t = x v = 18.4m 44.4m/s =0.414 s.

Transcription

1 1 Assuming he horizonal velociy of he ball is consan, he horizonal displacemen is x = v where x is he horizonal disance raveled, is he ime, and v is he (horizonal) velociy Convering v o meers per second, we have 160 km/h = 444 m/s Thus = x v = 184m 444m/s =0414 s The velociy-uni conversion implemened above can be figured from basics (1000 m = 1 km, 3600 s = 1 h) or found in Appendix D Convering o SI unis, we use Eq -3 wih d for disance (1106km/h) ( 1000 m/km 3600 s/h s avg = d ) = 000m which yields =6510 s We convered he speed km/h m/s by convering each uni (km m, h s) individually Bu we menion ha he one-sep conversion can be found in Appendix D (1 km/h = 0778 m/s) 3 We use Eq - and Eq -3 During a ime c when he velociy remains a posiive consan, speed is equivalen o velociy, and disance is equivalen o displacemen, wih x = v c (a) During he firs par of he moion, he displacemen is x 1 =40km and he ime inerval is 1 = (40 km) (30 km/h) =133 h During he second par he displacemen is x = 40 km and he ime inerval is (40 km) = (60 km/h) =067h Boh displacemens are in he same direcion, so he oal displacemen is x = x 1 + x = 40 km + 40 km = 80 km The oal ime for he rip is = 1 + =00 h Consequenly, he average velociy is (80 km) v avg = (0h) =40km/h (b) In his example, he numerical resul for he average speed is he same as he average velociy 40 km/h

2 (c) In he ineres of saving space, we briefly describe he graph (wih kilomeers and hours undersood): wo coniguous line segmens, he firs having a slope of 30 and connecing he origin o ( 1,x 1 ) = (133, 40) and he second having a slope of 60 and connecing ( 1,x 1 ) o (, x) =(00, 80) The average velociy, from he graphical poin of view, is he slope of a line drawn from he origin o (, x) 4 If he plane (wih velociy v) mainains is presen course, and if he errain coninues is upward slope of 43, hen he plane will srike he ground afer raveling x = h an θ = 35 m =4655 m 0465 km an 43 This corresponds o a ime of fligh found from Eq - (wih v = v avg since i is consan) = x v 0465 km = = h 13 s 1300 km/h This, hen, esimaes he ime available o he pilo o make his correcion 5 (a) Denoing he ravel ime and disance from San Anonio o Houson as T and D, respecively, he average speed is s avg 1 = D T = (55 km/h) T + (90 km/h) T T =75 km/h which should be rounded o 73 km/h (b) Using he fac ha ime = disance/speed while he speed is consan, we find s avg = D T = D D/ 55 km/h + D/ 90 km/h =683 km/h which should be rounded o 68 km/h (c) The oal disance raveled (D) mus no be confused wih he ne displacemen (zero) We obain for he wo-way rip s avg = D D 75km/h + D 683km/h =70km/h (d) Since he ne displacemen vanishes, he average velociy for he rip in is enirey is zero (e) In asking for a skech, he problem is allowing he suden o arbirarily se he disance D (he inen is no o make he suden go o an Alas o look i up); he suden can jus as easily arbirarily se T insead of D, as will be clear in he following discussion

3 In he ineres of saving space, we briefly describe he graph (wih kilomeers-per-hour undersood for he slopes): wo coniguous line segmens, he firs having a slope of 55 and connecing he origin o ( 1,x 1 )=(T/, 55T/) and he second having a slope of 90 and connecing ( 1,x 1 )o(t,d) where D = ( )T/ The average velociy, from he graphical poin of view, is he slope of a line drawn from he origin o (T,D) 6 (a) Using he fac ha ime = disance/velociy while he velociy is consan, we find v avg = 73m+73m 73m 1 m/s + 73m 305 m =174 m/s (b) Using he fac ha disance = v while he velociy v is consan, we find (1 m/s)(60 s) + (305 m/s)(60 s) v avg = =14 m/s 10 s (c) The graphs are shown below (wih meers and seconds undersood) The firs consiss of wo (solid) line segmens, he firs having a slope of 1 and he second having a slope of 305 The slope of he dashed line represens he average velociy (in boh graphs) The second graph also consiss of wo (solid) line segmens, having he same slopes as before he main difference (compared o he firs graph) being ha he sage involving higher-speed moion lass much longer x x Using x = wih SI unis undersood is efficien (and is he approach we will use), bu if we wished o make he unis explici we would wrie x =(3m/s) (4 m/s ) +(1m/s 3 ) 3 We will quoe our answers o one or wo significan figures, and no ry o follow he significan figure rules rigorously

4 (a) Plugging in = 1 s yields x = 0 Wih =swegex = m Similarly, = 3 s yields x =0and = 4 s yields x = 1 m For laer reference, we also noe ha he posiion a =0isx =0 (b) The posiion a = 0 is subraced from he posiion a =4so find he displacemen x =1m (c) The posiion a = s is subraced from he posiion a =4so give he displacemen x = 14 m Eq -, hen, leads o v avg = x = 14 =7m/s (d) The horizonal axis is 0 4 wih SI unis undersood No shown is a sraigh line drawn from he poin a (, x) = (, ) o he highes poin shown (a = 4 s) which would represen he answer for par (c) 8 Recognizing ha he gap beween he rains is closing a a consan rae of 60 km/h, he oal ime which elapses before hey crash is = (60 km)/(60 km/h) = 10 h During his ime, he bird ravels a disance of x = v =(60km/h)(10 h) = 60 km 9 Convering o seconds, he running imes are 1 = s and = s, respecively If he runners were equally fas, hen s avg 1 = s avg = L 1 1 = L From his we obain ( ) L L 1 = L m where we se L m in he las sep Thus, if L 1 and L are no differen han abou 135 m, hen runner 1 is indeed faser han runner However, if L 1 is shorer han L han 14 m hen runner is acually he faser 10 The velociy (boh magniude and sign) is deermined by he slope of he x versus curve, in accordance wih Eq -4

5 (a) The armadillo is o he lef of he coordinae origin on he axis beween =0 sand =40 s (b) The velociy is negaive beween =0and =30 s (c) The velociy is posiive beween =30 sand =70 s (d) The velociy is zero a he graph minimum (a =30 s) 11 We use Eq -4 (a) The velociy of he paricle is v = dx d = d ( ) = d Thus, a = 1 s, he velociy is v =( 1 + (6)(1)) = 6m/s (b) Since v<0, i is moving in he negaive x direcion a =1s (c) A =1s,hespeed is v = 6m/s (d) For 0 <<s, v decreases unil i vanishes For <<3s, v increases from zero o he value i had in par (c) Then, v is larger han ha value for >3s (e) Yes, since v smoohly changes from negaive values (consider he = 1 resul) o posiive (noe ha as +, wehavev + ) One can check ha v = 0 when =s (f) No In fac, from v = 1 + 6, we know ha v>0for>s 1 We use Eq - for average velociy and Eq -4 for insananeous velociy, and work wih disances in cenimeers and imes in seconds (a) We plug ino he given equaion for x for =00 s and =300 s and obain x =175 cm and x 3 =505 cm, respecively The average velociy during he ime inerval s is v avg = x = 505 cm 175 cm 300 s 00 s which yields v avg =85 cm/s (b) The insananeous velociy is v = dx d = 45, which yields v = (45)(00) =180 cm/s a ime =00 s (c) A = 300 s, he insananeous velociy is v = (45)(300) = 405 cm/s (d) A = 50 s, he insananeous velociy is v = (45)(50) = 81 cm/s (e) Le m sand for he momen when he paricle is midway beween x and x 3 (ha is, when he paricle is a x m =(x + x 3 )/ = 36cm) Therefore, x m = m = m =596 in seconds Thus, he insananeous speed a his ime is v =45(596) = 303 cm/s

6 (f) The answer o par (a) is given by he slope of he sraigh line beween =and = 3 in his x- vs- plo The answers x (cm) o pars (b), (c), (d) and (e) 60 correspond o he slopes of angen lines (no shown 40 (a) bu easily imagined) o he curve 0 a he appropriae poins 3 13 Since v = dx d (Eq -4), hen x = vd, which corresponds o he area under he v vs graph Dividing he oal area A ino recangular (base heigh) and riangular ( 1 base heigh) areas, we have A = A 0<< + A <<10 + A 10<<1 + A 1<<16 = 1 ()(8) + (8)(8) + ( ()(4) + 1 ()(4) ) + (4)(4) wih SI unis undersood In his way, we obain x = 100 m 14 From Eq -4 and Eq -9, we noe ha he sign of he velociy is he sign of he slope in an x-vs- plo, and he sign of he acceleraion corresponds o wheher such a curve is concave up or concave down In he ineres of saving space, we indicae sample poins for pars (a)-(d) in a single figure; his means ha all poins are no a = 1 s (which we feel is an accepable modificaion of he problem since he daum = 1 s is no used) Any change from zero o non-zero values of v represens increasing v (speed) Also, v a implies ha he paricle is going faser Thus, poins (a), (b) and (d) involve increasing speed 15 We appeal o Eq -4 and Eq -9 (a) This is v ha is, he velociy squared (b) This is he acceleraion a

7 (c) The SI unis for hese quaniies are (m/s) = m /s and m/s, respecively 16 Eq -9 indicaes ha acceleraion is he slope of he v-vs- graph Based on his, we show here a skech 10 of he acceleraion (in m/s ) as a funcion 0 a of ime The values along he acceleraion 10 axis should no be aken oo seriously 0 17 We represen is iniial direcion of moion as he +x direcion, so ha v 0 = +18 m/s and v = 30 m/s (when =4 s) Using Eq -7 (or Eq -11, suiably inerpreed) we find a avg = ( 30) (+18) 4 = 0 m/s which indicaes ha he average acceleraion has magniude 0 m/s and is in he opposie direcion o he paricle s iniial velociy 18 We use Eq - (average velociy) and Eq -7 (average acceleraion) Regarding our coordinae choices, he iniial posiion of he man is aken as he origin and his direcion of moion during 5 min 10 min is aken o be he posiive x direcion We also use he fac ha x = v when he velociy is consan during a ime inerval (a) Here, he enire inerval considered is =8 = 6min which is equivalen o 360 s, whereas he sub-inerval in which he is moving is only =8 5 = 3 min = 180 s His posiion a = min is x =0 and his posiion a = 8 min is x = v =()(180) = 396m Therefore, v avg = 396m 0 =110 m/s 360 s (b) The man is a res a = min and has velociy v =+ m/s a = 8 min Thus, keeping he answer o 3 significan figures, a avg = m/s 0 = m/s 360 s (c) Now, he enire inerval considered is =9 3 = 6min (360 s again), whereas he sub-inerval in which he is moving is =9 5 = 4 min = 40 s) His posiion a = 3 min is x = 0 and his posiion a = 9 min is x = v =()(40) = 58 m Therefore, v avg = 58 m 0 =147 m/s 360 s 5

8 (d) The man is a res a = 3 min and has velociy v =+ m/s a = 9 min Consequenly, a avg =/360 = m/s jus as in par (b) (e) The horizonal line near he boom of his x-vs- graph represens he man sanding a x = 0 for 0 < 300 s and he linearly rising line for s (c) represens his consan-velociy moion The doed lines represen he answers o par (a) and (c) in he sense ha heir slopes yield hose resuls x 0 (a) The graph of v-vs- is no shown here, bu would consis of wo horizonal seps (one a v =0for0 <300 s and he nex a v =m/s for s) The indicaions of he average acceleraions found in pars (b) and (d) would be doed lines conneced he seps a he appropriae values (he slopes of he doed lines represening he values of a avg ) 19 In his soluion, we make use of he noaion x() for he value of x a a paricular Thus, x() = wih SI unis (meers and seconds) undersood (a) The average velociy during he firs 3 s is given by v avg = x(3) x(0) = (50)(3) + (10)(3) 0 3 =80m/s (b) The insananeous velociy a ime is given by v = dx/d =50+0, in SI unis A =30 s,v = 50 + (0)(30) = 110 m/s (c) The insananeous acceleraion a ime is given by a = dv/d = 0 m/s I is consan, so he acceleraion a any ime is 0 m/s (d) and (e) The graphs below show he coordinae x and velociy v as funcions of ime, wih SI unis undersood The doed line marked (a) in he firs graph runs from =0,x =0o =30s, x = 40 m Is slope is he average velociy during he firs 3 s of moion The doed line marked (b) is angen o he x curve a =30 s Is slope is he insananeous velociy a =30 s

9 x (a) (b) Using he general propery d dx exp(bx) =b exp(bx), we wrie ( ( ) d (19) de v = dx d = d ) e + (19) 50 0 d v If a concern develops abou he appearance of an argumen of he exponenial ( ) apparenly having unis, hen an explici facor of 1/T where T = 1 second can be insered and carried hrough he compuaion (which does no change our answer) The resul of his differeniaion is v = 16(1 )e wih and v in SI unis (s and m/s, respecively) We see ha his funcion is zero when = 1 s Now ha we know when i sops, we find ou where i sops by plugging our resul = 1 ino he given funcion x =16e wih x in meers Therefore, we find x =59 m 1 In his soluion, we make use of he noaion x() for he value of x a a paricular The noaions v() and a() have similar meanings (a) Since he uni of c is ha of lengh, he uni of c mus be ha of lengh/ime, or m/s in he SI sysem Since b 3 has a uni of lengh, b mus have a uni of lengh/ime 3, or m/s 3 (b) When he paricle reaches is maximum (or is minimum) coordinae is velociy is zero Since he velociy is given by v = dx/d = c 3b, v = 0 occurs for = 0 and for = c 3b = (30m/s ) 3(0m/s 3 ) =10s For =0,x = x 0 = 0 and for =10s,x =10m >x 0 Since we seek he maximum, we rejec he firs roo ( = 0) and accep he second ( =1s)

10 (c) In he firs 4 s he paricle moves from he origin o x =10 m, urns around, and goes back o x(4 s) = (30m/s )(40s) (0m/s 3 )(40s) 3 = 80 m The oal pah lengh i ravels is 10m+10m+80m=8m (d) Is displacemen is given by x = x x 1, where x 1 =0andx = 80 m Thus, x = 80 m (e) The velociy is given by v =c 3b =(60m/s ) (60m/s 3 ) Thus v(1 s) = (60m/s )(10s) (60m/s 3 )(10s) =0 v( s) = (60m/s )(0s) (60m/s 3 )(0s) = 1 m/s v(3 s) = (60m/s )(30s) (60m/s 3 )(30s) = 360m/s v(4 s) = (60m/s )(40s) (60m/s 3 )(40s) = 7 m/s (f) The acceleraion is given by a = dv/d = c 6b = 60m/s (10m/s 3 ) Thus a(1 s) = 60m/s (10m/s 3 )(10s)= 60m/s a( s) = 60m/s (10m/s 3 )(0s)= 18 m/s a(3 s) = 60m/s (10m/s 3 )(30s)= 30 m/s a(4 s) = 60m/s (10m/s 3 )(40s)= 4 m/s For he auomobile v =55 5 = 30 km/h, which we conver o SI unis: ( ) a = v (30 km/h) 1000 m/km = 3600 s/h =08 m/s (050 min)(60 s/min) The change of velociy for he bicycle, for he same ime, is idenical o ha of he car, so is acceleraion is also 08 m/s 3 The consan-acceleraion condiion permis he use of Table -1 (a) Seing v =0andx 0 =0inv = v0 +a(x x 0 ), we find x = 1 v ( ) a = =0100 m Since he muon is slowing, he iniial velociy and he acceleraion mus have opposie signs (b) Below are he ime-plos of he posiion x and velociy v of he muon from he momen i eners he field o he ime i sops The compuaion in par (a) made no reference o, so ha oher equaions from Table -1 (such as v = v 0 + a and x = v a )are

11 10 75 x (cm) 50 used in making hese plos (ns) 80 v (Mm/s) (ns) 4 The ime required is found from Eq -11 (or, suiably inerpreed, Eq - 7) Firs, we conver he velociy change o SI unis: ( ) 1000 m/km v = (100 km/h) =78 m/s 3600 s/h Thus, = v/a =78/50 = 0556s 5 We use v = v 0 + a, wih = 0 as he insan when he velociy equals +96m/s (a) Since we wish o calculae he velociy for a ime before =0,wese = 5 sthus,eq-11gives ( v =(96m/s) + 3m/s ) ( 5s)=16 m/s (b) Now, =+5 sandwefind ( v =(96m/s) + 3m/s ) (5s)=18 m/s

12 6 The bulle sars a res (v 0 = 0) and afer raveling he lengh of he barrel ( x = 1 m) emerges wih he given velociy (v = 640 m/s), where he direcion of moion is he posiive direcion Turning o he consan acceleraion equaions in Table -1, we use x = 1 (v 0 + v) Thus, we find = s (abou 38 ms) 7 The consan acceleraion saed in he problem permis he use of he equaions in Table -1 (a) We solve v = v 0 + a for he ime: = v v ( m/s ) = a 98m/s = s which is equivalen o 1 monhs (b) We evaluae x = x 0 + v a, wih x 0 = 0 The resul is x = 1 ( 98m/s ) ( s ) = m 8 From Table -1, v v0 =a x is used o solve for a Is minimum value is a min = v v0 (360 km/h) = = km/h x max (180 km) which convers o 78 m/s 9 Assuming consan acceleraion permis he use of he equaions in Table -1 We solve v = v0 +a(x x 0 ) wih x 0 =0andx =0010 m Thus, ( a = v v ) ( ) = = m/s x (001) 30 The acceleraion is found from Eq -11 (or, suiably inerpreed, Eq -7) ( ) a = v (100 km/h) 1000 m/km = 3600 s/h = 04 m/s 14s In erms of he graviaional acceleraion g, his is expressed as a muliple of 98 m/s as follows: a = g =1g 31 We choose he posiive direcion o be ha of he iniial velociy of he car (implying ha a<0since i is slowing down) We assume he acceleraion is consan and use Table -1

13 (a) Subsiuing v 0 = 137 km/h =381m/s, v =90km/h =5m/s, and a = 5m/s ino v = v 0 + a, we obain = 5 m/s 38 m/s 5m/s =5 s (b) We ake he car o be a x = 0 when he brakes are applied (a ime = 0) Thus, he coordinae of he car as a funcion of ime is given by 80 x (m) 60 x = (38) + 1 ( 5) in SI unis This funcion is ploed from =0o = 5 s on he graph o he righ We have no shown he v-vs- graph here; i is a descending sraigh line from v 0 o v (s) 3 From he figure, we see ha x 0 = 0 m From Table -1, we can apply x x 0 = v a wih = 10 s, and hen again wih = 0 s This yields wo equaions for he wo unknowns, v 0 and a SI unis are undersood 00 ( 0) = v 0 (10) + 1 a(10) 60 ( 0) = v 0 (0) + 1 a(0) Solving hese simulaneous equaions yields he resuls v 0 =00 anda = 40 m/s The fac ha he answer is posiive ells us ha he acceleraion vecor poins in he +x direcion 33 The problem saemen (see par (a)) indicaes ha a = consan, which allows us o use Table -1 (a) We ake x 0 = 0, and solve x = v a (Eq -15) for he acceleraion: a =(x v 0 )/ Subsiuing x =40m, v 0 =560km/h = 1555 m/s and =00 s, we find a = (40m (1555 m/s)(00 s)) (00 s) = 356m/s The negaive sign indicaes ha he acceleraion is opposie o he direcion of moion of he car The car is slowing down

14 (b) We evaluae v = v 0 + a as follows: ( v =1555 m/s 356m/s ) (00 s) = 843 m/s which is equivalen o 303 km/h 34 We ake he momen of applying brakes o be = 0 The deceleraion is consan so ha Table -1 can be used Our primed variables (such as v o = 7 km/h = 0 m/s) refer o one rain (moving in he +x direcion and locaed a he origin when = 0) and unprimed variables refer o he oher (moving in he x direcion and locaed a x 0 = +950 m when = 0) We noe ha he acceleraion vecor of he unprimed rain poins in he posiive direcion, even hough he rain is slowing down; is iniial velociy is v o = 144 km/h = 40 m/s Since he primed rain has he lower iniial speed, i should sop sooner han he oher rain would (were i no for he collision) Using Eq -16, i should sop (meaning v =0)a x = (v ) (v o) a = 0 0 = 00 m The speed of he oher rain, when i reaches ha locaion, is v = v o +a x = ( 40) + (10)(00 950) = 100 = 10 m/s using Eq -16again Specifically, is velociy a ha momen would be 10 m/s since i is sill raveling in he x direcion when i crashes If he compuaion of v had failed (meaning ha a negaive number would have been inside he square roo) hen we would have looked a he possibiliy ha here was no collision and examined how far apar hey finally were A concern ha can be brough up is wheher he primed rain collides before i comes o res; his can be sudied by compuing he ime i sops (Eq -11 yields = 0 s) and seeing where he unprimed rain is a ha momen (Eq -18 yields x = 350 m, sill a good disance away from conac) 35 The acceleraion is consan and we may use he equaions in Table -1 (a) Taking he firs poin as coordinae origin and ime o be zero when he car is here, we apply Eq -17 (wih SI unis undersood): x = 1 (v + v 0) = 1 (15 + v 0)(6) Wih x =600m (which akes he direcion of moion as he +x direcion) we solve for he iniial velociy: v 0 =500 m/s (b) Subsiuing v = 15 m/s, v 0 = 5 m/s and =6sinoa =(v v 0 )/ (Eq -11), we find a =167 m/s

15 (c) Subsiuing v =0inv = v 0 +ax and solving for x, we obain x = v 0 a = 5 (167) = 750 m (d) The graphs require compuing he ime when v = 0, in which case, we use v = v 0 + a =0 Thus, = v 0 a = = 30 s indicaes he momen he car was a res SI unis are assumed 60 x 15 v We denoe he required ime as, assuming he ligh urns green when he clock reads zero By his ime, he disances raveled by he wo vehicles mus be he same (a) Denoing he acceleraion of he auomobile as a and he (consan) speed of he ruck as v hen ( ) 1 x = a =(v) ruck which leads o Therefore, car = v a = (95) =86 s x = v =(95)(86) = 8 m (b) The speed of he car a ha momen is v car = a =()(86) = 19 m/s

16 37 We denoe r as he reacion ime and b as he braking ime The moion during r is of he consan-velociy (call i v 0 ) ype Then he posiion of he car is given by x = v 0 r + v 0 b + 1 a b where v 0 is he iniial velociy and a is he acceleraion (which we expec o be negaive-valued since we are aking he velociy in he posiive direcion and we know he car is deceleraing) Afer he brakes are applied he velociy of he car is given by v = v 0 + a b Using his equaion, wih v = 0, we eliminae b from he firs equaion and obain x = v 0 r v 0 a + 1 v0 a = v 0 r 1 v0 a We wrie his equaion for each of he iniial velociies: x 1 = v 01 r 1 v01 a and x = v 0 r 1 v0 a Solving hese equaions simulaneously for r and a we ge r = v 0x 1 v 01x v 01 v 0 (v 0 v 01 ) and a = 1 v 0 v01 v 01 v0 v 0 x 1 v 01 x Subsiuing x 1 =567m, v 01 =805km/h =4m/s, x =44 mand v 0 =483km/h =134 m/s, we find r = 134 (567) 4 (44) (4)(134)(134 4) =074 s and a = 1 (134)4 (4)134 (134)(567) (4)(44) = 6 m/s The magniude of he deceleraion is herefore 6 m/s Alhough rounded off values are displayed in he above subsiuions, wha we have inpu ino our calculaors are he exac values (such as v 0 = m/s) 38 In his soluion we elec o wai unil he las sep o conver o SI unis Consan acceleraion is indicaed, so use of Table -1 is permied We sar wih Eq -17 and denoe he rain s iniial velociy as v and he locomoive s velociy as v l (which is also he final velociy of he rain, if he rear-end collision is barely avoided) We noe ha he disance x

17 consiss of he original gap beween hem D as well as he forward disance raveled during his ime by he locomoive v l Therefore, v + v l = x = D + v l = D + v l We now use Eq -11 o eliminae ime from he equaion Thus, leads o Hence, ( v + v l a = 1 a = (0676 km) v + v l = D (v l v ) /a + v l )( ) vl v v l = 1 D D (v l v ) ( 9 km h 161 km h ) = 1888 km/h which we conver as follows: ( a = 1888 km/h ) ( )( ) 1000 m 1h = 0994 m/s 1km 3600 s so ha is magniude is 0994 m/s A graph is shown below for he case where a collision is jus avoided (x along he verical axis is in meers and along he horizonal axis is in seconds) The op (sraigh) line shows he moion of he locomoive and he boom curve shows he moion of he passenger rain The oher case (where he collision is no quie avoided) would be similar excep ha he slope of he boom curve would be greaer han ha of he op line a he poin where hey mee x We assume he periods of acceleraion (duraion 1 ) and deceleraion (duraion ) are periods of consan a so ha Table -1 can be used Taking he direcion of moion o be +x hen a 1 = +1 m/s and a = 1 m/s We use SI unis so he velociy a = 1 is v = 305/6 0 = 508 m/s

18 (a) We denoe x as he disance moved during 1, and use Eq -16: v = v0 +a 1 x = x = 508 (1) which yields x = m (b) Using Eq -11, we have 1 = v v 0 = 508 a 1 1 =417 s The deceleraion ime urns ou o be he same so ha 1 + = 833 s The disances raveled during 1 and are he same so ha hey oal o (1059) = 118 m This implies ha for a disance of = 1688 m, he elevaor is raveling a consan velociy This ime of consan velociy moion is 3 = 1688 m 508 m/s =331 s Therefore, he oal ime is s 40 Neglec of air resisance jusifies seing a = g = 98 m/s (where down is our y direcion) for he duraion of he fall This is consan acceleraion moion, and we may use Table -1 (wih y replacing x) (a) Using Eq -16and aking he negaive roo (since he final velociy is downward), we have v = v0 g y = 0 (98)( 1700) = 183 in SI unis Is magniude is herefore 183 m/s (b) No, bu i is hard o make a convincing case wihou more analysis We esimae he mass of a raindrop o be abou a gram or less, so ha is mass and speed (from par (a)) would be less han ha of a ypical bulle, which is good news Bu he fac ha one is dealing wih many raindrops leads us o suspec ha his scenario poses an unhealhy siuaion If we facor in air resisance, he final speed is smaller, of course, and we reurn o he relaively healhy siuaion wih which we are familiar 41 We neglec air resisance, which jusifies seing a = g = 98 m/s (aking down as he y direcion) for he duraion of he fall This is consan acceleraion moion, which jusifies he use of Table -1 (wih y replacing x) (a) Saring he clock a he momen he wrench is dropped (v o = 0), hen v = v o g y leads o y = ( 4) (98) = 94 m

19 so ha i fell hrough a heigh of 94 m (b) Solving v = v 0 g for ime, we find: = v 0 v g = 0 ( 4) 98 =45 s (c) SI unis are used in he graphs, and he iniial posiion is aken as he coordinae origin In he ineres of saving space, we do no show he acceleraion graph, which is a horizonal line a 98m/s y v 4 We neglec air resisance, which jusifies seing a = g = 98 m/s (aking down as he y direcion) for he duraion of he fall This is consan acceleraion moion, which jusifies he use of Table -1 (wih y replacing x) (a) Noing ha y = y y 0 = 30 m, we apply Eq -15 and he quadraic formula (Appendix E) o compue : y = v 0 1 g = = v 0 ± v 0 g y g which (wih v 0 = 1 m/s since i is downward) leads, upon choosing he posiive roo (so ha >0), o he resul: = 1 + ( 1) (98)( 30) 98 =154 s (b) Enough informaion is now known ha any of he equaions in Table -1 can be used o obain v; however, he one equaion ha does no use our resul from par (a) is Eq -16: v = v0 g y =71 m/s where he posiive roo has been chosen in order o give speed (which is he magniude of he velociy vecor)

20 43 We neglec air resisance for he duraion of he moion (beween launching and landing ), so a = g = 98 m/s (we ake downward o be he y direcion) We use he equaions in Table -1 (wih y replacing x) because his is a = consan moion (a) A he highes poin he velociy of he ball vanishes Taking y 0 =0, we se v =0inv = v0 gy, and solve for he iniial velociy: v 0 = gy Since y = 50 m we find v 0 = 31 m/s (b) I will be in he air from he ime i leaves he ground unil he ime i reurns o he ground (y = 0) Applying Eq -15 o he enire moion (he rise and he fall, of oal ime >0) we have y = v 0 1 g = = v 0 g which (using our resul from par (a)) produces = 64 s Iis possible o obain his wihou using par (a) s resul; one can find he ime jus for he rise (from ground o highes poin) from Eq -16 and hen double i (c) SI unis are undersood in he x and v graphs shown In he ineres of saving space, we do no show he graph of a, which is a horizonal line a 98m/s y v There is no air resisance, which makes i quie accurae o se a = g = 98 m/s (where downward is he y direcion) for he duraion of he fall We are allowed o use Table -1 (wih y replacing x) because his is consan acceleraion moion; in fac, when he acceleraion changes (during he process of caching he ball) we will again assume consan acceleraion condiions; in his case, we have a =+5g = 45 m/s (a) The ime of fall is given by Eq -15 wih v 0 =0andy =0 Thus, y0 (145) = g = =544 s 98

21 (b) The final velociy for is free-fall (which becomes he iniial velociy during he caching process) is found from Eq -16(oher equaions can be used bu hey would use he resul from par (a)) v = v0 g (y y 0)= gy 0 = 533 m/s where he negaive roo is chosen since his is a downward velociy (c) For he caching process, he answer o par (b) plays he role of an iniial velociy (v 0 = 533 m/s) and he final velociy mus become zero Using Eq -16, we find y = v v 0 a = ( 533) (45) = 580 m where he negaive value of y signifies ha he disance raveled while arresing is moion is downward 45 Taking he +y direcion downward and y 0 =0,wehavey = v g which (wih v 0 = 0) yields = y/g (a) For his par of he moion, y = 50 m so ha (50) = =3 s 98 (b) For his nex par of he moion, we noe ha he oal displacemen is y = 100 m Therefore, he oal ime is (100) = =45 s 98 The difference beween his and he answer o par (a) is he ime required o fall hrough ha second 50 m disance: 45 3 =13s 46 We neglec air resisance, which jusifies seing a = g = 98 m/s (aking down as he y direcion) for he duraion of he moion We are allowed o use Table -1 (wih y replacing x) because his is consan acceleraion moion The ground level is aken o correspond o y =0 (a) Wih y 0 = h and v 0 replaced wih v 0, Eq -16leads o v = ( v 0 ) g (y y 0 )= v0 +gh The posiive roo is aken because he problem asks for he speed (he magniude of he velociy)

22 (b) We use he quadraic formula o solve Eq -15 for, wih v 0 replaced wih v 0, y = v 0 1 g = = v 0 + ( v 0 ) g y g where he posiive roo is chosen o yield >0 Wih y =0and y 0 = h, his becomes v = 0 +gh v 0 g (c) If i were hrown upward wih ha speed from heigh h hen (in he absence of air fricion) i would reurn o heigh h wih ha same downward speed and would herefore yield he same final speed (before hiing he ground) as in par (a) An imporan perspecive relaed o his is reaed laer in he book (in he conex of energy conservaion) (d) Having o ravel up before i sars is descen cerainly requires more ime han in par (b) The calculaion is quie similar, however, excep for now having +v 0 in he equaion where we had pu in v 0 in par (b) The deails follow: y = v 0 1 g = = v 0 + v 0 g y g wih he posiive roo again chosen o yield >0 Wih y =0and y 0 = h, we obain v = 0 +gh + v 0 g 47 We neglec air resisance, which jusifies seing a = g = 98 m/s (aking down as he y direcion) for he duraion of he moion We are allowed o use Table -1 (wih y replacing x) because his is consan acceleraion moion The ground level is aken o correspond o he origin of he y axis (a) Using y = v 0 1 g, wih y =0544 m and =000 s, we find v 0 = y + 1 g = (98)(000) 000 =370 m/s (b) The velociy a y =0544 m is v = v 0 g =370 (98)(000) = 174 m/s

23 (c) Using v = v0 gy (wih differen values for y and v han before), we solve for he value of y corresponding o maximum heigh (where v = 0) y = v 0 g = 37 (98) =0698 m Thus, he armadillo goes = 0154 m higher 48 We neglec air resisance, which jusifies seing a = g = 98 m/s (aking down as he y direcion) for he duraion of he moion We are allowed o use Table -1 (wih y replacing x) because his is consan acceleraion moion The ground level is aken o correspond o he origin of he y axis The oal ime of fall can be compued from Eq -15 (using he quadraic formula) y = v 0 1 g = = v 0 + v 0 g y g wih he posiive roo chosen Wih y =0,v 0 =0andy 0 = h =60m,we obain gh h = = =35 s g g Thus, 1 s earlier means we are examining where he rock is a =3s: y h = v 0 (3) 1 g(3) = y =34m where we again use he fac ha h =60mandv 0 =0 49 The speed of he boa is consan, given by v b = d/ Here, d is he disance of he boa from he bridge when he key is dropped (1 m) and is he ime he key akes in falling To calculae, we pu he origin of he coordinae sysem a he poin where he key is dropped and ake he y axis o be posiive in he downward direcion Taking he ime o be zero a he insan he key is dropped, we compue he ime when y =45m Since he iniial velociy of he key is zero, he coordinae of he key is given by y = 1 g Thus = Therefore, he speed of he boa is y g = (45 m) 98m/s =303 s v b = 1 m =40 m/s 303 s

24 50 Wih +y upward, we have y 0 =366 mandy =1 m Therefore, using Eq -18 (he las equaion in Table -1), we find y y 0 = v + 1 g = v = m/s a =00 s The erm speed refers o he magniude of he velociy vecor, so he answer is v =0 m/s 51 We firs find he velociy of he ball jus before i his he ground During conac wih he ground is average acceleraion is given by a avg = v where v is he change in is velociy during conac wih he ground and = s is he duraion of conac Now, o find he velociy jus before conac, we pu he origin a he poin where he ball is dropped (and ake +y upward) and ake = 0 o be when i is dropped The ball srikes he ground a y = 150 m Is velociy here is found from Eq -16: v = gy Therefore, v = gy = (98)( 150) = 171 m/s where he negaive sign is chosen since he ball is raveling downward a he momen of conac Consequenly, he average acceleraion during conac wih he ground is a avg = 0 ( 171) = 857 m/s The fac ha he resul is posiive indicaes ha his acceleraion vecor poins upward In a laer chaper, his will be direcly relaed o he magniude and direcion of he force exered by he ground on he ball during he collision 5 The y axis is arranged so ha ground level is y =0and+y is upward (a) A he poin where is fuel ges exhaused, he rocke has reached a heigh of y = 1 a = (400)(600) =70 m From Eq -11, he speed of he rocke (which had sared a res) a his insan is v = a =(400)(600) = 40 m/s The addiional heigh y 1 he rocke can aain (beyond y ) is given by Eq -16wih vanishing final speed: 0 = v g y 1 This gives y 1 = v g = (40) (98) =94 m

25 Recalling our value for y, he oal heigh he rocke aains is seen o be = 101 m (b) The ime of free-fall fligh (from y unil i reurns o y = 0) afer he fuel ges exhaused is found from Eq -15: y = v 1 g = 70 = (40) 980 Solving for (using he quadraic formula) we obain =700 s Recalling he upward acceleraion ime used in par (a), we see he oal ime of fligh is = 130 s 53 The average acceleraion during conac wih he floor is given by a avg = (v v 1 )/, where v 1 is is velociy jus before sriking he floor, v is is velociy jus as i leaves he floor, and is he duraion of conac wih he floor ( s) Taking he y axis o be posiively upward and placing he origin a he poin where he ball is dropped, we firs find he velociy jus before sriking he floor, using v 1 = v 0 gy Wih v 0 =0 and y = 400 m, he resul is v 1 = gy = (98)( 400) = 885 m/s where he negaive roo is chosen because he ball is raveling downward To find he velociy jus afer hiing he floor (as i ascends wihou air fricion o a heigh of 00 m), we use v = v g(y y 0 ) wih v =0, y = 00 m (i ends up wo meers below is iniial drop heigh), and y 0 = 400 m Therefore, v = g(y y 0 )= (98)( ) = 66m/s Consequenly, he average acceleraion is a avg = v v 1 = = m/s The posiive naure of he resul indicaes ha he acceleraion vecor poins upward In a laer chaper, his will be direcly relaed o he magniude and direcion of he force exered by he ground on he ball during he collision 54 The heigh reached by he player is y =076m (where we have aken he origin of he y axis a he floor and +y o be upward) (a) The iniial velociy v 0 of he player is v 0 = gy = (98)(076) = 386m/s This is a consequence of Eq -16where velociy v vanishes As he player reaches y 1 = = 061 m, his speed v 1 saisfies v0 v1 =gy 1, which yields v 1 = v0 gy 1 = (386) (980)(061) = 171 m/s

26 The ime 1 ha he player spends ascending in he op y 1 =015 m of he jump can now be found from Eq -17: y 1 = 1 (v 1 + v) 1 = 1 = (015) =0175 s which means ha he oal ime spend in ha op 15 cm (boh ascending and descending) is (017) = 035 s = 350 ms (b) The ime when he player reaches a heigh of 015 m is found from Eq -15: 015 = v 0 1 g =(386) 98, which yields (using he quadraic formula, aking he smaller of he wo posiive roos) =0041 s = 41 ms, which implies ha he oal ime spend in ha boom 15 cm (boh ascending and descending) is (41) = 8 ms 55 We neglec air resisance, which jusifies seing a = g = 98 m/s (aking down as he y direcion) for he duraion of he moion We are allowed o use Table -1 (wih y replacing x) because his is consan acceleraion moion The ground level is aken o correspond o he origin of he y axis The ime drop 1 leaves he nozzle is aken as =0and is ime of landing on he floor 1 can be compued from Eq -15, wih v 0 =0andy 1 = 00 m y 1 = 1 y g 1 = 1 = g ( 00) = =0639 s 98 A ha momen,he fourh drop begins o fall, and from he regulariy of he dripping we conclude ha drop leaves he nozzle a =0639/3 = 013 s and drop 3 leaves he nozzle a = (013) = 046s Therefore, he ime in free fall (up o he momen drop 1 lands) for drop is = = 046s and he ime in free fall (up o he momen drop 1 lands) for drop 3 is 3 = 1 046= 013 s Their posiions a ha momen are y = 1 g = 1 (98)(046) = 0889 m y 3 = 1 g 3 = 1 (98)(013) = 0 m, respecively Thus, drop is 89 cm below he nozzle and drop 3 is cm below he nozzle when drop 1 srikes he floor 56 The graph shows y = 5 m o be he highes poin (where he speed momenarily vanishes) The neglec of air fricion (or whaever passes for ha on he disan plane) is cerainly reasonable due o he symmery of he graph

27 (a) To find he acceleraion due o graviy g p on ha plane, we use Eq -15 (wih +y up) y y 0 = v + 1 g p = 5 0 = (0)(5) + 1 g p(5) so ha g p =80 m/s (b) Tha same (max) poin on he graph can be used o find he iniial velociy y y 0 = 1 (v 0 + v) = 5 0= 1 (v 0 +0)(5) Therefore, v 0 = 0 m/s 57 Taking +y o be upward and placing he origin a he poin from which he objecs are dropped, hen he locaion of diamond 1 is given by y 1 = 1 g and he locaion of diamond is given by y = 1 g( 1) We are saring he clock when he firs objec is dropped We wan he ime for which y y 1 = 10 m Therefore, 1 g( 1) + 1 g =10 = = (10/g)+05 =15 s 58 We neglec air resisance, which jusifies seing a = g = 98 m/s (aking down as he y direcion) for he duraion of he moion We are allowed o use Table -1 (wih y replacing x) because his is consan acceleraion moion When somehing is hrown sraigh up and is caugh a he level i was hrown from (wih a rajecory similar o ha shown in Fig -5), he ime of fligh is half of is ime of ascen a, which is given by Eq -18 wih y = H and v = 0 (indicaing he maximum poin) H = v a + 1 H g a = a = g Wriing hese in erms of he oal ime in he air = a we have H = 1 H 8 g = = g We consider wo hrows, one o heigh H 1 for oal ime 1 and anoher o heigh H for oal ime, and we se up a raio: H H 1 = 1 8 g 1 = 8 g 1 ( ) from which we conclude ha if = 1 (as is required by he problem) hen H = H 1 =4H 1 1

28 59 We neglec air resisance, which jusifies seing a = g = 98 m/s (aking down as he y direcion) for he duraion of he moion We are allowed o use Table -1 (wih y replacing x) because his is consan acceleraion moion We placing he coordinae origin on he ground We noe ha he iniial velociy of he package is he same as he velociy of he balloon, v 0 = +1 m/s and ha is iniial coordinae is y 0 = +80 m (a) We solve y = y 0 +v 0 1 g for ime, wih y = 0, using he quadraic formula (choosing he posiive roo o yield a posiive value for ) = v 0 + v 0 +gy 0 g = (98)(80) 98 =54 s (b) If we wish o avoid using he resul from par (a), we could use Eq -16, bu if ha is no a concern, hen a variey of formulas from Table -1 can be used For insance, Eq -11 leads o v = v 0 g = 1 (98)(54) = 41 m/s Is final speed is 41 m/s 60 We neglec air resisance, which jusifies seing a = g = 98 m/s (aking down as he y direcion) for he duraion of he moion We are allowed o use Eq -15 (wih y replacing x) because his is consan acceleraion moion We use primed variables (excep ) wih he firs sone, which has zero iniial velociy, and unprimed variables wih he second sone (wih iniial downward velociy v 0, so ha v 0 is being used for he iniial speed) SI unis are used hroughou y = 0() 1 g y = ( v 0 )( 1) 1 g( 1) Since he problem indicaes y = y = 439 m, we solve he firs equaion for (finding = 99 s) and use his resul o solve he second equaion for he iniial speed of he second sone: y which leads o v 0 =13 m/s 439 =( v 0 )(199) 1 (98)(199)

29 61 We neglec air resisance, which jusifies seing a = g = 98 m/s (aking down as he y direcion) for he duraion of he moion of he sho ball We are allowed o use Table -1 (wih y replacing x) because he ball has consan acceleraion moion We use primed variables (excep ) wih he consan-velociy elevaor (so v = 0 m/s), and unprimed variables wih he ball (wih iniial velociy v 0 = v +10 = 30 m/s, relaive o he ground) SI unis are used hroughou (a) Taking he ime o be zero a he insan he ball is sho, we compue is maximum heigh y (relaive o he ground) wih v = v 0 g(y y o ), where he highes poin is characerized by v =0 Thus, y = y o + v 0 g =76m where y o = y o + = 30 m (where y o = 8 m is given in he problem) and v 0 = 30 m/s relaive o he ground as noed above (b) There are a variey of approaches o his quesion One is o coninue working in he frame of reference adoped in par (a) (which reas he ground as moionless and fixes he coordinae origin o i); in his case, one describes he elevaor moion wih y = y o +v and he ball moion wih Eq -15, and solves hem for he case where hey reach he same poin a he same ime Anoher is o work in he frame of reference of he elevaor (he boy in he elevaor migh be oblivious o he fac he elevaor is moving since i isn acceleraing), which is wha we show here in deail: y e = v 0e 1 g = = v 0e + v 0e g y e g where v 0e = 0 m/s is he iniial velociy of he ball relaive o he elevaor and y e = 0 m is he ball s displacemen relaive o he floor of he elevaor The posiive roo is chosen o yield a posiive value for ; he resul is =4 s 6 We neglec air resisance, which jusifies seing a = g = 98 m/s (aking down as he y direcion) for he duraion of he sone s moion We are allowed o use Table -1 (wih x replaced by y) because he ball has consan acceleraion moion (and we choose y o = 0) (a) We apply Eq -16o boh measuremens, wih SI unis undersood v B = v 0 gy B = ( ) 1 v +g(y A +3)=v0 v A = v 0 gy A = v +gy A = v 0 We equae he wo expressions ha each equal v0 and obain 1 4 v +gy A +g(3) = v +gy A = g(3) = 3 4 v

30 which yields v = g(4) = 885 m/s (b) An objec moving upward a A wih speed v =885 m/s will reach a maximum heigh y y A = v /g =400 m above poin A (his is again a consequence of Eq -16, now wih he final velociy se o zero o indicae he highes poin) Thus, he op of is moion is 100 m above poin B 63 The objec, once i is dropped (v 0 = 0) is in free-fall (a = g = 98 m/s if we ake down as he y direcion), and we use Eq -15 repeaedly (a) The (posiive) disance D from he lower do o he mark corresponding o a cerain reacion ime is given by y = D = 1 g,or D = g / Thus for 1 =500ms D 1 = (98m/s )( s) =0013 m = 13 cm (b) For = 100 ms D = (98m/s )( s) for 3 = 150 ms D 3 = (98m/s )( s) for 4 = 00 ms =0049 m = 4D 1 ; =011 m = 9D 1 ; D 4 = (98m/s )( s) and for 4 = 50 ms D 5 = (98m/s )( s) =0196m = 16D 1 ; =0306m = 5D 1 64 During free fall, we ignore he air resisance and se a = g = 98 m/s where we are choosing down o be he y direcion The iniial velociy is zero so ha Eq -15 becomes y = 1 g where y represens he negaive of he disance d she has fallen Thus, we can wrie he equaion as d = 1 g for simpliciy (a) The ime 1 during which he parachuis is in free fall is (using Eq -15) given by d 1 =50m= 1 g 1 = 1 ( 980 m/s ) 1

31 which yields 1 =3s The speed of he parachuis jus before he opens he parachue is given by he posiive roo v1 =gd 1,or v 1 = gh 1 = ()(980 m/s )(50 m) = 31 m/s If he final speed is v, hen he ime inerval beween he opening of he parachue and he arrival of he parachuis a he ground level is = v 1 v a = 31 m/s 30m/s m/s =14s This is a resul of Eq -11 where speeds are used insead of he (negaive-valued) velociies (so ha final-velociy minus iniial-velociy urns ou o equal iniial-speed minus final-speed); we also noe ha he acceleraion vecor for his par of he moion is posiive since i poins upward (opposie o he direcion of moion which makes i a deceleraion) The oal ime of fligh is herefore 1 + =17s (b) The disance hrough which he parachuis falls afer he parachue is opened is given by d = v 1 v a = (31 m/s) (30m/s) ()(0m/s ) 40 m In he compuaion, we have used Eq -16wih boh sides muliplied by 1 (which changes he negaive-valued y ino he posiive d on he lef-hand side, and swiches he order of v 1 and v on he righhand side) Thus he fall begins a a heigh of h =50+d 90 m 65 The ime he po spends passing in fron of he window of lengh L = 0 mis05 s each way We use v for is velociy as i passes he op of he window (going up) Then, wih a = g = 98 m/s (aking down o be he y direcion), Eq -18 yields L = v 1 g = v = L 1 g The disance H he po goes above he op of he window is herefore (using Eq -16wih he final velociy being zero o indicae he highes poin) H = v (L/ g/) (00/05 (980)(05)/) = = =34 m g g () (980) 66 The ime being considered is 6 years and roughly 35 days, which is approximaely = s Using Eq -3, we find he average speed o be m =015 m/s s