Chapter 2 PROBLEM SOLUTIONS

Transcription

1 Chper PROBLEM SOLUTIONS. We ssume h you re pproximely m ll nd h he nere impulse rels uniform speed. The elpsed ime is hen Δ x m Δ = m s s. s.3 Disnces reled beween pirs of ciies re ( ) Δx = Δ = 8. km h.5 h = 4. km ( ) Δx = Δ = km h. h =. km ( ) Δx = Δ = 4. km h.75 h = 3. km Thus, he ol disnce reled Δx = ( ) km = 9. km, nd he elpsed ime is Δ =.5 h h +.5 h =.7 h). () Δ x 9. km Δ.7 h 5.9 km h Δx = 9. km (see boe).4 () Δ x Δ f yr m yr f m s 7 = 3.56 s or in priculrly windy imes, Δx Δ f yr m yr f m s 7 = 3.56 s The ime required mus he been Δ x 3 3 mi Δ mm yr 69 m 3 mm mi 5 8 yr = m Pge.

2 Chper.5 () Bo A requires. h o cross he lke nd. h o reurn, ol ime. h. Bo B requires. h o cross he lke which ime he rce is oer. Bo A wins, being 6 km hed of B when he rce ends Aerge elociy is he ne displcemen of he bo diided by he ol elpsed ime. The winning bo is bck where i sred, is displcemen hus being zero, yielding n erge elociy of zero..6 The erge elociy oer ny ime inerl is Δ x x f xi Δ f i () Δx. m Δ. s 5. m s Δx 5. m Δ 4. s.5 m s (c) Δx 5. m. m.5 m s Δ 4. s. s (d) Δx 5. m 5. m 3.33 m s Δ 7. s 4. s (e) Δx x x = Δ 8. s h Displcemen =Δ = 85. km h 35. min + 3 km = 8 km 6. min.7 () x The ol elpsed ime is Δ = h 35. min + 5. min +. h =.84 h 6. min Pge.

3 Chper so, Δ x 8 km Δ.84 h 63.4 km h.8 The erge elociy oer ny ime inerl is Δ x x f xi Δ f i () Δx 4. m + 4. m s Δ. s Δx. m.5 m s Δ 4. s (c) Δx 4. m. m s Δ 5. s. s (d) Δx Δ 5. s.9 The insnneous elociy ny ime is he slope of he x s. grph h ime. We compue his slope by using wo poins on srigh segmen of he cure, one poin on ech side of he poin of ineres. x x 4. m. s. s (). s.5 s 4. m/s. s x x.5 s. s 6. m.5 s. s.5 s = 4. m s (c) 3. s x x 4. s.5 s 4. s.5 s.5 s = Pge.3

4 Chper x x +. m. m/s 5. s 4. s. s (d) 5. s 4. s 4.5 s. () The ime for cr o mke he rip is = Δ x.thus, he difference in he imes for he wo crs o complee he sme mile rip is Δx Δx mi mi 6 min Δ = = = = 55 mi h 7 mi h h.3 min When he fser cr hs 5. min led, i is hed by disnce equl o h reled by he slower cr in ime of 5. min. This disnce is gien by Δx = υ (Δ) = (55 mi/h) (5 min). The fser cr pulls hed of he slower cr re of υ relie = 7 mi/h 55 mi/h = 5 mi/h Thus, he ime required for i o ge disnce Δx hed is relie ( 55 mi h)( 5 min) Δ x Δ = 55 min 5. mi h Finlly, he disnce he fser cr hs reled during his ime is h Δx = ( Δ ) = ( 7 mi h)( 55 min) = 64 mi 6 min. () (c) ( x) Δ Δ + L + L ( Δ x) Δ L L ol Δ x Δ x + Δ x ol + L L ol Δ (d) ( e. speed) rip ol ol disnce reled Δ x + Δ x + L + L L = Δ Pge.4

5 Chper.3 The ol ime for he rip is = +. min = h, where is he ime spen reling υ = 89.5 km/h. Thus, he disnce reled is Δ x = =, which gies Δ x = 89.5 km h = 77.8 km h h = 77.8 km h km or 89.5 km h 77.8 km h = 8.5 km From which, =.44 for ol ime of. = h =.8h Δ x = = 77.8 km h.8 h = 8km. Therefore,.4 () A he end of he rce, he oroise hs been moing for ime nd he hre for ime. min = s. The speed of he oroise is υ =. m/s, nd he speed of he hre is υ h = υ =. m/s. The oroise rels disnce x, which is. m lrger hn he disnce x h reled by he hre. Hence, x = x +. m h ` which becomes ( ) = s +. m h or ( ). m s =. m s s +. m This gies he ime of he rce s =.3 s x. m s.3 s = 3 m.5 The mximum llowed ime o complee he rip is ol ol disnce 6 m km h required erge speed 5 km h.78 m s = 3. s The ime spen in he firs hlf of he rip is Pge.5

6 Chper hlf disnce 8 m km h.5 s 3 km h.78 m s Thus, he mximum ime h cn be spen on he second hlf of he rip is = ol =.3 s.5 s =.5 s nd he required erge speed on he second hlf is hlf disnce 8 m km h 76. m s = 74 km h \.5 s.78 m s.6 () In order for he riling hlee o be ble o cch he leder, his speed (υ ) mus be greer hn h of he leding hlee (υ ), nd he disnce beween he leding hlee nd he finish line mus be gre enough o gie he riling hlee sufficien ime o mke up he deficien disnce, d. During ime he leding hlee will rel disnce d = υ nd he riling hlee will rel disnce d = υ. Only when d = d + d (where d is he iniil disnce he riling hlee ws behind he leder) will he riling hlee he cugh he leder. Requiring h his condiion be sisfied gies he elpsed ime required for he second hlee o oerke he firs: d = d + d or = + d giing = d or = d ( ) (c) In order for he riling hlee o be ble o les ie for firs plce, he iniil disnce D beween he leder nd he finish line mus be greer hn or equl o he disnce he leder cn rel in he ime clculed boe (i.e., he ime required o oerke he leder). Th is, we mus require h D d = = d ( ) or D d.7 The insnneous elociy ny ime is he slope of he x s. grph h ime. We compue his slope by using wo poins on srigh segmen of he cure, one poin on ech side of he poin of ineres. Pge.6

7 Chper () =. s. m. s 5. m s = 3. s ( ) ( ) 5.. m.5 m s 4.. s (c) = 4.5 s ( ) ( ) m s (d) = 7.5 s ( 5. m) ( ) s 5. m s.8 () A few ypicl lues re (s) x(m) We will use.4 s inerl cenered = 4. s. We find = 3.8 s, x = 6. m nd = 4. s, x = Therefore, Δ x 6.4 m Δ.4 s 4. ms / Using ime inerl of. s, we find he corresponding lues o be: = 3.9 s, x = 64. m nd = 4. s, x = 7. m. Thus, Δ x 8. m Δ. s 4. ms / For ime inerl of. s, he lues re: = 3.95 s, x = 66. m, nd = 4.5 s, x = 7. m Therefore, Pge.7

8 Chper Δ x 4. m Δ. s 4. ms / (c) A = 4. s, x = 68. m. Thus, for he firs 4. s, Δx 68. m 7. ms / Δ 4. s This lue is much less hn he insnneous elociy = 4. s..9 Choose coordine xis wih he origin he flgpole nd es s he posiie direcion. Then, using x = x + + wih = for ech runner, he x-coordine of ech runner ime is x = 4. mi + 6. mi h nd x = 3. mi + 5. mi h A B When he runners mee, x A = x B, giing 4. mi +(6. mi/h) = 3. mi + ( 5. mi/h), Or(6. mi/h + 5. mi/h) = 3. mi + 4. mi. This gies he elpsed ime when hey mee s 7. mi. mi h.64 h A his ime, x A = x =.8 mi. Thus, hey mee.8 mi wes of he flgpole. B. The erge speed during ime inerl is = disnce reled Δ During ny qurer mile segmen, he disnce reled is mi 5 8 f Δ x = 3 f 4 = mi () During he firs qurer mile segmen, Secreri s erge speed ws 3 f 5. s 5.4 f s Pge.8

9 Chper During he second qurer mile segmen, 3 f 4. s 55. f s For he hird qurer mile of he rce, 3 3 f 3.8 s 55.5 f s For he fourh finl qurer mile, 4 3 f 3. s 56.9 f s nd during he finl qurer mile, 5 3 f 3. s 57.4 f s Assuming h finl = 5 nd recognizing h υ =, he erge ccelerion for he enire rce ws 57.4 f s ol elpsed ime s finl ( ).48 f s. From υ = Δυ/Δ, he required ime is seen o be Δ 6. mi h g.447 m s Δ 7g 9.8 m s mi h.39 s.3 From = Δυ/Δ, we he Δ 6 55 mi h.447 m s Δ.6 m s = mi h 3.7 s..4 () From = o = 5. s, f i 8. m s 8. m s 5. s f i Pge.9

10 Chper From o = 5 s, ( ) 8. m s 8. m s 5 s 5. s.6 m s nd from = o = s, ( ) 8. m s 8. m s s.8 m s A ny insn, he insnneous ccelerion equls he slope of he line ngen o he υ s. grph h poin in ime. A =. s, he slope of he ngen line o he cure is. A = s, he slope of he ngen line is.6 m s, nd = 8 s, he slope of he ngen line is..6 We choose eswrd s he posiie direcion so he iniil elociy of he cr is gien by =+ 5. m s. () In his cse, he ccelerion is =+.75 m s nd he finl elociy will be or = + =+ 5. m s m s 8.5 s =+ 3.4 m s = 3.4 m s eswrd When he ccelerion is direced weswrd, =.75 m s, he finl elociy is = + = + 5. m s +.75 m s 8.5 s = m s, or = 8.6 m s eswrd..7 Choose he direcion of he cr s moion (eswrd) s he posiie direcion. Then, he iniil elociy of he cr is =+ 4. m s nd he finl elociy (fer n elpsed ime of Δ = 3.5 s ) is =+ 5. m s. () The cr s ccelerion is Δ 5. m s 4. m s = 4.9 m s or = 4.9 m s weswrd Δ Δ 3.5 s The disnce reled during he 3.5 s ime inerl is + 5. m s + 4. m s Δ x = ( Δ ) = Δ = ( 3.5 s) = 4 m Pge.

11 Chper.9 () Δ x = ( Δ ) = ( + ) Δ becomes 4. m ( 8.5 s ).8m s + =, which yields = 4. m.8 m s = 6.6 m s 8.5 s.8 m s 6.6 m s.448 m s Δ 8.5 s.3 () The known quniies re iniil elociy, finl elociy, nd displcemen. The kinemics equion h reles hese quniies o ccelerion is = + ( Δ ) f i x. (c) = f i ( Δ x ) (d) ( x ) ( 3. m s) (. m s) f i Δ. m.5 m s (e) Using =ΔΔ, we find h Δ f i 3. m s. m s Δ =.5 m s 8. s.3 () Wih = km h, = + ( Δx) yields ( x ) km h Δ 4 m km h.78 m s.3m/s The required ime is km h.78 m s Δ.3 m s = km h 4.4 s..3 () The ime for he ruck o rech m/s, sring from res, is found from: = + : Pge.

12 Chper m s s speed up. m s The ol ime for he rip is ol = speed + consn + brking = s + s + 5. s = 35 s up speed. The disnce reled during he firs s is + m s + Δ = s = m ( x) speed up speed speed speed up up up The disnce reled during he nex s (wih = ) is ( x) consn Δ = = m s s = 4 m speed consn speed The disnce reled in he ls 5. s is + + m s Δ = brking brking brking = brking = 5. s = 5 m f ( x) The ol displcemen is hen ( x) ( x) speed + ( x) consn + ( x) Δ = Δ Δ Δ = m + 4 m + 5 m = 55 m ol up speed brking nd he erge elociy for he enire rip is rip ( Δ x ) ol 55 m 35 s ol 6 m s.33 () 4. m s 8.4 m s Δ.95 s From =ΔΔ, he required ime is f i. m s. m s Δ 8.4 m s.3 s. (c) Yes. For uniform ccelerion, he chnge in elociy Δυ genered in ime Δ is gien by Δυ = (Δ). From his, i is seen h doubling he lengh of he ime inerl Δ will lwys double he chnge in elociy Δυ. A more precise wy of sing his is: When ccelerion is consn, elociy is liner funcion of Pge.

13 Chper ime..34 () The ime required o sop he plne is m s 5. m s. s The minimum disnce needed o sop is + + m s Δ x = = = (. s) = m =. km Thus, he plne requires minimum runwy lengh of. km. I cnno lnd sfely on.8 km runwy..36 The elociy he end of he firs inerl is = + = + (.77 m s) 5. s = 4.6 m s This is lso he consn elociy during he second inerl nd he iniil elociy for he hird inerl. () From Δ x = +, he ol displcemen is ( Δ x) = ( Δ x) + ( Δ x) + ( Δx) ol 3 = + (.77 m s )( 5. s) + ( 4.6 m s)( 3 s) ( 4.6 m s)( 4.39 s ) ( 9.47 m s )( 4.39 s) or ( x 3 3 ) Δ = 3 m + 5. m + 9. m = 5.5 m = 5.5 km ol ( Δ x ) 3 m 5. s.8 m s Pge.3

14 Chper ( x ) 3 Δ 5. m 3 s 4.6 m s 3 ( Δ x ) 3 9. m 4.39 s 3.8 m/s nd he erge elociy for he ol rip is ol ( x ) Δ 3 ol 5.5 m s ol ( + + ) 38.7 m s.37 Using he uniformly ccelered moion equion Δ x = + for he full 4 s inerl yields Δ x = m s 4 s +. m s 4 s =, which is obiously wrong. The source of he error is found by compuing he ime required for he rin o come o res. This ime is m s. m s s Thus, he rin is slowing down for he firs s nd is res for he ls s of he 4 s inerl. The ccelerion is no consn during he full 4 s. I is, howeer, consn during he firs s s he rin slows o res. Applicion of Δ x = + o his inerl gies he sopping disnce s Δ x = m s s +. m s s = m mi.447 m s.38 = nd f = 4. = 7.9 m s h mi h () To find he disnce reled, we use f m s + Δ x = = = (. s) = 7 m Pge.4

15 Chper The consn ccelerion is f 7.9 m s. s.49 m s.39 A he end of he ccelerion period, he elociy is = + ccel = +.5 m s 5. s = 7.5 m s This is lso he iniil elociy for he brking period. f = + brke = 7.5 m s +. m s 3. s =.5 m s. () Afer brking, The ol disnce reled is + + Δ x = ( Δ x) + ( Δ x) = ( ) + ( ) = + ccel brke ccel brke f ol ccel brke 7.5 ms+.5 ms+ 7.5 ms Δ xol = ( 5. s) + ( 3. s) = 3 m.4 For he ccelerion period, he prmeers for he cr re: iniil elociy = υ i =, ccelerion = =, elpsed ime = (Δ) = nd finl elociey = υ f. For he brking period, he prmeers re: iniil elociy = υ ib = finl el. of ccel. period = υ f, ccelerion = b =, nd elpsed ime = (Δ) b =. () To deermine he elociy of he cr jus before he brkes re engged, we pply = + ( Δ ) ccelerion period nd find f i o he or = = + Δ = + = ib f i ib We my use Δ x = Δ + Δ o deermine he disnce reled during he ccelerion period i (i.e., before he drier begins o brke). This gies ( Δ x ) = ( Δ ) + ( Δ ) = + or ( Δ x ) = i Pge.5

16 Chper (c) The displcemen occurring during he brking period is Δ x = Δ + Δ = + ib b b b b Thus, he ol displcemen of he cr during he wo inerls combined is ( Δ x ) = ( Δ x ) + ( Δ x ) = + + ol b.4 The ime he Thunderbird spends slowing down is ( Δ x ) Δ x 5 m Δ = m s 6.99 s The ime required o regin speed fer he pi sop is ( Δ x ) Δ x 35 m Δ = m s s Thus, he ol elpsed ime before he Thunderbird is bck up o speed is Δ = Δ + 5. s + Δ = 6.99 s + 5. s s =.8 s During his ime, he Mercedes hs reled ( consn speed) disnce Δ xm = Δ = 7.5 m s.8 s = 558 m nd he Thunderbird hs fllen behind disnce d =ΔxM Δx Δ x = 558 m 5 m 35 m = 958 m.4 The cr is disnce d from he dog nd hs iniil elociy when he brkes re pplied, giing i consn ccelerion. Apply =Δx Δ = + o he enire rip (for which Δx = d +4. m, Δ = s, nd υ = ) o obin Pge.6

17 Chper d + 4. m = + s or d + 4. m = [] 5. s Then, pplying ( = + Δx) o he enire rip yields = + ( d + 4. m ). Subsiue for from Equion [] o find h ( d + 4. m) d + 4. m = + ( d + 4. m ) nd = [] 5 s 5 s Finlly, pply Δ x = + o he firs 8. s of he rip (for which Δ x = d ). This gies d ( 8. s 64 s = + ) [3] Subsiue Equions [] nd [] ino Equion [3] o obin d d + 4. m d + 4. m = ( 8. s) ( 64 s ).96d 3.84 m 5.s + = 5s + which yields d = 3.84 m/.4= 96m..43 () Tke = he ime when he plyer srs o chse his opponen. A his ime, he opponen is disnce = ( m s) ( 3. ) d s = 36 m in fron of he plyer. A ime >, he displcemens of he plyers from heir iniil posiions re Δ xplyer = ( ) + plyer plyer = + ( 4. m s ) [] nd Δ x opponen = ( ) + opponen = ( m s) + [] opponen When he plyers re side-by-side, Pge.7

18 Chper Δ xplyer = Δ xopponen + 36 m [3] Subsiuing Equions [] nd [] ino Equion [3] gies 4. m s m s 36 m ( ) = + or + 6. s + 8 s = Applying he qudric formul o his resul gies = ( 6. s) ( 6. s) 4 ( 8 s ) ± which hs soluions of =. s nd = + 8. s Since he ime mus be greer hn zero, we mus choose = 8. s s he proper nswer. Δ xplyer = + plyer plyer = + 4. m s 8. s =.3 m.44 The iniil elociy of he rin is = 8.4 km h nd he finl elociy is = 6.4 km h for he 4 m rin o pss he crossing is found from Δ x = = + s. The ime required ( x) + ( + ) Δ.4 km s ( 8. h) = 9. s km h h.45 () From = + Δy) wih =, we he ( ( y) ( ) ( ) 5. m s Δ mx 9.8 m s 3.9 m The ime o rech he highes poin is 5. m s 9.8 m s up.55 s Pge.8

19 Chper (c) The ime required for he bll o fll 3.9 m, sring from res, is found from Δ y = ( ) + s ( Δy) ( ) 3.9 m 9.8 m s.55 s (d) The elociy of he bll when i reurns o he originl leel (.55 s fer i srs o fll from res) is = + = m s.55 s = 5. m/s.46 () For he upwrd fligh of he rrow, = + m s, = g = 9.8 m s, nd he finl elociy is =. Thus, = + Δ y yields ( y) ( ) ( ) m s Δ mx 9.8 m s 5 m The ime for he upwrd fligh is up ( Δy) ( Δy) ( 5 m) up mx o mx =. s + m s + Δ y = Δ y = 5 m, =, nd = 9.8 m s Thus, For he downwrd fligh, mx Δ y = + gies ( Δy) ( ) 5 m. s 9.8 m s down nd he ol ime of he fligh is ol = down + down =. s+. s =.4 s..47 The elociy of he objec when i ws 3. m boe he ground cn be deermined by pplying o he ls.5 s of he fll. This gies Δ y = + m 3. m = (.5 s) (.5 s s ) or =.7 m s The displcemen he objec mus he undergone, sring from res, o chiee his elociy poin 3. m s boe he ground is gien by = + ( Δy) Pge.9

20 Chper ( y) ( ) ( ).7 m s Δ 8.3 m 9.8 m s The ol disnce he objec drops during he fll is ( y) ( y) Δ = Δ + 3. m = 38. m ol.48 () Consider he rock s enire upwrd fligh, for which = m s,, = g = 9.8 m s, y i = (king y = ground leel), nd y f = h mx = mximum liude reched by rock. Then pplying.55 m = + f i ( Δy) o his upwrd fligh gies f = ( h ) = 7.4 m s m s.55 m mx nd soling for he mximum liude of he rock gies ( 7.4 m s) h mx =.55 m + = 4.34 m 9.8 m s Since h mx > 3.65 m (heigh of he wll).he rock does rech he op of he wll. To find he elociy of he rock when i reches he op of he wll, we use f = i + ( Δy) nd sole for when y f = 3.65 m (sring wih = m s y =.55 m ). This yields f i i ( y ) = + y = 7.4 m s m s 3.65 m.55 m = 3.69 m s f i f i (c) A rock hrown downwrd speed of 7.4 m s ( 7.4 m s) i = from he op of he wll undergoes displcemen of (Δy) = y f y i =.55 m 3.65 m =. m before reching he leel of he cker. Is elociy when i reches he cker is 7.4 m s 9.8 m s f = i + Δ y = +. m = 9.79 m s so he chnge in speed of his rock s i goes beween he poins loced he op of he wll nd he cker is gien by Pge.

21 Chper ( speed) down 9.79 m s 7.4 m s.39 m s Δ = = = f i (d) Obsere h he chnge in speed of he bll hrown upwrd s i wen from he cker o he op of he wll ws ( speed) up 3.69 m s 7.4 m s 3.7 m s Δ = = = f i Thus, he wo rocks do no undergo he sme mgniude chnge in speeds. As he wo rocks rel beween he leel of he cker nd he leel of he op of he wll, he rock hrown upwrd undergoes greer chnge in speed hn does he rock hrown downwrd. The reson for his is h he rock hrown upwrd hs smller erge speed beween hese wo leels: up i + up f up 7.4 m s m s 5.55 m s nd down i + down f down 7.4 m s m s 8.6 m s Thus, he rock hrown upwrd spends more ime relling beween he wo leels. wih griy chnging is speed by 9.8 m/s for ech second h psses..5 () Afer. s, he elociy of he milbg is bg = + =.5 m s m s. s =. m s The negie sign ells h he bg is moing downwrd nd he mgniude of he elociy gies he speed s. m/s. The displcemen of he milbg fer. s is ( y) +. m s +.5 m s Δ = = (. s) =.6 m bg During his ime, he helicoper, moing downwrd wih consn elociy, undergoes displcemen of Pge.

22 Chper Δ = coper + =.5 m s. s + = 3. m ( y) The disnce sepring he pckge nd he helicoper his ime is hen d = Δy Δ y =.6 m 3. m = 9.6 m = 9.6 m p h = = +.5 m s nd (c) Here, bg coper bg = 9.8 m s while coper =. Afer. s, he elociy of he milbg is m m = (. s) 8. s = s bg m s nd is speed is bg = 8. m s In his cse, he displcemen of he helicoper during he. s inerl is Δ y coper = ( +.5 m s)(. s) + = + 3. m Menwhile, he milbg hs displcemen of + 8. m s +.5 m s Δ. s = 6.6 m bg bg ( y) The disnce sepring he pckge nd he helicoper his ime is hen d = Δy Δ y = 6.6 m + 3. m = 9.6 m = 9.6 m p h.5 () From he insn he bll lees he plyer s hnd unil i is cugh, he bll is freely flling body wih n ccelerion of = g = = 9.8 m s 9.8 m s downwrd A is mximum heigh, he bll comes o res momenrily nd hen begins o fll bck downwrd. Thus, mx heigh =. Pge.

23 Chper (c) Consider he relion Δ y = + wih = g. When he bll is he hrower s hnd, he displcemen is Δy =, giing = g This equion hs wo soluions, = which corresponds o when he bll ws hrown, nd = g corresponding o when he bll is cugh. Therefore, if he bll is cugh =. s, he iniil elociy mus he been ( 9.8 m s )(. s) g 9.8 m/s ( (d) From = + Δy), wih = he mximum heigh, ( y) ( ) ( ) 9.8 m s Δ mx 9.8 m s 4.9 m.5 () Le = be he insn he pckge lees he helicoper, so he pckge nd he helicoper he common iniil elociy of =. (choosing upwrd s posiie). i A imes >, he elociy of he pckge (in free-fll wih consn ccelerion p = g) is gien by = i + s p = g = ( o + g ) nd speed = p = o + g Afer n elpsed ime, he downwrd displcemen of he pckge from is poin of relese will be Δ y = p i + p = g = + g nd he downwrd displcemen of he helicoper (moing wih consn elociy, or ccelerion h = ) from he relese poin his ime is Δ y = i + h = + = h The disnce sepring he pckge nd he helicoper his ime is hen d ( y ) ( y ) p h g = Δ Δ = + = g (c) If he helicoper nd pckge re moing upwrd he insn of relese, hen he common iniil elociy is Pge.3

24 Chper i =+. The ccelerions of he helicoper (moing wih consn elociy) nd he pckge ( freely flling objec) remin unchnged from he preious cse ( p = g nd h = ). In his cse, he pckge speed ime > is p = i + p = g = g A his ime, he displcemens from he relese poin of he pckge nd he helicoper re gien by Δ y = p i + p = g nd Δ y = i + h = + = + h The disnce sepring he pckge nd helicoper ime is now gien by d ( y ) ( y ) p h g = Δ Δ = = g (he sme s erlier!).53 () Afer is engines sop, he rocke is freely flling body. I coninues upwrd, slowing under he influence of griy unil i comes o res momenrily is mximum liude. Then i flls bck o Erh, gining speed s i flls. When i reches heigh of 5 m, he speed of he rocke is ( y) = + Δ = 5. m s +. m s 5 m = 55.7 m s Afer he engines sop, he rocke coninues moing upwrd wih n iniil elociy of = 55.7 m s nd ccelerion = g = 9.8 m/s. When he rocke reches mximum heigh, = he rocke boe he poin where he engines sopped (h is, boe he 5-m leel) is ( ) ( ) 55.7 m s Δ y 58 m 9.8 m s. The displcemen of The mximum heigh boe ground h he rocke reches is hen gien by h mx = 5 m + 58 m = 38 m. (c) The ol ime of he upwrd moion of he rocke is he sum of wo inerls. The firs is he ime for he rocke o go from = 5. m s he ground o elociy of = 55.7 m s n liude of 5 m. This ime is gien by ( Δy) ( Δy) ( + ) ( 5 m) ( + ) = m s.84 s Pge.4

25 Chper The second inerl is he ime o rise 58 m sring wih = 55.7 m s nd ending wih =. This ime is ( Δy) ( Δy) ( + ) 58 m = m s 5.67 s The ol ime of he upwrd fligh is hen up = + = s = 8.5s (d) The ime for he rocke o fll 38 m bck o he ground, wih = nd ccelerion = g = 9.8 m/s, is found from Δ y = + s ( Δy) ( ) 38 m 9.8 m s down 7.93 s so he ol ime of he fligh is fligh = up + down = s = 6.4 s..54 () The cmer flls 5 m wih free-fll ccelerion, sring wih = m s. Is elociy when i reches he ground is ( y) = + Δ = + = ms 9.8 ms 5 m 33 ms The ime o rech he ground is gien by 33 m s m s 9.8 m s.3 s This elociy ws found o be = 33 m s in pr () boe..55 During he.6 s required for he rig o pss compleely ono he bridge, he fron bumper of he rcor moes disnce equl o he lengh of he rig consn elociy of = km h. Therefore, he lengh of he rig is L rig km.78 m s = = (.6 s) 6.7 m h = km h Pge.5

26 Chper While some pr of he rig is on he bridge, he fron bumper moes disnce Δ x= L + L = 4 m m bridge rig Wih consn elociy of, he ime for his o occur is Lbridge + Lrig 4 m m km h 5. s km h.78 m s.56 () From, we he m ( 3. m s) ( 3.5 m s ) Δ x = + = +. This reduces o ( 6. s) + ( s ) =, nd he qudric formul gies = ( 6. s) ( 6. s) 4 ( 3.5)( s ) ( 3.5) ± The desired ime is he smller soluion of = 4.53 s. The lrger soluion of =.6 s is he ime when he bo would pss he buoy moing bckwrds, ssuming i minined consn ccelerion. The elociy of he bo when i firs reches he buoy is = + = 3. m s m s 4.53 s = 4.m/s.57 () The ccelerion of he bulle is ( Δ x ) ( 3 m/s) ( 4 m/s) 5 υ υ 3.5 m/s. m The ime of conc wih he bord is ( ) 3 4 m s.86 4 s m s.58 We ssume h he bulle begins o slow jus s he fron end ouches he firs surfce of he bord, nd h i reches is exi elociy jus s he fron end emerges from he opposie fce of he bord. () The ccelerion is Pge.6

27 Chper ( Δx ) ( 8 m s) ( 4 m s) exi m s. m The erge elociy s he fron of he bulle psses hrough he bord is exi + 8 m s + 4 m s 35 m s nd he ol ime of conc wih he bord is he ime for he fron of he bulle o pss hrough plus he ddiionl ime for he riling end o emerge ( speed exi ), ( Δ x ) bord L bulle. m. m 4 = + = + = 3.57 s 35 m s 8 m s exi ( ) (c) From = + Δx, wih =, gies he required hickness is ( 4 m s) 5 ( ) Δ x =.8 m = 8. cm 4.9 m s.59 () The keys he ccelerion = g = 9.8 m/s from he relese poin unil hey re cugh.5 s ler. Thus, Δ y = + gies ( + ) ( ) Δy 4. m 9.8 m s.5 s +. m s.5 s or =. m s upwrd The elociy of he keys jus before he cch ws = + =. m s m s.5 s = 4.68 m s or = 4.68 m s downwrd Pge.7

28 Chper.6 () The keys, moing freely under he influence of griy ( = g), undergo ericl displcemen of Δ y = + h in ime. We use i Δ y = + o find he iniil elociy s h = i + g giing h + g h g i + The elociy of he keys jus before hey were cugh ( ime ) is gien by = + s i h g = + + ( g) = h + g g = h g ( ).6 () From = + Δy, he insec s elociy fer srighening is legs is = + Δ y = + 4 m s. 3 m = 4. m s nd he ime o rech his elociy is 4. m s. 3 s =. ms 4 m s The upwrd displcemen of he insec beween when is fee lee he ground nd i comes o res momenrily mximum liude is ( g) ( 4. m s) ( ) Δ y = 9.8 m s.8 m.63 The flling bll moes disnce of (5 m h) before hey mee, where h is he heigh boe he ground where hey mee. Apply Δ y = +, wih = g, o obin ( 5 m h ) = g or h = 5 m g [] Pge.8

29 Chper Applying Δ y = + o he rising bll gies h ( 5 m s ) = g [] Combining equions [] nd [] gies ( 5 m s ) g = g 5 m or 5 m 5 m s.6 s.64 The consn speed he suden hs minined for he firs minues, nd hence her iniil speed for he finl 5 yrd dsh, is ( 5 8 f 5 f) Δ x. mi 5 yrds m =.9 m s Δ min 6 s 3.8 f Wih n iniil speed of =.9 m s nd consn ccelerion of =.5 m/s, he mximum disnce he suden cn rel in he remining. min ( s) of her lloed ime is m m Δ 3. = mx + mx =.9 s +.5 s =.3 m s s ( x ) or 3.8 f yrd Δ. mx = = m 3f ( x ).3 3 m.4 3 yrds Since ( Δ x. ) mx is considerbly greer hn he 5 yrds she mus sill run, she cn esily mee he requirmen of running. miles in minues..65 We sole Pr of his problem firs. When he eiher bll reches he ground, is displcemen from he blcony is Δy = 9.6 m (king upwrd Pge.9

30 Chper s posiie). The iniil elociies of he wo blls were = 4.7 m s nd =+ 4.7 m s, so hs he lue of (4.7 m/s) for eiher bll. Also, = g for ech bll, giing he downwrd elociy of eiher bll when i reches he ground s ( y) = + Δ = 4.5 m s m s 9.6 m = 4.5 m s eiher bll () The ime for eiher bll o rech he ground (nd hence chiee he elociy compued boe) is gien by eiher bll 4.5 m s 4.5 m s + g 9.8 m s where is he iniil elociy of he priculr bll of ineres. For bll, = 4.7 m s, giing 4.5 m s 4.7 m s. s 9.8 m s For bll, =+ 4.7 m s, nd 4.5 m s m s 9.8 m s 4. s The difference in he ime of fligh for he wo blls is seen o be Δ = = 4.. s = 3. s (c) A =.8 s, he displcemen of ech bll from he blcony ( heigh h boe ground) is y h = g = ( 4.7 m s)(.8 s) ( 4.9 m s )(.8 s) y h = g = + ( 4.7 m s)(.8 s) ( 4.9 m s )(.8 s) These gie he liudes of he wo blls =.8 s s y = h 4.9 m nd y = h m. Therefore he disnce sepring he wo blls his ime is Pge.3

31 Chper d = y y = h m h 4.9 m = 3.5 m.66 () While in he ir, boh blls he ccelerion = = g (where upwrd is ken s posiie). Bll (hrown downwrd) hs iniil elociy =, while bll (hrown upwrd) hs iniil elociy = +. Tking y = ground leel, he iniil y-coordine of ech bll is y = y = +h. Applying i i Δ y = y y = + o ech bll gies heir y-coordines ime s y h = υ + g or y = h g Bll : y h = + + g or y = h + g Bll : A ground leel, y =. Thus, we eque ech of he equions found boe o zero nd use he qudric formul o sole for he imes when ech bll reches he ground. This gies he following: = g + + = Bll : h g ( ) ( h ) so ± 4 g h h ± + g g g g Using only he posiie soluion gies h g g g = + + = + g + + = Bll : h g ( ) ( h ) nd ± 4 g h h + ± + g g g g Pge.3

32 Chper Agin, using only he posiie soluion h g g g = + + Thus, he difference in he imes of fligh of he wo blls is h h g g g g g g g Δ = = = (c) Relizing h he blls re going downwrd ( < ) s hey ner he ground, we use = + ( Δ ) Δy = h o find he elociy of ech bll jus before i srikes he ground: y wih f i Bll : = + h = + g h = + gh f i Bll : = + h = + + g h = + gh f i (d) While boh blls re sill in he ir, he disnce sepring hem is d = y y = h+ g h g = o.67 () The firs bll is dropped from res (ν = ) from he heigh h of he window. Thus, = + ( Δ ) y gies he speed of his bll s i reches he ground (nd hence he iniil elociy of he second bll) s = + Δ y = + g h = gh f. When bll is hrown upwrd he sme ime h bll f is dropped, heir y-coordines ime during he flighs re gien by y y = + o s y h = + g or Bll : y = h g Bll : y = ( gh) + ( g) or y = ( gh) g When he wo blls pss, y = y, or Pge.3

33 Chper h g = ( gh) g giing = h 8.7 m. s = h gh g = 9.8 m s = When he blls mee, = h g nd h h 3h y = h g = = h g 4 4 Thus, he disnce below he window where his een occurs is 3h h 8.7 m d = h y = h 7.8 m We do no know eiher he iniil elociy nor he finl elociy (h is, elociy jus before impc) for he ruck. Wh we do know is h he ruck skids 6.4 m in 4. s while ccelering 5.6 m/s. We he ν = ν + nd Δ x = = [( + ) / ]. Applied o he moion of he ruck, hese yield nd = or = 5.6 m s 4. s = 3.5 m s [] ( Δx) 6.4 m m s 4. s [] Adding equions [] nd [] gies he elociy jus before impc s = m s, or = 3. m s Pge.33

34 Chper.69 When relesed from res (ν = ), he bill flls freely wih downwrd ccelerion due o griy ( = g = 9.8 m/s ). Thus, he mgniude of is downwrd displcemen during Did s. s recion ime will be 9.8 m s. s. m cm Δ y = + = + This is oer wice he disnce from he cener of he bill o is op edge ( 8 cm), so.did will be unsuccessful..7 () The elociy wih which he firs sone his he wer is m m = + ( Δ y) = ( 5. m) = 3.4 s s m s The ime for his sone o hi he wer is 3.4 m s. m s = 9.8 m s 3. s Since hey hi simulneously, he second sone which is relesed. s ler will hi he wer fer n fligh ime of. s. Thus, Δy 5. m 9.8 m s. s 5. m s. s (c) From pr (), he finl elociy of he firs sone is = 3.4 m s. The finl elociy of he second sone is = + = 5. m s m s. s = 34.8 m s.7 () The sled s displcemen, Δx, while ccelering =+ for ime is 4 f s Δ = ( ) + = ( f s ) x or Δ ( ) = f s x [] A he end of ime, he sled hd chieed elociy of ( = + = + 4f s ) or = ( 4 f s ) The displcemen of he sled while moing consn elociy for ime is [] Pge.34

35 Chper ( 4 f s ) x or ( 4 f s ) Δ Δ x = [3] I is known h Δ x +Δ x = 7 5 f, nd subsiuions from Equions [] nd [3] gie f s + 4 f s = 75 f or + = 875 s [4] Also, i is known h + = 9 s [5] Soling Equions [4] nd [5] simulneously yields + 9 s = 875 s or + 8 s s = The qudric formul hen gies = ( 8 s) ( 8 s) 4( 875 s ) ± () wih soluions = 5. s nd = 9 s 5. s = 85 s or = 75 s ( nd = 85 s). Since i is necessry h >, he lid soluions re = 5. s nd = 85 s. = 4 f s = 4 f s 5. s = f s. From Equion [] boe, (c) The displcemen Δx 3 of he sled s i comes o res (wih ccelerion 3 = f/s ) is ( f s) Δ x f ( ) 3 3 f s Thus, he ol displcemen for he rip (mesured from he sring poin) is Δ xol = Δ x +Δ x +Δ x 3 = 7 5 f + f = 8 5 f (d) The ime required o come o res from elociy (wih ccelerion 3 ) is Pge.35

36 Chper f s 3 3 f s s so he durion of he enire rip is = = 5. s + 85 s + s = s. ol.7 () From Δ = + y wih ν =, we he ( Δy) ( ) 3 m 9.8 m s. s = ms ms. s = m/s. The finl elociy is (c) The ime i kes for he sound of he impc o rech he specor is sound Δy 3 m 34 m s sound 6.8 s so he ol elpsed ime is. ol =. s s.3 s.73 () Since he sound hs consn elociy, he disnce i reled is 3 Δ x = = f s 5. s = 5.5 f sound The plne rels his disnce in ime of 5. s + s = 5 s, so is elociy mus be plne 3 Δ x 5.5 f 5 s 3.7 f s (c) The ime he ligh ook o rech he obserer ws ligh 3 Δ x 5.5 f m s 8 3. m s 3.8 f s ligh s Pge.36

37 Chper During his ime he plne would only rel disnce of. f. l d d.74 The disnce he glider moes during he ime Δ d is gien by Δ ( Δ ) + ( Δ ) x, where ν is he glider s elociy when he flg firs eners he phooge nd is he glider s ccelerion. Thus, he erge elociy is d ( Δ ) + ( Δ ) l d d + Δ Δ Δ d d ( d ) () The glider s elociy when i is hlfwy hrough he phooge in spce ( i.e., when Δ = ) = + Δ x s ( l ) l = + = + = + d Δ d = + d Δd x l is found from Noe h his is no equl o d unless =, in which cse ν = ν d = ν. The speed ν when he glider is hlfwy hrough he phooge in ime (i.e., when he elpsed ime is = Δ d /) is gien by ν = ν + s = + = + Δ d = + Δ d which is equl o ν d for ll possible lues of ν =..75 The ime required for he sun mn o fll 3. m, sring from res, is found from Δ y = + s 3. m = + ( 9.8 m s ) so 3. m.78 s 9.8 m s () Wih he horse moing wih consn elociy of. m/s, he horizonl disnce is (. m s)(.78 s) 7.8 m Δ x = horse The required ime is =.78 s s clculed boe. Pge.37