALt=4 ss. the distance borween Á and B is Asas = 19240 = Ans 12-18. A car starts from rest and moves along a straight line with an acceleration of a = (3571/%) m/s, where s is in meters, Determine the car's acceleration when £= 43. 12-19, A stone A is dropped from rest down a well, and in 1 s another stone B is dropped from rest. Determine the distance between the stones another second later. asas ads=vdv 397 de = [yde Ja a=p dadd ¿Os y vadsh de E =30t are Sas [3d f* £ le ; 1=3 s=04 sas =(2(4)3 =22.62 =22.6m alima = 322,627 = 1.06m/s? Ana 1 Hess rmtrzal 1 z 5 20404 5 = 544n 1 a d9 =0+04 6220) 4 = 1618 As= 644 — 16.1 = 483 12-22, The acceleration of a rocket traveling upward is : given by a = (6 + 0.025) m/s?, where s is in meters, Determine the rocket's velocity when s = 2 km añd the time needed to reach this altitude, Initially, v = 0 and s=0whent= ads =vde [seo as Evo 1 es.omá 3" y EE Wbsas = 2000 m, | | i y = 322 ms Ano 5 h . as e Lia S, ny Bl AUR TP ORDA + 5 (00% + 5, Sets = 2000 m 1= 1935 Ane 12-23, The acceleration of a rocket traveling upward is given by a = (6 + 0.025) m/s?, where s is in meters. Determine the time needed for the rocket to reach an altitude of s = 100 m. Initially, v = 0 and s = 0 when 1=0 [ lo ads=vde , E (6+0.025) de = f vd 6s+o0é = 12 2 ERE de = vde pa de =fa a fs+00m5s % a[va 00237 + 540.02 + P= 5628 An *2-24, At1=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When t = 3 s, bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time :, after A is fired, as to when bullet La B passes bullet 4. At what altitude does this occur? +7 sa = de + do ta EN = 045019 98 e la +7 2 de Or 1 39 =0 + 9001 3) + ¿CIDE Requires, = Sa 7 4501 49057 = 6001 — 1800 — 4905É + 2943 1 — 44.145 1=1038 An sa = 5 = 411 km Ans "12.25, A particle moves along a straight line with an acceleration of a =5/(351/* + 55/2) m/s?, where s is in meters. Determine the particle's velocity when s=2 m, if it starts from rest when s = 1 m. Use Simpsorv's rule to ads =vdv evaluate the integral. ás 1 . =P 0.8351 =3 v=1.29m/s Ans 12-26. Ball A is released from rest at a height of 40 fe at the same time that a second ball B is thrown upward 5 ft from the ground. 1f the balls pass one another at a height of 20 ft, determine the speed at which ball B was thrown upward. Forbali $ 1: (bs=5p +v1+ ha 20=0+0+402% += 111465 Por ball $2: ñ (sms rorrrda 1504 va (1.1146) + $(-322101.1140)* Ue =3LA fs Ans 12-29, A particle is moving along a straight Eine such that: vhen itis at the origin it has a velocity of 4 m/s. I£it begins to decelerate at the rate of a = (—L.5012) m/s?, where v is in m/s, determine the distance it travels before it stops. 2, = 1.54) A). v=(2-0.75% m/s (1) [ten [00758 den [ (4-s0+0.5025%) de sde 1.5 +0.1875% (2) FromEg. (1), the paricle will stop when 0=(2- 0.75% M 122.667 9 sliazasr = 62, 667) --1.5(2,6677 +0.1875(2.667P = 3.56 m Ans : 12-30. A particle moves along a straight fine with an aceeleration of a = 5/(35% + 52) m/s?, where s is in meters. Determine the particle's velocity when s =2m, ; if it starts from rest when s = 1 m. Use Simpson 's rule to evaluate the integral. h ads =vdu ! a= 2 Sds d rt a+) 1 0.8351 254 v=1.29m/s Ans 12-31, Determine the time required for a car to travel 1 km along a road if the car starts from rest, reaches a . maximum speed at some intermediate point, and then stops at the end of the road. The car can accelerate át 1,5 m/s? 7. S m/s” and decelerate at 2 m/s Utlag formulas of acc o vz E 15%, e E x Te0p-x 1000 — x = Vat jo Combining equations; 113% v<=2e x= 13% 1000-1334 =24-¿ tl = 207025 3% = 276035 124 +24 =34835 Ana *12-32, When two cars Á and B are next to one another, they are traveling in the same direction with speeds Y4 and va, respectively. 1£ B maintains its constant speed, while A begins to decelerate at a y, determine the distance d between the cars at the instant A stops. Motion of carÁ : : v=W +at Ya Ora 1 * % Yavi +24 (s=40) Ord +20 (8,0) A 2 Motion of earB + Ya Ye dy moot [22] 24d 2.» 0) a The distance between carA and B is Ya an a das =l59 =841= Ans 12-35. Accounting for the variation of gravitational acceleration a with respect to altitude y (see Prob. 12-34), ' derive an equation that relates the velocity of a freely ] falling particle to its altitude. Assume that the particle is released from rest at an altitude yo from the earts ! surface. With what velocity does the particle strike the 1 earth iFitís released from rest at an altitude yo = 500 km? Use the numerical data in Prob. 12-34, i From Prob. 12-34, > AD : When yo = 500 km, y=0, Since g dy = vdv then y => 035600 / SITO o ) S356(6336+ 500,109) RN 2 = vado 9 (RAYA do y = —3016 m/s = 3.02 km/s 4 Ane e AU sor ES 7 | A | e ST | Thus ECO A SS +12,36, When a particle falls through the air, its initial acocleration a=8 diminishes until it is zero, and La (¿Jun ! thereafter it falls at a constant or terminal velocity 0; 1€ d v this variation of the acceleration can be expressed as a = » ¡ (e/A(v)— v?), determine the time: needed for the [ de e velocity to become v < vy. Initially the particle falls from rest, 1 +v 12-37. An airplane starts from rest, travels 5000 ft down a runway, and after uniform acceleration, takes off with a speed of 162 mi/h. It then climbs in a straight line with a uniform acceleration of 3 ft/s? until it reaches a constant speed of 220 mi/h. Draw the s-+, v-f, and a-t graphs that describe the motion. t) 5280 8 e 10 50D Y a Yo + 2alo - $) (231.5)? = 0 + 24 (5000-0) 30 12 943 q = 564538 1018 +ql Seo ” =>» "2 Y Et) 237.6 = 0+ 5.64538 1 T= 42,09 = 42.15 3as EEES = 322.67 Ms 238 » +24 60030 a 314246) -s:) 11 (222,67)" == (237.6)? + 2(3)(s — 5000) alte) s= 1203348 i Y =»+taqt 322.67 = 2376 + 31 12848 12-39. A freight train starts from rest and travels with a constant acceleration of 0,5 ft/s?, After a time £ it maintains a constant speed so that when t = 160 s it has traveled 2000 ft. Determine the time £' and draw the v+ graph for the motion, Total Distance Traveled : The distance for part óns of the motion can be relased to timer ="" by applying Eq. 125 with sy =0and dp =0. (5) sos torrjar s 0404305117 =025(£7* The velociiy at timo + can be obiained by applying Eq. 12=4 with 19 "0. (5) D=0)+0,1=0+0.51=0.5 “The time for (be second stage of motion is yz = 160—£' and the train is sravelling as a constant veloctiy of w=0.51” (Eq.[1D. Thos, the distance for this part of motion is (5) 1, = Dl = 0.5 (160—1) = 801 -0.5(1? Tf tho total distance traveled is S,, = 2000, then Eo 5 0 2000=0.25(4)?+808 0.5(0Y 0.25(4)? - 801 +2000 =0 Chose a root that is less than 160 s, then 1=2145=2735 AÁns u-1 Graph : The equation for the velocity is given by Eq.[1). When 12 fu 27,345, v =0.5(27.34) = 13.7 ft/s. *12-40, If the position of a particle is defined by s = E sin (7/5): + 4] m, where £ is in seconds, construct the 5-t,v-t, and a-t graphs for 0 <=1=<108, (y vrs) alme) 12-41. The v-1 graph for a particle moving through an electric field from one plate to another has the shape shown in the figure. The acceteration and decéeleratior that occur are constant and both have a magnitude of 4 m/s. If the plates are spaced 200 mm apart, determine the maximum velocity %nax and the time £* for the particle lo travel from one plate to the other. Also draw the s-+ graph. When + = 1/2 the particle is at s = 100 mm. 4 =4m/s 2 = 100 mm =0lm 2 Y = Y + 244s-8) Va = 0 + KANOI0) Yana = 0.89442 n/a Ans v=v+2r 089442 =0+ « Y = 046721 5 Ans sms tvtrzaf s=0+0+ ja 1-28 viven 6 2 02236 02240, s <= 01m Fo oa=-f +0 9.894 9.1239 va 474 1.788 . . UN ds Í,,, 504 1588) de 3-28 + 1788102 When» 04475, 5=02m 
