A : + B v 2 = v a c (s - s 0 )

Transcription

1 91962_01_s12-p /8/09 8:05 M Page car sars from res and wih consan acceleraion achieves a velociy of 15 m>s when i ravels a disance of 200 m. Deermine he acceleraion of he car and he ime required. Kinemaics: v 0 = 0, v = 15 m>s, s 0 = 0, and s = 200 m. : + B v 2 = v a c (s - s 0 ) 15 2 = a c (200-0) a c = m>s 2 : + B v = v 0 + a c 15 = = 26.7 s rain sars from res a a saion and ravels wih a consan acceleraion of 1 m>s 2. Deermine he velociy of he rain when = 30 s and he disance raveled during his ime. Kinemaics: a c = 1 m>s 2, v 0 = 0, s 0 = 0, and = 30 s. : + B v = v 0 + a c = 0 + 1(30) = 30 m>s : + B s = s 0 + v a c 2 = (1)302 B = 450 m 1

2 91962_01_s12-p /8/09 8:05 M Page n elevaor descends from res wih an acceleraion of 5 f>s 2 unil i achieves a velociy of 15 f>s. Deermine he ime required and he disance raveled. Kinemaics: a c = 5 f>s 2, v 0 = 0, v = 15 f>s, and s 0 = 0. +TB v = v 0 + a c 15 = = 3s +TB v 2 = v a c (s - s 0 ) 15 2 = (5)(s - 0) s = 22.5 f *12 4. car is raveling a 15 m>s, when he raffic ligh 50 m ahead urns yellow. Deermine he required consan deceleraion of he car and he ime needed o sop he car a he ligh. Kinemaics: v 0 = 0, s 0 = 0, s = 50 m and v 0 = 15 m>s. : + B v 2 = v a c (s - s 0 ) 0 = a c (50-0) a c = m>s 2 = 2.25 m>s 2 ; : + B v = v 0 + a c 0 = 15 + (-2.25) = 6.67 s 2

3 91962_01_s12-p /8/09 8:05 M Page paricle is moving along a sraigh line wih he acceleraion a = (12 3 1/2 ) f>s 2, where is in seconds. Deermine he velociy and he posiion of he paricle as a funcion of ime. When = 0, v = 0 and s = 15 f. Velociy: : + B dv = a d v dv = >2 Bd L0 L 0 v v = >2 B 2 0 v = >2 Bf>s Posiion: Using his resul and he iniial condiion s = 15 f a = 0 s, : + B ds = v d L s 15 f ds = >2 Bd L s s = 15 f a >2 b s = a >2 + 15bf ball is released from he boom of an elevaor which is raveling upward wih a velociy of 6 f>s. If he ball srikes he boom of he elevaor shaf in 3 s, deermine he heigh of he elevaor from he boom of he shaf a he insan he ball is released. lso, find he velociy of he ball when i srikes he boom of he shaf. Kinemaics: When he ball is released, is velociy will be he same as he elevaor a he insan of release. Thus, v 0 = 6 f>s. lso, = 3 s, s 0 = 0, s = -h, and a c = f>s 2. + c B s = s 0 + v a c 2 -h = 0 + 6(3) (-32.2)32 B h = 127 f + c B v = v 0 + a c v = 6 + (-32.2)(3) = f>s = 90.6 f>s T 3

4 91962_01_s12-p /8/09 8:05 M Page car has an iniial speed of 25 m>s and a consan deceleraion of 3 m>s 2. Deermine he velociy of he car when = 4 s.wha is he displacemen of he car during he 4-s ime inerval? How much ime is needed o sop he car? v = v 0 + a c v = 25 + (-3)(4) = 13 m>s s = s - s 0 = v a c 2 s = s - 0 = 25(4) (-3)(4)2 = 76 m v = v 0 + a c 0 = 25 + (-3)() = 8.33 s *12 8. If a paricle has an iniial velociy of v 0 = 12 f>s o he righ, a s 0 = 0, deermine is posiion when = 10 s, if a = 2 f>s 2 o he lef. : + B s = s 0 + v a c 2 = (10) (-2)(10)2 = 20 f The acceleraion of a paricle raveling along a sraigh line is a = k>v, where k is a consan. If s = 0, v = v 0 when = 0, deermine he velociy of he paricle as a funcion of ime. Velociy: : + B d = dv a v dv d = L0 L k>v L0 v 0 d = L v v 0 1 k vdv 2 = 1 v 0 2k v2 2 v 0 = 1 2k v2 - v 0 2 B v = 22k + v 0 2 4

5 91962_01_s12-p /8/09 8:05 M Page Car sars from res a = 0 and ravels along a sraigh road wih a consan acceleraion of 6 f>s 2 unil i reaches a speed of 80 f>s. ferwards i mainains his speed. lso, when = 0, car B locaed 6000 f down he road is raveling owards a a consan speed of 60 f>s. Deermine he disance raveled by car when hey pass each oher f B 60 f/s Disance Traveled: Time for car o achives applying Eq : + B y = y 0 + a c 80 = = s y = 80 f>s can be obained by The disance car ravels for his par of moion can be deermined by applying Eq : + B y 2 = y a c (s - s 0 ) For he second par of moion, car ravels wih a consan velociy of y = 80 f>s and he disance raveled in =( ) s ( is he oal ime) is : + B s 2 = y =80( ) Car B ravels in he opposie direcion wih a consan velociy of y = 60 f>s and he disance raveled in is : + B s 3 = y 1 = 60 1 I is required ha 80 2 = 0 + 2(6)(s 1-0) The disance raveled by car is 1 s 1 = f s 1 + s 2 + s 3 = ( ) = = s 1 s = s 1 + s 2 = ( ) = 3200 f 5

6 91962_01_s12-p /8/09 8:05 M Page paricle ravels along a sraigh line wih a velociy v = ( ) m>s, where is in seconds. When = 1 s, he paricle is locaed 10 m o he lef of he origin. Deermine he acceleraion when = 4 s, he displacemen from = 0 o = 10 s, and he disance he paricle ravels during his ime period. v = a = dv d = -6 2 = -24 m>s 2 = 4 (1) L s -10 ds = v d = Bd L L 1 s + 10 = s = s = 0 = -21 s = 10 = s = (-21) = -880 m From Eq. (1): v = 0 when = 2s s = 2 = 12(2) - (2) 3-21 = -5 s T = (21-5) + (901-5) = 912 m 6

7 91962_01_s12-p /8/09 8:05 M Page 7 * sphere is fired downwards ino a medium wih an iniial speed of 27 m>s. If i experiences a deceleraion of a = (-6) m>s 2, where is in seconds, deermine he disance raveled before i sops. Velociy: y 0 = 27 m>s a 0 = 0 s. pplying Eq. 12 2, we have +TB dy = ad y dy = -6d L27 L 0 y = B m>s [1] y = 0, from Eq.[1] 0 = = 3.00 s Disance Traveled: s a. Using he resul y = = 0 m 0 = 0 s and applying Eq. 12 1, we have +TB ds = yd s ds = B d L0 L 0 s = 27-3 B m [2] = 3.00 s, from Eq. [2] s = 27(3.00) = 54.0 m paricle ravels along a sraigh line such ha in 2 s i moves from an iniial posiion s = +0.5 m o a posiion s B = -1.5 m. Then in anoher 4 s i moves from s B o s C = +2.5 m. Deermine he paricle s average velociy and average speed during he 6-s ime inerval. s = (s C - s ) = 2 m s T = ( ) = 6 m = (2 + 4) = 6 s v avg = s = 2 6 = m>s (v sp ) avg = s T = 6 6 = 1 m>s 7

8 91962_01_s12-p /8/09 8:05 M Page paricle ravels along a sraigh-line pah such ha in 4 s i moves from an iniial posiion s = -8 m o a posiion s B = +3 m.then in anoher 5 s i moves from s B o s C = -6 m. Deermine he paricle s average velociy and average speed during he 9-s ime inerval. verage Velociy: The displacemen from o C is = 2 m. y avg = s = 2 = m>s s = s C - S = -6 - (-8) verage Speed: The disances raveled from o B and B o C are s : B = = 11.0 m and s B : C = = 9.00 m, respecively. Then, he oal disance raveled is s To = s : B + s B : C = = 20.0 m. y sp B avg = s To = 20.0 = 2.22 m>s Tess reveal ha a normal driver akes abou 0.75 s before he or she can reac o a siuaion o avoid a collision. I akes abou 3 s for a driver having 0.1% alcohol in his sysem o do he same. If such drivers are raveling on a sraigh road a 30 mph (44 f>s) and heir cars can decelerae a 2 f>s 2, deermine he shores sopping disance d for each from he momen hey see he pedesrians. Moral: If you mus drink, please don drive! v 1 44 f/s d Sopping Disance: For normal driver, he car moves a disance of d =y = 44(0.75) = 33.0 f before he or she reacs and deceleraes he car. The sopping disance can be obained using Eq wih s 0 = d =33.0 f and y = 0. : + B y 2 = y a c (s - s 0 ) 0 2 = (-2)(d ) d = 517 f For a drunk driver, he car moves a disance of d =y = 44(3) = 132 f before he or she reacs and deceleraes he car. The sopping disance can be obained using Eq wih s 0 = d =132 f and y = 0. : + B y 2 = y a c (s - s 0 ) 0 2 = (-2)(d - 132) d = 616 f 8

9 91962_01_s12-p /8/09 8:05 M Page 9 * s a rain acceleraes uniformly i passes successive kilomeer marks while raveling a velociies of 2 m>s and hen 10 m>s. Deermine he rain s velociy when i passes he nex kilomeer mark and he ime i akes o ravel he 2-km disance. Kinemaics: For he firs kilomeer of he journey, v 0 = 2 m>s, v = 10 m>s, s 0 = 0, and s = 1000 m. Thus, : + B v 2 = v a c (s - s 0 ) 10 2 = a c (1000-0) a c = m>s 2 For he second kilomeer, v 0 = 10 m>s, s 0 = 1000 m, s = 2000 m, and m>s 2. Thus, : + B v 2 = v a c (s - s 0 ) v 2 = (0.048)( ) v = 14 m>s For he whole journey, v,, and m>s 2 0 = 2 m>s v = 14 m>s. Thus, : + B v = v 0 + a c 14 = = 250 s ball is hrown wih an upward velociy of 5 m>s from he op of a 10-m high building. One second laer anoher ball is hrown verically from he ground wih a velociy of 10 m>s. Deermine he heigh from he ground where he wo balls pass each oher. Kinemaics: Firs, we will consider he moion of ball wih (v ) 0 = 5 m>s, (s, s,, and a c = m>s 2 ) 0 = 0 = (h - 10) m =. Thus, + c B s = (s ) 0 + (v ) a c 2 h - 10 = (-9.81)( )2 h = ( ) (1) Moion of ball B is wih (v B ) 0 = 10 m>s, (s B ) 0 = 0, s B = h, B = -1 and a c = m>s 2. Thus, + c B s B = (s B ) 0 + (v B ) 0 B a c 2 B h = ( -1) + 1 (-9.81)( -1)2 2 h = ( ) (2) Solving Eqs. (1) and (2) yields h = 4.54 m =1.68 m 9

10 91962_01_s12-p /8/09 8:06 M Page car sars from res and moves wih a consan acceleraion of 1.5 m>s 2 unil i achieves a velociy of 25 m>s. I hen ravels wih consan velociy for 60 seconds. Deermine he average speed and he oal disance raveled. Kinemaics: For sage (1) of he moion, v 0 = 0, s 0 = 0, v = 25 m>s, and a c = 1.5 m>s 2. : + B v = v 0 + a c 25 = = s : + B v 2 = v a c (s - s 0 ) 25 2 = 0 + 2(1.5)(s 1-0) s 1 = m For sage (2) of he moion, s 0 = f, v 0 = 25 f>s, = 60 s, and a c = 0. Thus, : + B s = s 0 + v a c 2 s = (60) + 0 = f = 1708 m The average speed of he car is hen v avg = s = = 22.3 m>s car is o be hoised by elevaor o he fourh floor of a parking garage, which is 48 f above he ground. If he elevaor can accelerae a 0.6 f>s 2, decelerae a 0.3 f>s 2, and reach a maximum speed of 8 f>s, deermine he shores ime o make he lif, saring from res and ending a res. + c v 2 = v a c (s - s 0 ) v 2 max = 0 + 2(0.6)(y - 0) 0 = v 2 max + 2(-0.3)(48 - y) 0 = 1.2 y - 0.6(48 - y) y = 16.0 f, v max = f>s 6 8 f>s + c v = v 0 + a c = = s 0 = = s = = 21.9 s 10

11 91962_01_s12-p /8/09 8:06 M Page 11 * paricle is moving along a sraigh line such ha is speed is defined as v = (-4s 2 ) m>s, where s is in meers. If s = 2 m when = 0, deermine he velociy and acceleraion as funcions of ime. v = -4s 2 ds d = -4s2 s s - 2 ds = -4 d L2 L 0 -s - 1 s 2 = -4 0 = 1 4 (s ) s = v = -4a b 16 = a - (8 + 1) 2 b m>s a = dv d 16(2)(8 + 1)(8) 256 = (8 + 1) 4 = a (8 + 1) 3 b m>s2 11

12 91962_01_s12-p /8/09 8:06 M Page Two paricles and B sar from res a he origin s = 0 and move along a sraigh line such ha a and a B = (12 2-8) f>s 2 = (6-3) f>s 2, where is in seconds. Deermine he disance beween hem when = 4 s and he oal disance each has raveled in = 4 s. Velociy: The velociy of paricles and B can be deermined using Eq dy = a d L0 L0 y B = The imes when paricle sops are = 0 = 0 s and = 1 s The imes when paricle B sops are = 0 = 0 s and = 22 s Posiion: The posiion of paricles and B can be deermined using Eq ds = y d L0 L0 y y B s s B dy = (6-3)d L y = dy B = a B d dy B = (12 2-8)d L ds = (3 2-3)d L s = ds B = y B d 0 ds B = (4 3-8)d L s B = The posiions of paricle a = 1 s and 4 s are s = 1 s = (12 ) = f Paricle has raveled s = 4 s = (42 ) = 40.0 f d = 2(0.5) = 41.0 f The posiions of paricle B a = 22 s and 4 s are s B = 12 = (22) 4-4(22) 2 = -4 f s B = 4 = (4) 4-4(4) 2 = 192 f Paricle B has raveled d B = 2(4) = 200 f = 4 s he disance beween and B is s B = = 152 f 12

13 91962_01_s12-p /8/09 8:06 M Page paricle moving along a sraigh line is subjeced o a deceleraion a = (-2v 3 ) m>s 2, where v is in m>s. If i has a velociy v = 8 m>s and a posiion s = 10 m when = 0, deermine is velociy and posiion when = 4 s. Velociy: The velociy of he paricle can be relaed o is posiion by applying Eq s - 10 = 1 2y y = [1] 16s Posiion: The posiion of he paricle can be relaed o he ime by applying Eq When = 4 s, L s 10m ds = ydy a ds = L y d = ds y s d = L0 L 10m 8m>s - dy 2y 2 8 = 8s 2-159s (4) = 8s 2-159s s 2-159s = 0 1 (16s - 159) ds 8 Choose he roo greaer han 10 m s = m = 11.9 m Subsiue s = m ino Eq. [1] yields y = 8 = m>s 16(11.94)

14 91962_01_s12-p /8/09 8:06 M Page paricle is moving along a sraigh line such ha is acceleraion is defined as a = (-2v) m>s 2, where v is in meers per second. If v = 20 m>s when s = 0 and = 0, deermine he paricle s posiion, velociy, and acceleraion as funcions of ime. a = -2v dv d = -2v L 20 v dv v = -2 d L0 ln v 20 = -2 v = (20e - 2 )m>s a = dv d = (-40e - 2 )m>s 2 s ds = v d = (20e - 2 )d L0 L 0 L 0 s = -10e = -10(e - 2-1) s = 10(1 - e - 2 )m 14

15 91962_01_s12-p /8/09 8:06 M Page 15 * paricle sars from res and ravels along a sraigh line wih an acceleraion a = (30-0.2v) f>s 2, where v is in f>s. Deermine he ime when he velociy of he paricle is v = 30 f>s. Velociy: : + B d = dv a v dv d = L0 L v 0 = - 1 v 0.2 ln(30-0.2v) 2 30 = 5ln v 0 30 = 5ln (50) = 1.12 s When a paricle is projeced verically upwards wih an iniial velociy of v 0, i experiences an acceleraion a = -(g + kv 2 ), where g is he acceleraion due o graviy, k is a consan and v is he velociy of he paricle. Deermine he maximum heigh reached by he paricle. Posiion: + c B ds = v dv a s v ds = L0 L v 0 - vdv g + kv 2 s s 0 = -c 1 v 2k lng + kv2 B d 2 s = 1 2k ln g + kv 2 0 g + kv 2 The paricle achieves is maximum heigh when v = 0. Thus, h max = 1 2k ln g + kv 2 0 g v 0 = 1 2k ln 1 + k g v

16 91962_01_s12-p /8/09 8:06 M Page The acceleraion of a paricle raveling along a sraigh line is a = (0.02e ) m>s 2, where is in seconds. If v = 0, s = 0 when = 0, deermine he velociy and acceleraion of he paricle a s = 4 m. Velociy: a = 0.02e = 4.13 m>s 2 : + B dv = a d v dv = 0.02e d L0 L 0 v v 0 = 0.02e 2 v = C0.02e - 1BD m>s 0 (1) Posiion: : + B ds = v d s ds = 0.02e - 1Bd L0 L 0 s s 0 = 0.02e - B 2 0 s = 0.02e - - 1B m When s = 4 m, 4 = 0.02e - - 1B e = 0 Solving he above equaion by rial and error, = s Thus, he velociy and acceleraion when s = 4 m ( = s) are v = 0.02e B = 4.11 m>s a = 0.02e = 4.13 m>s paricle moves along a sraigh line wih an acceleraion of a = 5>(3s 1>3 + s 5>2 ) m>s 2, where s is in meers. Deermine he paricle s velociy when s = 2 m, if i sars from res when s = 1 m. Use Simpson s rule o evaluae he inegral. a = a ds = v dv 2 5 3s s 5 2 B 5 ds L1 3s = 1 5 v dv 3 + s 2 B L 0 v = 1 2 v2 v = 1.29 m>s 16

17 91962_01_s12-p /8/09 8:06 M Page 17 * If he effecs of amospheric resisance are accouned for, a falling body has an acceleraion defined by he equaion a = 9.81[1 - v 2 (10-4 )] m>s 2, where v is in m>s and he posiive direcion is downward. If he body is released from res a a very high aliude, deermine (a) he velociy when = 5 s, and (b) he body s erminal or maximum aainable velociy (as : q). Velociy: The velociy of he paricle can be relaed o he ime by applying Eq ( +T) d = dy a = c L0 y dy d = L0 L [1 - (0.01y) 2 ] y dy 2( y) + L = 50lna y y b y dy 2(1-0.01y) d y = 100(e ) e [1] a) When = 5 s, hen, from Eq. [1] y = 100[e0.1962(5) - 1] e (5) + 1 = 45.5 m>s e b) If : q - 1,. Then, from Eq. [1] e : 1 y max = 100 m>s 17

18 91962_01_s12-p /8/09 8:06 M Page The posiion of a paricle along a sraigh line is given by s = ( ) f, where is in seconds. Deermine he posiion of he paricle when = 6 s and he oal disance i ravels during he 6-s ime inerval. Hin: Plo he pah o deermine he oal disance raveled. Posiion: The posiion of he paricle when = 6 s is s = 6s = 1.5(6 3 ) (6 2 ) (6) = f Toal Disance Traveled: The velociy of he paricle can be deermined by applying Eq y = ds d = The imes when he paricle sops are = 0 = 1 s and = 5 s The posiion of he paricle a = 0 s, 1 s and 5 s are s = 0s = 1.5(0 3 ) (0 2 ) (0) = 0 s = 1s = 1.5(1 3 ) (1 2 ) (1) = 10.5 f s = 5s = 1.5(5 3 ) (5 2 ) (5) = f From he paricle s pah, he oal disance is s o = = 69.0 f 18

19 91962_01_s12-p /8/09 8:06 M Page The velociy of a paricle raveling along a sraigh line is v = v 0 - ks, where k is consan. If s = 0 when = 0, deermine he posiion and acceleraion of he paricle as a funcion of ime. Posiion: : + B d = ds y s ds d = L0 L 0 v 0 - ks 0 = - 1 s k ln (v 0 - ks) 2 0 = 1 k ln v 0 v 0 - ks e k = v 0 v 0 - ks s = v 0 k 1 - e - k B Velociy: v = ds d = d d c v 0 k 1 - e - k B d v = v 0 e - k cceleraion: a = dv d = d d v 0e - k B a = -kv 0 e - k The acceleraion of a paricle as i moves along a sraigh line is given by a = m>s 2, where is in seconds. If s = 1 m and v = 2 m>s when = 0, deermine he paricle s velociy and posiion when = 6 s. lso, deermine he oal disance he paricle ravels during his ime period. v dv = (2-1) d L2 L 0 v = s ds = ( ) d L1 L 0 s = When = 6 s, Since v Z 0 hen v = 32 m>s s = 67 m d = 67-1 = 66 m 19

20 91962_01_s12-p /8/09 8:06 M Page 20 * Ball is hrown verically upward from he op of a 30-m-high-building wih an iniial velociy of 5 m>s. he same insan anoher ball B is hrown upward from he ground wih an iniial velociy of 20 m>s. Deermine he heigh from he ground and he ime a which hey pass. Origin a roof: Ball : + c B s = s 0 + v a c 2 -s = (9.81)2 Ball B: + c B s = s 0 + v a c 2 -s = (9.81)2 Solving, = 2 s s = 9.62 m Disance from ground, d = ( ) = 20.4 m lso, origin a ground, s = s 0 + v a c 2 s = (-9.81)2 s B = (-9.81)2 Require s = s B (-9.81)2 = (-9.81)2 = 2 s s B = 20.4 m 20

21 91962_01_s12-p /8/09 8:06 M Page moorcycle sars from res a = 0 and ravels along a sraigh road wih a consan acceleraion of 6 f>s 2 unil i reaches a speed of 50 f>s. ferwards i mainains his speed. lso, when = 0, a car locaed 6000 f down he road is raveling oward he moorcycle a a consan speed of 30 f>s. Deermine he ime and he disance raveled by he moorcycle when hey pass each oher. Moorcycle: : + B v = v 0 + a c 50 = =8.33 s v 2 = v a c (s - s 0 ) (50) 2 = 0 + 2(6)(s -0) s = f In =8.33 s car ravels s =v 0 =30(8.33) = 250 f Disance beween moorcycle and car: = f When passing occurs for moorcycle, s = v 0 ; x = 50( ) For car: s = v 0 ; x = 30( ) Solving, x = f =69.27 s Thus, for he moorcycle, = = 77.6 s s m = = 3.67(10) 3 f 21

22 91962_01_s12-p /8/09 8:07 M Page paricle moves along a sraigh line wih a velociy v = (200s) mm>s, where s is in millimeers. Deermine he acceleraion of he paricle a s = 2000 mm. How long does he paricle ake o reach his posiion if s = 500 mm when = 0? cceleraion: : + B dv ds = 200s Thus, a = v dv ds = (200s)(200) = Bs mm>s 2 When s = 2000 mm, Posiion: : + B d = ds v 2 0 s = 2000 mm, a = B(2000) = B mm>s 2 = 80 km>s 2 s d = L L mm s = lns 2 = ln s 500 ds 200s 500 mm = ln = B s = 6.93 ms 22

23 91962_01_s12-p /8/09 8:07 M Page paricle has an iniial speed of 27 m>s. If i experiences a deceleraion of a = 1-62 m>s 2, where is in seconds, deermine is velociy, afer i has raveled 10 m. How much ime does his ake? Velociy: : + B dv = a d dv = (-6)d L27 L 0 v v 2 27 = -3 2 B 2 0 v = ( ) m>s : + B ds = v d s ds = Bd L0 L 0 s s 2 0 = 27-3 B 2 0 s = (27-3 ) m>s When s = 100 m, = s v = 26.6 m>s * The acceleraion of a paricle raveling along a sraigh line is a = (8-2s) m>s 2, where s is in meers. If v = 0 a s = 0, deermine he velociy of he paricle a s = 2 m, and he posiion of he paricle when he velociy is maximum. Velociy: : + B v dv = a ds s = 2 m, v s vdv = (8-2s) ds L0 L 0 v 2 v 2 ` 0 s = 8s - s 2 B 2 n = 216s - 2s 2 m>s 0 v s = 2 m = 216(2) B = ;4.90 m>s dv When he velociy is maximum. Thus, ds = 0 dv ds = 16-4s = 0 s = 4 m 16-4s 2216s - 2s 2 = 0 23

24 91962_01_s12-p /8/09 8:07 M Page Ball is hrown verically upwards wih a velociy of v 0. Ball B is hrown upwards from he same poin wih he same velociy seconds laer. Deermine he elapsed ime 6 2v 0>g from he insan ball is hrown o when he balls pass each oher, and find he velociy of each ball a his insan. Kinemaics: Firs, we will consider he moion of ball wih (v ) 0 = v 0, (s ) 0 = 0, s = h, =, and (a c ) = -g. + c B s = (s ) 0 + (v ) (a c) 2 h = 0 + v (-g)( )2 h = v 0 - g 2 2 (1) + c B v = (v ) 0 + (a c ) v = v 0 + (-g)( ) v = v 0 - g (2) The moion of ball B requires (v B ) 0 = v 0, (s B ) 0 = 0, s B = h, B = -, and (a c ) B = -g. + c B s B = (s B ) 0 + (v B ) 0 B (a c) B 2 B h = 0 + v 0 ( - ) + 1 (-g)( - )2 2 h = v 0 ( - ) - g ( - )2 2 (3) + c B v B = (v B ) 0 + (a c ) B B v B = v 0 + (-g)( - ) v B = v 0 - g( -) (4) Solving Eqs. (1) and (3), v 0 - g 2 2 = v 0 ( - ) - g ( - )2 2 = 2v 0 + g 2g Subsiuing his resul ino Eqs. (2) and (4), v = v 0 - ga 2v 0 + g b 2g 1 = - 2 g = 1 2 gt v B = v 0 - ga 2v 0 + g 2g = 1 2 g c - b 24

25 91962_01_s12-p /8/09 8:07 M Page s a body is projeced o a high aliude above he earh s surface, he variaion of he acceleraion of graviy wih respec o aliude y mus be aken ino accoun. Neglecing air resisance, his acceleraion is deermined from he formula a = -g 0 [R 2 >(R + y) 2 ], where g 0 is he consan graviaional acceleraion a sea level, R is he radius of he earh, and he posiive direcion is measured upward. If g 0 = 9.81 m>s 2 and R = 6356 km, deermine he minimum iniial velociy (escape velociy) a which a projecile should be sho verically from he earh s surface so ha i does no fall back o he earh. Hin: This requires ha v = 0 as y : q. 0 v dv = a dy v dv = -g 0 R 2 Ly L0 v y = g 0 R 2 R + y 2 q v = 22g 0 R 0 q dy (R + y) 2 = 22(9.81)(6356)(10) 3 = m>s = 11.2 km>s ccouning for he variaion of graviaional acceleraion a wih respec o aliude y (see Prob ), derive an equaion ha relaes he velociy of a freely falling paricle o is aliude.ssume ha he paricle is released from res a an aliude y 0 from he earh s surface.wih wha velociy does he paricle srike he earh if i is released from res a an aliude y 0 = 500 km? Use he numerical daa in Prob From Prob , (+ c) a = -g 0 Since a dy = v dv hen -g 0 R 2 L g 0 R 2 y 1 c R + y d g 0 R 2 1 [ R + y - 1 ] = v2 R + y 0 2 Thus y 0 y dy (R + y) 2 = v dv L0 y 0 R 2 (R + y) 2 = v 2 2 v = -R 2g 0 (y 0 - y) (R + y)(r + y 0 ) When y 0 = 500 km, y = 0, v v = -6356(10 3 ) 2(9.81)(500)(10 3 ) 6356( )(10 6 ) v = m>s = 3.02 km>s T 25

26 91962_01_s12-p /8/09 8:07 M Page 26 * When a paricle falls hrough he air, is iniial acceleraion a = g diminishes unil i is zero, and hereafer i falls a a consan or erminal velociy v f. If his variaion of he acceleraion can be expressed as a = 1g>v 2 f21v 2 f - v 2 2, deermine he ime needed for he velociy o become v = v f >2. Iniially he paricle falls from res. dv d = a = g v 2 v 2 f - v 2 B f v dy = g L0 v 2 f - v 2 v 2 d f L0 1 ln v f + v 2v f v f - v ` y = v f 2g ln v f + v v f - v = v f 2g ln v f + v f>2 v f - v f>2 0 = g v 2 f = 0.549a v f g b 26

27 91962_01_s12-p /8/09 8:07 M Page paricle is moving along a sraigh line such ha is posiion from a fixed poin is s = ( ) m, where is in seconds. Deermine he oal disance raveled by he paricle from = 1 s o = 3 s. lso, find he average speed of he paricle during his ime inerval. Velociy: : + B v = ds d = d d B v = m>s The velociy of he paricle changes direcion a he insan when i is momenarily brough o res. Thus, v = = 0 ( ) = 0 = 0 and 2 s Posiion: The posiions of he paricle a = 0 s, 1 s, 2 s, and 3 s are s = 0 s = B B = 12 m s = 1 s = B B = 2 m s = 2 s = B B = -8 m s = 3 s = B B = 12 m Using he above resuls, he pah of he paricle is shown in Fig. a. From his figure, he disance raveled by he paricle during he ime inerval = 1 s o = 3 s is s To = (2 + 8) + (8 + 12) = 30 m The average speed of he paricle during he same ime inerval is v avg = s To = 30 = 15 m>s

28 91962_01_s12-p /8/09 8:07 M Page The speed of a rain during he firs minue has been recorded as follows: (s) v ( m>s) Plo he v- graph, approximaing he curve as sraigh-line segmens beween he given poins. Deermine he oal disance raveled. The oal disance raveled is equal o he area under he graph. s T = 1 2 (20)(16) (40-20)( ) + 1 (60-40)( ) = 980 m 2 28

29 91962_01_s12-p /8/09 8:07 M Page wo-sage missile is fired verically from res wih he acceleraion shown. In 15 s he firs sage burns ou and he second sage B ignies. Plo he v- and s- a (m/s 2 ) graphs which describe he wo-sage moion of he missile for 0 20 s. B Since v = a d, he consan lines of he a graph become sloping lines for he v graph. L The numerical values for each poin are calculaed from he oal area under he a graph o he poin. = 15 s, v = (18)(15) = 270 m>s = 20 s, v = (25)(20-15) = 395 m>s (s) Since s = v d, he sloping lines of he v graph become parabolic curves for he s graph. L The numerical values for each poin are calculaed from he oal area under he v graph o he poin. = 15 s, s = 1 (15)(270) = 2025 m 2 = 20 s, s = (20-15) + 1 ( )(20-15) = m = 3.69 km 2 lso: 0 15: a = 18 v = v 0 + a c = s = s 0 + v a c 2 = = 15: v = 18(15) = 270 s = 9(15) 2 = : a = 25 v = v 0 + a c = ( - 15) s = s 0 + v a c 2 = ( - 15) + 1 (25)( - 15)2 2 When = 20: v = 395 m>s s = m = 3.69 km 29

30 91962_01_s12-p /8/09 8:08 M Page 30 * freigh rain sars from res and ravels wih a consan acceleraion of 0.5 f>s 2. fer a ime i mainains a consan speed so ha when = 160 s i has raveled 2000 f. Deermine he ime and draw he v graph for he moion. Toal Disance Traveled: The disance for par one of he moion can be relaed o ime = by applying Eq wih s 0 = 0 and y 0 = 0. : + B s = s 0 + y a c 2 s 1 = (0.5)( )2 = 0.25( ) 2 The velociy a ime can be obained by applying Eq wih y 0 = 0. : + B y = y 0 + a c = = 0.5 [1] The ime for he second sage of moion is 2 = and he rain is raveling a a consan velociy of y = 0.5 (Eq. [1]). Thus, he disance for his par of moion is : + B s 2 = y 2 = 0.5 (160 - ) = ( ) 2 If he oal disance raveled is s To = 2000, hen s To = s 1 + s = 0.25( ) ( ) ( ) = 0 Choose a roo ha is less han 160 s, hen =27.34 s = 27.3 s Y Graph: The equaion for he velociy is given by Eq. [1]. When = =27.34 s, y = 0.5(27.34) = 13.7 f>s. 30

31 91962_01_s12-p /8/09 8:08 M Page If he posiion of a paricle is defined by s = [2 sin (p>5) + 4] m, where is in seconds, consruc he s-, v-, and a- graphs for 0 10 s. 31

32 91962_01_s12-p /8/09 8:08 M Page rain sars from saion and for he firs kilomeer, i ravels wih a uniform acceleraion. Then, for he nex wo kilomeers, i ravels wih a uniform speed. Finally, he rain deceleraes uniformly for anoher kilomeer before coming o res a saion B. If he ime for he whole journey is six minues, draw he v graph and deermine he maximum speed of he rain. For sage (1) moion, : + B v 1 = v 0 + (a c ) 1 v max = 0 + (a c ) 1 1 v max = (a c ) 1 1 (1) : + B v 1 2 = v (a c ) 1 (s 1 - s 0 ) v max 2 = 0 + 2(a c ) 1 (1000-0) (a c ) 1 = v max Eliminaing (a c ) 1 from Eqs. (1) and (2), we have 1 = 2000 v max For sage (2) moion, he rain ravels wih he consan velociy of = ( 2-1 ). Thus, : + B s 2 = s 1 + v (a c) = v max ( 2-1 ) = 2000 v max For sage (3) moion, he rain ravels for = Thus, : + B v 3 = v 2 + (a c ) 3 0 = v max - (a c ) 3 (360-2 ) v max = (a c ) 3 (360-2 ) : + B v 3 2 = v (a c ) 3 (s 3 - s 2 ) 0 = v max 2 + 2C -(a c ) 3 D( ) (a c ) 3 = v max Eliminaing (a c ) 3 from Eqs. (5) and (6) yields = 2000 v max Solving Eqs. (3), (4), and (7), we have 1 = 120 s 2 = 240 s v max (2) (3) for (4) (5) (6) (7) v max = 16.7 m>s Based on he above resuls he v- graph is shown in Fig. a. 32

33 91962_01_s12-p /8/09 8:08 M Page The paricle ravels along a sraigh line wih he velociy described by he graph. Consruc he a-s graph. v (m/s) v 2s 4 v s 7 a s Graph: For 0 s 6 3 m, 3 6 s (m) : + B a = v dv ds s = 0 m and 3 m, For 3m 6 s 6 m, : + B a = v dv ds s = 3 m and 6 m, a s = 0 m = 4(0) + 8 = 8 m>s 2 a s = 3 m = 4(3) + 8 = 20 m>s 2 a s = 3 m = = 10 m>s 2 a s = 6 m = = 13 m>s 2 The a-s graph is shown in Fig. a. = (2s + 4)(2) = (4s + 8) m>s2 = (s + 7)(1) = (s + 7) m>s2 33

34 91962_01_s12-p /8/09 8:08 M Page 34 * The a s graph for a jeep raveling along a sraigh road is given for he firs 300 m of is moion. Consruc he v s graph. s = 0, v = 0. a (m/s 2 ) s (m) a s Graph: The funcion of acceleraion a in erms of s for he inerval 0 m s m is a - 0 s - 0 = a = (0.01s) m>s 2 For he inerval 200 m 6 s 300 m, a - 2 s = 0-2 a = (-0.02s + 6) m>s Y s Graph: The funcion of velociy y in erms of s can be obained by applying ydy = ads. For he inerval 0 m s<200 m, s = 200 m, y = 0.100(200) = 20.0 m>s For he inerval 200 m 6 s 300 m, L y 20.0m>s ydy = ds y s ydy = 0.01sds L0 L 0 y = (0.1s) m>s ydy = ads ydy = L s 200m (-0.02s + 6)ds y = s s B m>s s = 300 m, y = (300 2 ) + 12(300) = 24.5 m>s 34

35 91962_01_s12-p /8/09 8:08 M Page paricle ravels along a curve defined by he equaion s = ( ) m. where is in seconds. Draw he s -, v -, and a - graphs for he paricle for 0 3 s. s = v = ds d = a = dv d = 6-6 v = 0 a 0 = = s, and = s, s = = m s = = m ruck is raveling along he sraigh line wih a velociy described by he graph. Consruc he a-s graph for 0 s 1500 f. a s Graph: For 0 s f, v (f/s) 75 v 0.6 s 3/4 : + dv B a = v ds = 0.6s3>4 B c 3 4 (0.6)s - 1>4 d = 0.27s 1>2 Bf >s 2 s = 625f, a s = 625 f = >2 B = 6.75f>s 2 For 625 f 6 s f, : + B a = v dv ds = 75(0) = 0 The a-s graph is shown in Fig. a s(f) 35

36 91962_01_s12-p /8/09 8:09 M Page car sars from res and ravels along a sraigh road wih a velociy described by he graph. Deermine he oal disance raveled unil he car sops. Consruc he s and a graphs. v(m/s) 30 v v s Graph: For he ime inerval s, he iniial condiion is s = 0 when = 0 s. : + B ds = vd s ds = d L0 L (s) s = 2 2 m When = 30 s, For he ime inerval 30 s 6 90 s, he iniial condiion is s = 450 m when = 30 s. : + B ds = vd L s 450 m ds = ( )d L 30s s = = 450 m s = a b m When = 90 s, s = 90 s = B + 45(90) = 1350 m The s - graph shown is in Fig. a. a Graph: For he ime inerval s, For he ime inerval 30 s 6 90 s, The a- graph is shown in Fig. b. a = dv d = d () = 1 m>s2 d a = dv d = d ( ) = -0.5 m>s2 d Noe: Since he change in posiion of he car is equal o he area under he graph, he oal disance raveled by he car is v- s = L vd s = 90 s - 0 = 1 2 (90)(30) s = 90 s = 1350 s 36

37 91962_01_s12-p /8/09 8:09 M Page 37 * car ravels up a hill wih he speed shown. Deermine he oal disance he car ravels unil i sops ( = 60 s). Plo he a- graph. v (m/s) 10 Disance raveled is area under v- graph (s) s = (10)(30) + 1 (10)(30) = 450 m The snowmobile moves along a sraigh course according o he v graph. Consruc he s and a graphs for he same 50-s ime inerval. When = 0, s = 0. s Graph: The posiion funcion in erms of ime can be obained by applying y = ds. For ime inerval 0 s 6 30 s, y = = a 2 b m>s d 5 ds = yd v (m/s) 12 s 2 ds = L0 L 0 5 d (s) s = a b m = 30 s, s = B = 180 m For ime inerval 30 s< 50 s, ds = yd s = (12-180) m = 50 s, s = 12(50) = 420 m L s 180 m ds = 12d L 30 s a Graph: The acceleraion funcion in erms of ime can be obained by applying a = dy. For ime inerval 0 s <30 s and 30 s< 50 s, a = dy = 0.4 and a = dy d = 2 d 5 m>s2, respecively. d = 0 37

38 91962_01_s12-p /8/09 8:09 M Page moorcyclis a is raveling a 60 f>s when he wishes o pass he ruck T which is raveling a a consan speed of 60 f>s. To do so he moorcyclis acceleraes a 6 f>s 2 unil reaching a maximum speed of 85 f>s. If he hen mainains his speed, deermine he ime needed for him o reach a poin locaed 100 f in fron of he ruck. Draw he v- and s- graphs for he moorcycle during his ime. (v m ) 1 60 f/s (v m ) 2 85 f/s v 60 f/s T 40 f 55 f 100 f Moorcycle: Time o reach 85 f>s, v = v 0 + a c 85 = = s v 2 = v a c (s - s 0 ) Disance raveled, (85) 2 = (60) 2 + 2(6)(s m - 0) s m = f In = s, ruck ravels s = 60(4.167) = 250 f Furher disance for moorcycle o ravel: = f Moorcycle: s = s 0 + v 0 (s ) = Truck: s = Thus =5.717 s = = 9.88 s Toal disance moorcycle ravels s T = (5.717) = 788 f 38

39 91962_01_s12-p /8/09 8:09 M Page n airplane raveling a 70 m>s lands on a sraigh runway and has a deceleraion described by he graph. Deermine he ime and he disance raveled for i o reach a speed of 5 m>s. Consruc he v and s- graphs for his ime inerval, 0. a(m/s 2 ) 5 (s) 4 N Graph: For he ime inerval s, he iniial condiion is v = 70 m>s when = 0 s. 10 : + B dv = ad L v 70 m>s dv = -10d L 0 v = ( ) m>s When = 5 s, v = 5 s = -10(5) + 70 = 20 m>s For he ime inerval 5 s 6, he iniial condiion is v = 20 m>s when = 5 s. : + B dv = ad L v 20 m>s dv = -4d L 5 s v = ( ) m>s When v = 5 m>s, 5 = =8.75 s lso, he change in velociy is equal o he area under he a graph. Thus, v = L ad =8.75s The v graph is shown in Fig. a = -C5(10) + 4( -5)D 39

40 91962_01_s12-p /8/09 8:09 M Page Coninued s Graph: For he ime inerval s, he iniial condiion is s = 0 when = 0 s. : + B ds = vd s ds = ( )d L0 L 0 s = Bm When = 5 s, For he ime inerval 5 6 =8.75 s he iniial condiion is s = 225 m when = 5 s. : + B ds = vd L s 225 m s = 5 s = B + 70(5) = 225 m ds = ( )d L s = Bm When = =8.75 s, 5 s = 8.75 s = B + 40(8.75) + 75 = m = 272 m lso, he change in posiion is equal o he area under he v graph. Referring o Fig. a, we have s = L vd s = 8.75 s - 0 = 1 2 ( )(5) (20 + 5)(3.75) = m = 272 m The s graph is shown in Fig. b. 40

41 91962_01_s12-p /8/09 8:10 M Page 41 * The posiion of a cyclis raveling along a sraigh road is described by he graph. Consruc he v and a graphs. s (m) N Graph: For he ime inerval s, s : + B v = ds d = d d B = B m>s When = 0 s and 10 s, v = 0 = B = 0 v = 10 s = B = 15 m/s For he ime inerval 10 s 6 20 s, 50 s (s) : + B v = ds d = d d B = ( ) m>s When = 10 s and 20 s, v = 10 s = -1.25(10) = 15 m>s v = 20 s = -1.25(20) = 2.5 m>s The v graph is shown in Fig. a. a Graph: For he ime inerval s, : + B a = dv d = d d B = (0.3) m>s 2 When = 0 s and 10 s, a = 0 s = 0.3(0) = 0 a = 10 s = 0.3(10) = 3 m>s2 For he ime inerval 10 s 6 20 s, : + B a = dv d = d ( ) = m>s2 d he a graph is shown in Fig. b. 41

42 91962_01_s12-p /8/09 8:10 M Page The dragser sars from res and ravels along a sraigh rack wih an acceleraion-deceleraion described by he graph. Consruc he v-s graph for 0 s s, and deermine he disance s raveled before he dragser again comes o res. a(m/s 2 ) 25 a 0.1s 5 N s Graph: For 0 s m, he iniial condiion is v = 0 a s = s s(m) : + B vdv = ads 15 v s vdv = (0.1s + 5)ds L0 L 0 v 2 v s = 0.05s 2 + 5sB 2 0 v = a 20.1s sb m>s s = 200 m, v s = 200 m = B + 10(200) = m>s = 77.5 m>s For 200 m 6 s s, he iniial condiion is v = m>s a s = 200 m. : + B vdv = ads L v m>s v dv = L s 200 m -15ds v 2 v m>s s = -15s 200 m v = 2-30s B m>s When v = 0, 0 = 2-30s s =400 m The v s graph is shown in Fig. a. 42

43 91962_01_s12-p /8/09 8:10 M Page spors car ravels along a sraigh road wih an acceleraion-deceleraion described by he graph. If he car sars from res, deermine he disance s he car ravels unil i sops. Consruc he v-s graph for 0 s s. a(f/s 2 ) s s(f) 4 N s Graph: For 0 s f, he iniial condiion is v = 0 a s = 0. : + B vdv = ads When s = 1000 f, For 1000 f 6 s s, he iniial condiion is v = f>s a s = 1000 f. : + B vdv = ads When v = 0, v s vdv = 6ds L0 L 0 v 2 2 = 6s v = 212s 1>2 B f>s L v f>s v 2 v 2 2 v = 212(1000) 1>2 = f>s = 110 f>s f>s vdv = L s 1000 f s = -4s 1000 f v = sB f>s -4ds 0 = s s =2500 f The v s graph is shown in Fig. a. 43

44 91962_01_s12-p /8/09 8:10 M Page missile saring from res ravels along a sraigh rack and for 10 s has an acceleraion as shown. Draw he v- graph ha describes he moion and find he disance raveled in 10 s. a (m/s 2 ) a 2 20 For 5 s, a = 6 dv = a d a 6 v dv = 6 d L0 L 0 v = (s) When = 5 s, v = 75 m>s For s, a = dv = a d v dv = (2 + 20) d L75 L 5 v - 75 = v = When = 10 s, v = 250 m>s Disance a = 5 s: ds = v d s 5 ds = 3 2 d L0 L 0 s = (5) 3 = 125 m Disance a = 10 s: L s 125 ds = v dv 10 ds = Bd L 5 s = d 5 s = 917 m 44

45 91962_01_s12-p /8/09 8:10 M Page 45 * moorcyclis saring from res ravels along a sraigh road and for 10 s has an acceleraion as shown. Draw he v- graph ha describes he moion and find he disance raveled in 10 s. a (m/s 2 ) For dv = a d v 1 dv = L0 L d v = ds = v d s 1 ds = L0 L d s = When = 6 s, v = 12 m>s s = 18 m For dv = a d v dv = 6 d L12 L 6 v = 6-24 ds = v d s ds = (6-24) d L18 L 6 s = When = 10 s, v = 36 m>s 6 1 a (s) s = 114 m 45

46 91962_01_s12-p /8/09 8:19 M Page The v- graph of a car while raveling along a road is shown. Draw he s- and a- graphs for he moion. v (m/s) a = v = 20 5 = 4 m>s (s) 5 20 a = v = = 0 m>s a = v From he v graph a 1 = 5 s, 2 = 20 s, and 3 = 30 s, s 1 = 1 = 1 (5)(20) = 50 m 2 = 0-20 = -2 m>s s 2 = = (20-5) = 350 m s 3 = = (30-20)(20) = 450 m 2 The equaions defining he porions of he s graph are s 0 5 s v = 4; ds = v d; ds = 4 d; s = 2 2 L L 0 0 s 5 20 s v = 20; ds = v d; ds = 20 d; s = L L 50 5 s s v = 2(30 - ); ds = v d; ds = 2(30 - ) d; s = L L

47 91962_01_s12-p /8/09 8:20 M Page The boa ravels in a sraigh line wih he acceleraion described by he a-s graph. If i sars from res, consruc he v-s graph and deermine he boa s maximum speed. Wha disance s does i ravel before i sops? a(m/s 2 ) 6 a 0.02s 6 N s Graph: For 0 s m, he iniial condiion is v = 0 a s = 0. 3 : + B vdv = ads 150 s s(m) v s vdv = (-0.02s + 6)ds L0 L 0 4 v 2 v = -0.01s 2 s + 6sB 0 v = a s sb m>s The maximum velociy of he boa occurs a changes sign. Thus, s = 150 m, where is acceleraion v max = v s = 150 m = B + 12(150) = m>s = 36.7 m>s For 150 m 6 s 6 s, he iniial condiion is v = m>s a s = 150 m. : + B vdv = ads L v 2 v 2 2 Thus, when v = 0, v m>s m>s vdv = L s 150 m s = -4s m v = 2-8s m>s -4ds 0 = 2-8s s = m = 319 m The v s graph is shown in Fig. a. 47

48 91962_01_s12-p /8/09 8:20 M Page The rocke has an acceleraion described by he graph. If i sars from res, consruc he v- and s- graphs for he moion for he ime inerval 0 14 s. a(m/s 2 ) N Graph: For he ime inerval s, he iniial condiion is v = 0 a s = 0. + c B dv = ad v dv = 6 1>2 d L0 L a 2 36 a 4 18 v = 4 3>2 Bm>s When = 9 s, 9 14 (s) v = 9 s = 49 3>2 B = 108 m>s The iniial condiion is v = 108 m>s a = 9 s. + c B dv = ad L v 108 m>s dv = (4-18)d L v = B m>s When = 14 s, 9 s v = 14 s = B - 18(14) = 248 m>s The v graph is shown in Fig. a. s Graph: For he ime inerval s, he iniial condiion is s = 0 when = 0. + c B ds = vd s ds = 4 3>2 d L0 L 0 s = 8 5 5>2 When = 9 s, s = 9 s = >2 B = m For he ime inerval 9 s 6 14 s, he iniial condiion is s = m when = 9 s. + c B ds = vd L s m ds = Bd L 9 s s = a b m When = 14 s, s = 14 s = B B + 108(14) = 1237 m The s graph is shown in Fig. b. 48

49 91962_01_s12-p /8/09 8:20 M Page 49 * The je bike is moving along a sraigh road wih he speed described by he v-s graph. Consruc he a-s graph. v(m/s) v 5s 1/2 75 v 0.2s 120 a s Graph: For 0 s m, : + B a = v dv ds = 5s1>2 B a 5 2 s - 1>2 b = 12.5 m>s s(m) For 225 m 6 s 525 m, : + B a = v dv ds = (-0.2s + 120)(-0.2) = (0.04s - 24) m>s2 s = 225 m and 525 m, a s = 225 m = 0.04(225) - 24 = -15 m>s 2 a s = 525 m = 0.04(525) - 24 = -3 m>s 2 The a s graph is shown in Fig. a. 49

50 91962_01_s12-p /8/09 8:20 M Page The acceleraion of he speed boa saring from res is described by he graph. Consruc he v-s graph. a(f/s 2 ) 10 a 0.04s 2 N s Graph: For 0 s f, he iniial condiion is v = 0 a s = s(f) : + B vdv = ads v s vdv = (0.04s + 2)ds L0 L 0 v 2 v = 0.02s 2 s + 2s 0 v = 20.04s 2 + 4s f>s s = 200 f, v s = 200 f = B + 4(200) = f>s = 49.0 f >s For 200 f 6 s 500 f, he iniial condiion is v = f>s a s = 200 f. : + B vdv = ads L v f>s v 2 v f>s vdv = L s 200 f s = 10s 200 f 10ds v = 220s f>s s = 500 f, v s = 500 f = 220(500) = f>s = 91.7 f>s The v s graph is shown in Fig. a. 50

51 91962_01_s12-p /8/09 8:21 M Page The boa ravels along a sraigh line wih he speed described by he graph. Consruc he s- and a-s graphs. lso, deermine he ime required for he boa o ravel a disance s = 400 m if s = 0 when = 0. v(m/s) 80 s Graph: For 0 s m, he iniial condiion is s = 0 when = 0 s. : + B d = ds v s ds d = L0 L 0 = s 1>2 s = 2 B m 2s 1>2 20 v 2 4s v 0.2s s(m) When s = 100 m, 100 = 2 = 10 s For s 400 m, he iniial condiion is s = 100 m when = 10 s. : + B d = ds v L 10 s - 10 = 5ln 5-2 = ln s 100 e >5-2 = e >5 d = L s s 100 e 2 = s m s 100 ds 0.2s s = 13.53e >5 B m When s = 400 m, 400 = 13.53e >5 = s = 16.9 s The s graph is shown in Fig. a. a s Graph: For 0 m s m, For 100 m 6 s 400 m, a = v dv ds = 2s1>2 Bs - 1>2 B = 2 m>s 2 a = v dv ds When s = 100 m and 400 m, a s = 100 m = 0.04(100) = 4 m>s 2 a s = 400 m = 0.04(400) = 16 m>s 2 The a s graph is shown in Fig. b. = (0.2s)(0.2) = 0.04s 51

52 91962_01_s12-p /8/09 8:21 M Page The s graph for a rain has been deermined experimenally. From he daa, consruc he v- and a graphs for he moion. s (m) 600 s s Y Graph: The velociy in erms of ime can be obained by applying y = ds. d For ime inerval 0 s 30 s, When = 30 s, y = 0.8(30) = 24.0 m>s For ime inerval 30 s 6 40 s, y = ds d = 0.8 y = ds d = 24.0 m>s a Graph: The acceleraion in erms of ime can be obained by applying a = dy. d For ime inerval 0 s 6 30 s and 30 s 6 40 s, a = dy = m>s2 and a = dy d, respecively. d = (s) 52

53 91962_01_s12-p /8/09 8:21 M Page 53 * The airplane lands a 250 f>s on a sraigh a(f/s 2 ) runway and has a deceleraion described by he graph. Deermine he disance s raveled before is speed is decreased o 25 f>s. Draw he s- graph s s(f) N s Graph: For 0 s f, he iniial condiion is v = 250 f>s a s = 0 s. : + B vdv = ads L v 250 f>s v 2 v f>s s vdv = -15ds L 0 s = -15s s = 1750 f, For 1750 f 6 s 6 s, he iniial condiion is v = 100 f>s a s = 1750 f. : + B vdv = ads L v 2 v 2 2 When v = 25 f>s v = sB f>s v s = 1750 f = (1750) = 100 f>s v 100 f>s 100 f>s vdv = L s 1750 f s = (-7.5s) 1750 f v = s 25 = s -7.5ds s =2375 f The v s graph is shown in Fig. a. 53

54 91962_01_s12-p /8/09 8:22 M Page The airplane ravels along a sraigh runway wih an acceleraion described by he graph. If i sars from res and requires a velociy of 90 m>s o ake off, deermine he minimum lengh of runway required and he ime for ake off. Consruc he v- and s- graphs. a(m/s 2 ) 8 v graph: For he ime inerval s, he iniial condiion is v = 0 when = 0 s. : + B dv = ad v dv = 0.8d L0 L 0 a (s) v = B m>s When = 10 s, v = B = 40 m>s For he ime inerval 10 s 6, he iniial condiion is v = 40 m>s when = 10 s. : + B dv = ad L v v 40 m>s = 8 10 s v = (8-40) m>s Thus, when v = 90 m>s, v 40 m>s dv = 8d L 10 s 90 = 8-40 = s lso, he change in velociy is equal o he area under he a- graph. Thus, v = ad L 90-0 = 1 (8)(10) + 8( -10) 2 =16.25 s The v graph is shown in Fig. a. 54

55 91962_01_s12-p /8/09 8:22 M Page 55 s Graph: For he ime inerval s, he iniial condiion is s = 0 when = 0 s. : + B ds = vd When = 10 s, For he ime inerval 10 s 6 =16.25 s, he iniial condiion is s = m when = 10 s. : + B ds = vd L s s When = =16.25 s s ds = d L0 L 0 s = B m s = 10 s = B = m s m ds = (8-40)d L 10s m = 42-40B 2 s = Bm 10 s s = s = 4(16.25)2-40(16.25) = m = 540 m The s graph is shown in Fig. b. 55

56 91962_01_s12-p /8/09 8:22 M Page The a graph of he bulle rain is shown. If he rain sars from res, deermine he elapsed ime before i again comes o res. Wha is he oal disance raveled during his ime inerval? Consruc he v and s graphs. a(m/s 2 ) N Graph: For he ime inerval s, he iniial condiion is v = 0 when = 0 s. : + B dv = ad v dv = 0.1d L0 L 0 3 a a ( ) 5 15 (s) v = B m>s When = 30 s, v = 30 s = B = 45 m>s For he ime inerval 30 s 6, he iniial condiion is v = 45 m>s a = 30 s. : + B dv = ad L v 45 m>s 1 dv = - L d 30 s v = m>s Thus, when v = 0, 0 = Choosing he roo 775 s, = s = 133 s lso, he change in velociy is equal o he area under he a graph. Thus, v = L ad 0 = 1 2 (3)(75) B ( -75)R 15 0 =

57 91962_01_s12-p /8/09 8:22 M Page 57 This equaion is he same as he one obained previously. The slope of he v graph is zero when = 75 s, which is he insan a = dv.thus, d = 0 The v graph is shown in Fig. a. s Graph: Using he resul of v, he equaion of he s graph can be obained by inegraing he kinemaic equaion ds = vd. For he ime inerval s, he iniial condiion s = 0 a = 0 s will be used as he inegraion limi. Thus, : + B ds = vd When = 30 s, s = a b m For he ime inerval 30 s 6 = s, he iniial condiion is s = 450 m when = 30 s. : + B ds = vd v = 75 s = B + 5(75) - 75 = m>s s ds = d L0 L 0 L s 450 m ds = L 30 s s = a b m When = 75 s and = s, s = 30 s = B = 450 m a bd s = 75 s = B B - 75(75) = 4500 m s = s = B B - 75(133.09) = 8857 m The s graph is shown in Fig. b. 57

58 91962_01_s12-p /8/09 8:23 M Page The posiion of a paricle is r = 5(3 3-2)i (4 1/2 + )j + (3 2-2)k6 m, where is in seconds. Deermine he magniude of he paricle s velociy and acceleraion when = 2 s. Velociy: v = dr d = d d c 33-2Bi - 4 1>2 + Bj Bk d = c 9 2-2Bi - 2-1>2 + 1Bj + (6)k d m>s When = 2 s, v = B c92 2 B - 2 di - c22-1>2 B + 1 dj + 6(2)kR m>s = [34i j + 12k] m>s Thus, he magniude of he paricle s velociy is v = 2v x 2 + v y 2 + v z 2 = (-2.414) = 36.1 m>s cceleraion: a = dv d = d d c 92-2Bi - 2-1>2 + 1Bj + (6)k d m>s = C(18)i + - 3>2 j + 6kD m>s 2 When = 2 s, a = C18(2)i + 2-3>2 j + 6kD m>s 2 = [36i j + 6k] m>s 2 Thus, he magniude of he paricle s acceleraion is a = 2a x 2 + a y 2 + a z 2 = = 36.5 m>s 2 * The velociy of a paricle is v = 53i + (6-2)j6 m>s, where is in seconds. If r = 0 when = 0, deermine he displacemen of he paricle during he ime inerval = 1 s o = 3 s. Posiion: The posiion r of he paricle can be deermined by inegraing he kinemaic equaion dr = vd using he iniial condiion r = 0 a = 0 as he inegraion limi. Thus, dr = vd When = 1 s and 3 s, r dr = C3i + (6-2)jDd L0 L 0 r = c3i Bj dm r = 1 s = 3(1)i + C6(1) Dj = [3i + 5j] m>s r = 3 s = 3(3)i + C6(3) - 32 Dj = [9i + 9j] m>s Thus, he displacemen of he paricle is r = r = 3 s - r = 1 s = (9i + 9j) - (3i + 5j) = [6i + 4j] m 58

59 91962_01_s12-p /8/09 8:23 M Page paricle ravels along he parabolic pah y = bx 2. If is componen of velociy along he y axis is v y = c 2, deermine he x and y componens of he paricle s acceleraion. Here b and c are consans. Velociy: Subsiuing he resul of y ino y = bx 2, Thus, he x componen of he paricle s velociy can be deermined by aking he ime derivaive of x. cceleraion: dy = v y d y dy = c 2 d L0 L 0 y = c 3 3 c 3 3 = bx 2 x = c 3b 3>2 v x = x # = d d B c 3b 3>2 R = 3 c 2 3b 1>2 a x = v # x = d d 3 c 2 3b 1>2 = 3 c 4 3b - 1>2 = 3 c 4 3b a y = v # y = d d c2 B = 2c