CHAPTER 2. Answer to Checkpoint Questions

Transcription

1 CHAPTER MOTION ALONG A STRAIGHT LINE CHAPTER Answer o Checkpoin Quesions. (b) and (c). zero 3. (a) () and (4); (b) () and (3); (c) (3) 4. (a) plus; (b) minus; (c) minus; (d) plus 5. () and (4) 6. (a) plus; (b) minus; (c) a g 9:8 m/s. Answer o Quesions. (a) yes; (b) no; (c) yes; (d) yes. (a) minus; (b) posiive; (c) zero; (d) negaive; (e) wice (a s and 3 s) 3. (a), 3; (b), 3; (c) 4 4. (a) negaive direcion of x; (b) posiive direcion of x; (c) yes (when graphed line crosses he axis); (e) posiive; (f) consan 5. all ie (see Eq. -6) 6. a and c 7. (a) g; (b) m/s upward 9. same. 4 mi/h, no. x and x 8( ) + (:5)( ). and ie, hen 3 3. increase

2 CHAPTER MOTION ALONG A STRAIGHT LINE Soluions o Exercises & Problems E (a) Use v x o nd he average speed. For Carl Lewis v L m s m/s; and for Bill Rodgers v R 6 mi 395 yd h min (6 mi)(69 m/mi) + (395 yd)(:944 m/yd) 78 s 5:4 m/s : (b) The ime i would have aken for Lewis o nish he marahon is x v L 486 m m/s 49 s h min 9 s : E The car has moved a disance of s v (9 km/h)(:5 s) (9 km/h)( m/km)( h36 s)(:5 s) 3 m : 3E The disance he gher ravels in a blink is 36 s/h x v ( mi/h)( ms) 94:3 m 39 f : 69 m/mi 4E (a) Assume ha he ball ravels a a consan velociy and use x v, where x is he horizonal disance raveled, is he ime, and v is he velociy. Conver v o meers per second. According o Appendix D, km/h :778 m/s, so 6 km/h (6)(:778 m/s) 44:45 m/s. Thus This may also be wrien 44 ms. x v 8:4 m 44:5 m/s :44 s :

3 CHAPTER MOTION ALONG A STRAIGHT LINE 3 5E The average speed can be found from Fig. -8 of he ex: 8 km 5 cm v cm/y 9 6 : y km 6E Calculae he ime required o make he rip a each of he wo speeds and compue he dierence. Use x v. A he slower speed he ime required is x v 7 km 88:5 km/h 7:9 h ; and a he faser one The dierence is 7:9 h x v 7 km 5 km/h 6:67 h : 6:67 h :4 h h 4 min. 7E The speed of ligh c in various unis is c 3 8 m/s (3 8 m/s)(6:4 4 mi/m)(36 s h) 6:7 8 mi/h (3 8 m/s)(3:8 f/m) 9:84 8 f/s : ly/y : 8E The average velociy during some ime inerval is v x, where x is he displacemen. In his case he inerval is divided ino wo pars. During he rs par we have x 4 km and (4 km)(3 km/h) :33 h. During he second par we have x 4 km and (4 km)(6 km/h) :67 h. Boh displacemens are in he same direcion so he oal displacemen is x x + x km. The oal ime inerval is + :33 h + :67 h : h. The average velociy is v (8 km)(: h) 4 km/h. (b) The average speed is he oal disance raveled divided by he ime. In his case he oal disance is he magniude of he oal displacemen, so he average speed is 4 km/h. (c) Assume he auomobile sars a he origin a ime. Then is coordinae as a funcion of ime is as shown as he solid x (km) (h)

4 4 CHAPTER MOTION ALONG A STRAIGHT LINE lines on he graph in he las page. The average velociy is he slope of he doed line. 9P (a) (b) (c) v x v x (4 + 4) f (44 + 4) s [(6)(4) + (6)()] f (6 + 6) s 5:7 f/s : 7: f/s : x (f) x (f) slopev v 4 slopev v v (s) 6 84 v 6 P Le he lengh of he hill be D. Then he oal ime of ravel is The average speed is hus given by D 4 km/h + D 6 km/h : v D D D(4 km/h) + D(6 km/h) 48 km/h :

5 CHAPTER MOTION ALONG A STRAIGHT LINE 5 P (a) Le he disance beween San Anonio and Houson be L, and he ime i akes o ravel o Houson be T. Then on he way o Houson, he average speed is x (f) v L T T (35 mi/h) + T (55 mi/h) (b) On he way back from Houson, he average speed is (c) The average speed for he enire rip is T L v 43 mi/h : L (35 mi/h) + L (55 mi/h) v L L(45 mi/h) + L(43 mi/h) 44 mi/h : x (f) 45 mi/h : (d) Since here is no ne displacemen in a round rip, he average velociy is zero. (e) slopev v 4 slopev v v (s) 6 84 v 6 Refer o he gure above. The average velociy is given by he slope of he dashed line, wih he posiive sign indicaing ha he moion is in he posiive x direcion. P (a) Subsiue, in urn,,, 3, and 4 s ino he expression x() : x( s) (3)() (4)() + 3 x( s) (3)() (4)() + 3 m x(3 s) (3)(3) (4)(3) x(4 s) (3)(4) (4)(4) m : (b) The displacemen during an inerval is he coordinae a he end of he inerval minus he coordinae a he beginning. For he inerval from o 4 s, he displacemen is x x(4 s) x() m + m. The displacemen is in he posiive x direcion.

6 6 CHAPTER MOTION ALONG A STRAIGHT LINE (c) The average velociy during an inerval is dened as he displacemen over he inerval divided by he inerval: v x. For he inerval from s o 4 s he displacemen is x x(4 s) x( s) ( m) ( m) 4 m and he ime inerval is 4 s s s. Thus v x 4 m 7 m/s : s (d) The solid curve on he graph below shows he coordinae x as a funcion of ime. The slope of he doed line is he average velociy beween : s and 4: s. x (m) P (a) The average velociy during he ime inerval beween : s and 3: s is v s!3 s x x( 3: s) x( : s) 3: s : s :5[(3:)3 (:) 3 ] cm 8:5 cm/s : : s (b) The insananeous velociy a ime is given by v() dxd 4:5, which gives (c) (d) v( : s) (4:5)(:) cm/s 8: cm/s : v( 3: s) (4:5)(3:) cm/s 4:5 cm/s : v( :5 s) (4:5)(:5) cm/s 8: cm/s : (e) Suppose ha, a ime m, he paricle is midway beween is posiions a : s and 3: s. Then x( m ) 9:75 + :5 3 x( : s) + x( 3: s) m [9:75 + :5(:)3 ] + [9:75 + :5(3:) 3 ] ;

7 CHAPTER MOTION ALONG A STRAIGHT LINE 7 yielding m :596 s. Thus he insananeous speed a his insan is v( m ) 4:5(:596) 3:3 cm/s: (f) The answers o he preceeding quesions are given by he slopes in he gure below. x() x (cm) v m v(3s) v(s) v(.5s) P θ v p Le he speed of he je plane be v p. From he gure above i is obvious ha he maximum ime ha he pilo has before a correcion has o be made saisies hv p an, which gives h v p an 35 m (3 km/h)( m/km)( h36 s)(an 4:3 ) :3 s :

8 8 CHAPTER MOTION ALONG A STRAIGHT LINE 5P (a) Neglec he size of he bird. Le D n be he separaion beween he wo rains when he bird begins is nh rip, hen 3 km/h D n+ D n D n 3 km/h + 6 km/h 3 D n ; which gives D n (3) n D, where D 6 km. The oal number of rips N ha he bird makes before he crash hus saises N D N D ; 3 which gives N : (b) The oal ime elapsed before he wo rains crash is 6 km (3 + 3) km/h h ; during which he bird has raveled a disance of x v (6 km/h)( h) 6 km: 6E (a) The velociy of he paricle is v() dx d d d (4 + 3 ) + 6 : Thus v( s) [ + (6)()] m/s 6 m/s. (b) Since v <, i is moving in he negaive x direcion. (c) The speed is jv( s)j 6 m/s. (d) For < < s, jv()j decerases wih ime. A s, v() is zero. Aferwards, jv()j increases wih ime. (e) Yes. Since v crosses over from negaive (a ) o posiive (a large ), i has o pass hrough zero. This acually happens a s. (f) In fac, from v() + 6, we know ha v() > for > s. Thus he paricle canno move lefward on he x axis afer 3 s. 7E The sign of he velociy is deermined by wheher he local slope of he x vs curve is posiive or negaive. (a) The animal is o he lef of he origin on he axis beween : s and 4: s. (b) The velociy is negaive beween and 3: s. (c) The velociy is posiive beween 3: s and 7: s. (d) The velociy is zero a 3: s.

9 CHAPTER MOTION ALONG A STRAIGHT LINE 9 8E The x vs curve is shown below. The seepes par corresponds o he highes speed, while he leas seep par corresponds o he lowes speed. x () () seepes (3) leas seep (4) 9P The disance D ha he runner ravels is represened by he area beween he v vs curve and he ime axis. Thus D ()(8) + ( )(8) + ( )(8 + 4) + (6 )(4) m/s m : E (a) If v is he velociy a he beginning of a ime inerval (a ime ) and v is he velociy a he end (a ) hen he average acceleraion in he inerval is given by a (v v )( ). Take, v 8 m/s, :4 s, and v 3 m/s. Then a ( 3 8) m/s (:4 ) s m/s : The minus sign indicaes ha he acceleraion is opposie o he original direcion of ravel. (b) v (m/s)

10 CHAPTER MOTION ALONG A STRAIGHT LINE E a E (a) Use v dxd and a d xd. For he inerval AB, v is posiive and a. Similarly, for BC: v is posiive and a is negaive; for CD: v and a ; and for DE: v is negaive and a is posiive. (b) No. (c) No, becuase neiher v nor a will be aeced by shifing he origin of he x axis. 3E See he previous problem. For he inerval AB, v is posiive and a is negaive; for BC, v and a ; for CD, v is posiive and a is posiive; and for DE, v is posiive and a. (b) No. (c) No, becuase neiher v nor a will be aeced by shifing he origin of he x axis. 4E v v

11 CHAPTER MOTION ALONG A STRAIGHT LINE 5E x x (s ) (a) (b) x x (s ) (c) (d) (e) For he speed o increase, he velociy and he acceleraion mus be in he same direcion, which is he case for siuaions (a), (b) and (d) above.

12 CHAPTER MOTION ALONG A STRAIGHT LINE 6E (a) No. The former is he square of he rs derivaive of x wih respec o, while he laer is he second derivaive of x wih respec o. (b) The SI uni for he former is m s, while i is m/s for he laer. 7E (a) The average velociy during he rs 3 s is given by v [x(3 s) x()]. Evaluae he expression x 5+ using and 3 s: x() and x(3 s) (5)(3)+()(3) 4 m. Thus v (4 ) m 8 m/s : 3 s (b) The insananeous velociy a ime is given by v dx d 5 +, in m/s. A 3: s, v 5 + ()(3:) m/s. (c) The insananeous acceleraion a ime is given by a dv d m/s. consan, so he acceleraion a any ime is m/s. (d) and (e) The graph on he lef below shows he coordinae x as a funcion of ime (solid curve). The doed line marked (a) runs from, x o 3: s, x 4 m. Is slope is he average velociy during he rs 3 s of moion. The doed line marked (b) is angen o x vs. a 3: s. Is slope is he insananeous velociy a 3: s. The gure on he righ shows he insananeous velociy as a funcion of ime. Is slope is he acceleraion. I is x (m) v (m/s) 4 3 (a) (b)

13 CHAPTER MOTION ALONG A STRAIGHT LINE 3 8E (a) Le v() dxd 5. Then : s. (b) Le a() dvd 3. Then. (c) A >, he acceleraion is always negaive. (d) x (m) x (m) x (m)

14 4 CHAPTER MOTION ALONG A STRAIGHT LINE 9P (a) During he inerval from : min o 8: min, v x (: m/s)(8: min 5: min) : m/s (8: min : min) and a v : m/s (8: min : min) 6: mm/s : (b) During he inerval from 3: min o 9: min, v x (: m/s)(9: min 5: min) :47 m/s (9: min 3: min) and a v : m/s (9: min 3: min) 6: mm/s : (c) The slopes of he lines labeled hrough in he wo gures below give he velociies and he acceleraions previously calculaed. In paricular, he slope of is he average velociy beween : min and 8: min, he slope of is he average velociy beween 3: min and 9: min; he slope of is he average acceleraion beween : min and 8: min, and he slope of is he average acceleraion beween 3: min and 9: min. x (m) 66. v (m/s) α β (min) γ δ (min) 3P (a) The insananeous velociy and acceleraion are given by v() dxd 6:, and a() dvd, respecively, where v is in m/s and a is in m/s. Beween : s and : s, v x (:)[(:)3 (:) 3 ] m 4 m/s (: :) s

15 CHAPTER MOTION ALONG A STRAIGHT LINE 5 and (b) a v (3)(:)[(:) (:) ] m (: :) m 8 m/s : v( : s) 6:(:) 6: m/s ; a( : s) (:) m/s ; v( : s) 6:(:) 4 m/s ; a( : s) (:) 4 m/s : (c) We see ha v( : s) < v < v( : s). This is because boh he acceleraion and he velociy are in he same direcrion (so he speed is increasing wih ime). Also a( : s) < a < a( : s), which is consisen wih he fac ha a() :, which increases wih. (d) The various answers obained in (a) hrough (c) are represened by he slopes in he gures below. x (m) 5 v v 5 v v (m/s) 5 a a -.5 a.5.5

16 6 CHAPTER MOTION ALONG A STRAIGHT LINE 3P (a) The velociy of he spo is given by v() dx d d d (9: :753 ) 9: :5 : Se v() in he equaion above o obain : s. (b) x( s) (9:)() :75() 3 : cm: (c) a() dvd d d (9: :5 ) 4:5 : A : s, a() 9: m/s. (d) and (e) Since v( s) and a( s) <, i mus be moving in he posiive x direcion jus before i sops, and in he negaive x direcion jus afer ha. Since v() > for > : s, we know ha x max x( : s) : cm: Thus he spo never really reaches x 5: cm. The nex ime i reaches an edge of he screen, is coordinae is zero. Thus we le x() 9: :75 3 and ge 3:46 s. 3P (a) Since he unis of e are hose of lengh and he unis of are hose of ime, so he unis of e mus be hose of (lengh)/(ime), or f/s. Since b 3 has unis of lengh, b mus have unis of (lengh)/(ime) 3, or f/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 e 3b, v occurs for and for e3b ()(3:)[3(:)] : s. For, x and for : s; x 4: f. Rejec he rs soluion and accep he second. (c) In he rs 4 s he paricle moves from he origin o x 4: f, urns around, and goes back o x(4 s) (3:)(4) (:)(4) 3 6 f. The oal pah lengh i ravels is 4: + 4: f. (d) Is displacemen is given by x x x, where x and x 6 f. Thus x 6 f. (e) The velociy is given by v 6: 3:. Thus v( s) (6:)(:) (3:)(:) 3: f/s v( s) (6:)(:) (3:)(:) : v(3 s) (6:)(3:) (3:)(3:) 9: f/s v(4 s) (6:)(4:) (3:)(4:) 4 f/s : (f) The acceleraion is given by a dv d 6: a( s) 6: (6:)(:) : 6:. Thus a( s) 6: (6:)(:) 6: f/s a(3 s) 6: (6:)(3:) f/s a(4 s) 6: (6:)(4:) 8 f/s :

17 CHAPTER MOTION ALONG A STRAIGHT LINE 7 33E The ime i akes is v a km/h 5 m/s (Quesion: How does his compare wih your Ferrari?) ( km/h)( h36 s)( m/km) 5 m/s :556 s : 34E Use v v + a, an equaion ha is valid for moion wih consan acceleraion. Take o be he ime when he velociy is +9:6 m/s. Then v 9:6 m/s. (a) Since we wish o calculae he velociy for a ime before, he value we use for is negaive: :5 s. Thus (b) Now +:5 s and v (9:6 m/s) + (3: m/s )( :5 s) :6 m/s : v (9:6 m/s) + (3: m/s )(:5 s) 8 m/s : 35E For he auomobile a v (55 5)( km/h)( h36 s)( m/km) (:5 min)(6 s/min) :8 m/s ; and for he bicycle a v (3 )( km/h)( h36 s)( m/km) (:5 min)(6 s/min) :8 m/s : 36E Solve v v + a for : (v v )a. Subsiue v (:)(3: 8 m/s) 3: 7 m/s, v, and a 9:8 m/s. The resul is 3: 6 s. (b) Evaluae x x + v + a, wih x. The resul is x (9:8 m/s )(3: 6 s) 4:6 3 m.

18 8 CHAPTER MOTION ALONG A STRAIGH 37E Use v v ax: o solve for a. Is minimum value is 38E a min v v x max (36 km/h) (:8 km) :8 m/s : (a) The velociy is given by v v + a. The ime o sop can be found by seing v and solving for : v a. Subsiue his expression ino x v + a o obain x v a. [Alernaively, you migh use v v + a(x x ).] Use v 5: 6 m/s and a :5 4 m/s. Noice ha since he muon slows he iniial velociy and he acceleraion mus have opposie signs. The resul is x (5: 6 m/s) ( :5 4 m/s ) : m : (b) Here are he graphs of he coordinae x and velociy v of he muon from he ime i eners he eld o he ime i sops: x (cm) v ( 6 m/s) E (ns) 3 4 Solve v v + a(x x ) for a. Take x. Then a (v v )x. Use v (ns) 3 4 :5 5 m/s, v 5:7 6 m/s, and x : cm : m. The resul is a (5:7 6 m/s) (:5 5 m/s) (: m) :6 5 m/s :

19 CHAPTER MOTION ALONG A STRAIGHT LINE 9 4E (a) Using v v ax, and noing ha v, we nd jaj v x (6 mi/h) [(69 m/mi)(36 s/h)] ()(43) m 8:3 m/s 8:3 m/s 9:8 m/s g :85g : (b) The sopping ime is v a 3: s; which corresponds o T 3: s4 ms 8 \reacion imes". 4E The acceleraion is a v ( km/h)( m/km)( h36 s) :4 s :4 m/s : In erms of g, his is :4 m/s a 9:8 m/s g g : 4E (a) The ime i akes o sop is v a 4:6 m/s 4:9 m/s 5: s : (b) The disance raveled is (c) x v a (4:6 m/s) ()(4:9 m/s ) 6:5 m/s : 6.5 x (m) 4.6 v (m/s) 5 5

20 3 CHAPTER MOTION ALONG A STRAIGHT LINE 43E The acceleraion of he sled is a v In erms of g, his is (6 km/h)( m/km)( h36 s) :8 s :5 m/s a 9:8 m/s g 5g : :5 m/s : (b) The disance raveled is s a (:5 m/s )(:8 s) 4: m: 44E (a) Assume he acceleraion is uniform and solve v v + a for : (v v )a. Subsiue v 85 mi/h 5 f/s, v 55 mi/h 8:7 f/s, and a 7 f/s. Noe ha since he car is slowing down he iniial velociy and he acceleraion mus have opposie signs. The resul is (8:7 5) f/s :6 s ( 7 f/s : ) (b) Suppose he coordinae of he car s x when he brakes are applied (a ime ). Then he coordinae of he car as a funcion of ime is given by x (5 f/s) (7 f/s ). This funcion is ploed from o :6 s on he graph o he righ. x (f) E Firs, calculae he deceleraion a of he moorcycle: a v (5 m/s 3 m/s)3: s 5: m/s : The disance x raveled by he moorcycle before i sops can hen be obained from v v + ax: x v v a (3 m/s) ( 5: m/s ) 9 m : 46P (a) a v (6 km/h)( m/km)( h36 s) 5:4 s (b) x a (3: m/s )(5:4 s) 45 m : 3: m/s :

21 CHAPTER MOTION ALONG A STRAIGHT LINE 3 (c) r s x a (5 m) 3: m/s 3 s : 47P (a) From v v ax, we nd he acceleraion o be (b) The ime required is a v v x (5 3 ) m s ()(6) m 5: m/s : v a (5 3) m/s 5 m/s 4: s : (c) The ime i akes for v o reach 3 m/s is v a 3 m/s 5: m/s 6: s : (d) The disance moved is (e) x a (5: m/s )(6: s) 9 m : 5 x (m ) 6 v (m/s)

22 3 CHAPTER MOTION ALONG A STRAIGHT LINE 48P (a) Take x, and solve x v + a for a: a (x v ). Subsiue x 4: m, v 56: km/h 5:55 m/s, and : s. The resul is a [4: m (5:55 m/s)(: s)] 3:56 m/s (: s) : The minus sign indicaes ha he acceleraion is opposie he direcion of moion of he car. The car is slowing down. (b) Evaluae v v + a. You should ge v 5:55 m/s (3:56 m/s )(: s) 8:43 m/s 3:3 km/h. 49P (a) Le x a he rs poin and ake he ime o be when he car is here. Use v v + a o eliminae a from x v + a. The rs equaion yields a (v v ), so x v + (v v ) (v + v ). Solve for v : v (x v). Subsiue x 6: m, v 5: m/s, and 6: s. Your resul should be v (6: m) (5: m/s)(6: s) 6: s 5: m/s : (b) Subsiue v 5: m/s, v 5: m/s, and 6: s ino a (v is a (5: m/s 5: m/s)(6: s) :67 m/s. v ). The resul (c) Subsiue v, v 5: m/s, and a :67 m/s ino v v + ax. Solve for x: x v a (5: m/s) (:67 m/s ) 7:5 m : (d) To draw he graphs you need o know he ime a which he car is a res. Solve v v + a for : v a (5: m/s)(:67 m/s ) 3: s. Your graphs should look like his: x (m) 6 v (m/s)

23 CHAPTER MOTION ALONG A STRAIGHT LINE 33 5P (a) The ime i akes o cover he rs half of he disance is given by a s, which yields r s s a m : m/s 3:3 s : Since i akes an equal amoun of ime o slow down o zero speed, he oal ime of ravel is T 6:6 s. (b) v max a (: m/s )(3:3 s) 36:3 m/s: (c) x (m) v (m/s) a (m/s ) (s ) -.

24 34 CHAPTER MOTION ALONG A STRAIGHT LINE 5P Le r be he reacion ime and b be he braking ime. Then he oal disance moved by he car is given by x v r + v b + a b, where v is he iniial velociy and a is he acceleraion. Afer he brakes are applied he velociy of he car is given by v v + a b. Use his equaion, wih v, o eliminae b from he rs equaion. According o he second equaion b v a, so x v r v a + v a v r v a. Wrie his equaion wice, once for each of he wo dieren iniial velociies: x v r v a and x v r v a. Solve hese equaions simulaneously for r and a. You should ge and r v x v x v v (v v ) a v v v v v x v x : Subsiue x 86 f, v 5 mi/h 73:4 f/s, x 8 f, and v 3 mi/h 44: f/s. The resuls are r (44: f/s) (86 f) (73:4 f/s) (8 f) (73:4 f/s)(44: f/s)(73:4 f/s 44: f/s) :74 s and a (44: f/s)(73:4 f/s) (73:4 f/s)(44: f/s) (44: f/s)(86 f) (73:4 f/s)(8 f) f/s : 5P (a) The iniial speed of he car, in m/s, is v (35 mi/h)(69 m/mi)( h36 s) 5:6 m/s. Denoe L 4 m and T :8 s. If you decide o brake o a sop, hen by he ime you sar applying he brake he car has already raveled furher owards he inersecion by as much as d v r, where r is your reacion ime. Thus a he momen you sar o apply he brake he disance beween he car and he inersecion is D L d 4 m (5:6 m/s)(:75 s) 8 m : A his poin, he car sars o decelerae. Suppose i will ravel furher by a disance x before coming o a complee sop, hen v ax, where a is he magniude of he deceleraion. Solve for x: x v a (5:6 m/s) (7 f/s )(:348 m/f) 4 m : Since D x 8 m 4 m 4 m >, he car will come o a sop before enering he inersecion.

25 CHAPTER MOTION ALONG A STRAIGHT LINE 35 If you choose o coninue o move, hen you will be able o reach he inersecion in Lv 4 m(5:6 m/s) :6 s. Since he yellow ligh will las for T :8 s, you can make i ino he inersecion while he ligh is sill yellow. Thus eiher sraegy will work. (b) In his case you can follow he same lines of argumen as par (a) above, wih he new values for L and T being L 3 m and T :8 s. As a resul, you will nd ha, if you choose o apply he brake, you wouldn' be able o sop your car unill is fron has already raveled 4 m ino he inersecion. And if you choose o coninue moving, he rac ligh would have already urned yellow while your car is sill abou 4 m shor of reaching he inersecion. So neiher sraegy will work in his case! 53P (a) Le he reacion ime of he driver be r. The reacion disance is hen D r v r, where v is he iniial speed of he car. The braking disance D b saises v jajd b ; where a is he deceleraion of he car. From he above equaions, we nd, for v m/s, r D r v 7:5 m m/s :75 s and jaj v D b ( m/s) (5: m) m/s : You can easily check ha r and jaj are he same for v m/s and v 3 m/s. (b) For v 5m/s, he sopping disance D s is D s D r + D b v r + v jaj (5 m/s) (5 m/s)(:75 s) + ( m/s ) 5 m : 54P (a) Since he maximum acceleraion and deceleraion have he same magniude, from symmery consideraion we know ha he maximum speed is aained when he rain is mideway beween he wo saions, which are separaed by a disance D. Thus v max v v max ax a(d), which gives v max p Da p (86 m)(:34 m/s) 3:9 m/s : (b) The ime i akes for he rain o accelerae o v max is given by v max a 3:9 m/s :34 m/s 4:5 s :

26 36 CHAPTER MOTION ALONG A STRAIGHT LINE Since i akes an equal amoun of ime o accelerae and decelerae, he oal ravel ime beween he wo saions is 49: s. (c) The maximum average speed is (d) v max D + s 86 m :7 m/s : (49 + ) s x (m ) v (m/s) 86 sop 3.9 v (s ) a( m/s ).34 a (s ) -.34

27 CHAPTER MOTION ALONG A STRAIGHT LINE 37 55P (a) Use v max ax o solve for x: x v max a ( f6 s) ()(4 f/s ) 34:7 f : (b) The ime i akes o accelerae and decelerae are boh given by v max a, while i remains a he op speed for (D x)v max, where x is given in (a). Thus he oal ime i akes is + v max a + D x D + v max v max v max a 64 f f6 s + 4:6 s f6 s 4: f/s : 56P (a) Le he required ime be. By his ime, he disances he wo vehicles have raveled mus be he same. Le he acceleraion of he auomobile be a and he speed of he ruck be v. Then s a v ; which gives Thus (b) The speed of he car is v a (9:5 m/s) : m/s 8:6 s : s v (9:5 m/s)(8:6 s) 8 m : v c a (: m/s )(8:6 s) 9 m/s : 57P (a) From he poin of view of he locomoive (refered o as ), he rain will be closing in on i as long as he speed of he rain (refered o as ) is greaer han ha of he locomoive. So he closes disance beween he wo will be reached he momen when hey have idenical speeds. Le he ime of his insan be. Then v v a, or v v a ;

28 38 CHAPTER MOTION ALONG A STRAIGHT LINE where v is he iniial speed of he rain. To ensure ha no collision akes place, he separaion beween he wo mus no be negaive a ime, i.e., v (v a )+:4 mi ; or v (v v ) v (v v ) + (v v ) a a a (v v ) + :4 mi ; a + :4 mi which gives (b) a (v v ) (:4 mi) ( mi/h 8 mi/h) (:4 mi/h) 3:6 f/s : x (mi) locomoive rain (collision).4 rain (acciden avoided) 58P The sopping disance for he rs rain is given by Similarly, for he second rain x v a [(7 km/h)(3 m/km)( h36 s)] ()(:) m/s m : x v a [(44 km/h)(3 m/km)( h36 s)] ()(:) m/s 8 m : Since x + x m > 95 m, here will indeed be a collision.

29 CHAPTER MOTION ALONG A STRAIGHT LINE 39 59P v 6E (a) Take he y axis o be posiive in he upward direcion and ake and y a he poin from which he wrench was dropped. If h is he heigh from which i was dropped and he y axis is upward hen he ground is a y h. Solve v v + gh for h: h (v v)g. Subsiue v, v 4 m/s, and g 9:8 m/s : h (4 m/s) (9:8 m/s ) 9:4 m : (b) Solve v v g for : (v v)g (4 m/s)(9:8 m/s ) :45 s. (c) y (m) -3 v (m/s) a(m/s )

30 4 CHAPTER MOTION ALONG A STRAIGHT LINE 6E (a) A he highes poin he velociy of he ball is insananeously zero. Take he y axis o be upward, se v in v v gy, and solve for v : v p q gy. Subsiue g 9:8 m/s and y 5 m o ge v (9:8 m/s )(5 m) 3 m/s. (b) I will be in he air unil y again. Solve y v g for. Since y he wo soluions are and v g. Rejec he rs and accep he second: v g (3 m/s)(9:8 m/s ) 6:4 s. (c) y (m) v (m/s) a(m/s ) 6E From v ax gh, we nd v p gh p ()(9:8 m/s )(7 m) 83 m/s : This is a much higher speed han ha of acual raindrops. I's hard o say wheher i can kill, bu if I were you, I'd deniely say under a surdy roof in a rainsorm like his!

31 CHAPTER MOTION ALONG A STRAIGHT LINE 4 63E (a) Use v v + gy o solve for v: v p gy p (9:8 m/s )( m) 48:5 m/s : (b) The ime of igh saises y g, so (c) The speed is r s y g ( m) 9:8 m/s 4:95 s : s v y g p gy p (9:8 m/s )( m) 34:3 m/s : (d) Use v g. The ime is hen v g 34:3 m/s 9:8 m/s 3:5 s : 64E The ime ha i akes for he sone o reach he ground saises y 3: m v + g (: m/s) + (9:8 m/s ) ; which gives :54 s. (b) The speed a impac is given by v v + g : m/s + (9:8 m/s )(:54 s) 7: m/s : 65E (a) r s y g (45 m) 9:8 m/s 5:44 s : (b) v g (9:8 m/s )(5:44 s) 53:3 m/s: (c) The disance y saises v ay (5g)y, which gives y v 5g (53:3 m/s) 5(9:8 m/s ) 5:8 m :

32 4 CHAPTER MOTION ALONG A STRAIGHT LINE 66E Assume ha he rocke is being propelled by a consan force F, in which case he acceleraion a of he rocke increases as more and more fuel ge burn, since he mass m of he rocke is decreasing (Noe ha a Fm). The gures are shown below. y v a y max fuel exhaused his ground fuel exhaused fuel exhaused his ground 67E Take he y axis o be upward, se v, and solve y v (a) For his par y 5 m, so s ( 5 m) 9:8 m/s 3: s : g for : p yg. (b) For his par se y m, so s ( m) 9:8 m/s 4:5 s : The dierence is he ime aken o fall he second 5 m: 4:5 s 3: s :3 s. 68P (a) Take he y axis o be posiive upward and use y v : s. Solve for v : v y + g (:544 m) + (9:8 m/s )(: s) : s (b) Use v v g (3:7 m/s) (9:8 m/s )(: s) :74 m/s. g, wih y :544 m and 3:7 m/s :

33 CHAPTER MOTION ALONG A STRAIGHT LINE 43 (c) Use v v gy, wih v. Solve for y: y v g (3:7 m/s) (9:8 m/s ) :698 m : I goes (:698 m :544 m) :54 m higher. 69P Assume ha i akes ime for he objec o fall. Then from h g we nd s s h g (45 m) 9:8 m/s 3: s : In he mean ime, he boa has moved a disance D m. Thus he speed of he boa is v b D ( m)(3: s) 4: m/s: 7P (a) A he poin where is fuel ges exhaused, he rocke has reached a heigh of y a (4: m/s )(6: s) 7: m : The speed of he rocke a his insan is v a (4: m/s )(6: s) 4: m/s : The addiional heigh y he rocke can aain on op of y is hen given by v gy, which gives y v g (4: m/s) (9:8 m/s ) 9: m : The oal heigh he rocke aains is hus 7: m + 9: m m. (b) The ime of igh afer he fuel ges exhaused is given by y v g ; or 7: m (4: m/s) (9:8 m/s ) : Solve for o obain 7: s. The oal ime of igh is herefore 7: s + 6: s 3: s.

34 44 CHAPTER MOTION ALONG A STRAIGHT LINE 7P (a) Le he heigh reached by he player be y. The iniial speed v of he player is v p gy p (9:8 m/s )(:76 m) 3:86 m/s: As he player reaches y 76: cm 5: cm 6: cm, his speed v saises v v gy, which gives v q v gy p (3:86 m/s) (9:8 m/s )(:6 m) :7 m/s : The ime ha he player spends in he op 5: cm of his jump hen saises v g, or v g (:7 m/s) :35 s 35 ms 9:8 m/s ; where he facor of in he expression for accouns for he ime for him o fall. (b) The ime for he player o jump o a heigh of 5: cm saises (9:8 m/s ) 5: cm v g (3:86 m/s) ; which yields :8 s 8: ms. 7P Le he speeds of he ball a levels L and U be v L and v U, respecively. Then since i akes T U o go from h U o he maximum heigh h max, h max h U v U TU TU : Similarly, h max h L v L TL Taking he dierence of he above wo equaions, we ge vl T L H h U h L v U T U TL g TL : T U : 8 Now subsiue v U g(t U ) and v L g(t L ) ino he expression for H above o obain H g(tl TU )8, or g 8H TL TU :

35 CHAPTER MOTION ALONG A STRAIGHT LINE 45 73P The speed of he ball jus before hiing he ground is given by v p gy, where y is he iniial heigh from which he ball is dropped. The average acceleraion is hen a v p gy The direcion of he acceleraion is upward. p (9:8 m/s )(5: m) : ms 857 m/s : 74P (a) For he ball which sars wih a speed v and reaches a speed of v jus before hiing he ground, we may wrie v v gh, where H is he disance i raveled. Thus v p v + gh: The direcion of he velociy is downward. (b) The ime i akes is given by v v g p v + gh v g : (c) If he ball is hrown upward, he answer o (a) will remain he same, while he answer o (b) will be greaer, wih he new ime-of-igh being + v g p v + gh + v g : 75P (a) The disance D from he lower do o he mark corresponding o a cerain reacion ime is given by D g. Thus for 5: ms D (9:8 m/s )(5: ms) :3 cm : (b) For ms for 3 5 ms for 4 ms and for 4 5 ms D (9:8 m/s )( ms) D 3 (9:8 m/s )(5 ms) D 4 (9:8 m/s )( ms) D 5 (9:8 m/s )(5 ms) 4D ; 9D ; 6D ; 5D :

36 46 CHAPTER MOTION ALONG A STRAIGHT LINE 76P For he rs hrow H g, while for he second hrow H g g( ) 4H. Thus one has o hrow four imes as high o make he ball say in he air wice as long. 77P (a) Use v v + a(x x ): Thus v v + ( g)(x x ) v + ( 9:8 m/s )(3: m) ; which gives v 8:85 m/s. (b) The heigh reached above B is given by y (v) g (8:85 m/s) 8(9:8 m/s ) : m : 78P The average acceleraion during he ime when ball is in conac wih he oor is given by a (v v ), where v is is velociy jus before sriking he oor and v is is velociy jus afer leaving he oor. Take he y axis o upward and place he origin a he poin where he ball is dropped. To nd he velociy jus before hiing he oor, use v v gy, where v and y 4: m. The resul is v p q gy (9:8 m/s )( 4: m) 8:85 m/s : The negaive square roo is used because he ball is raveling downward. To nd he velociy jus afer hiing he oor, use v v g(y y ), wih v, y : m, and y 4: m. The resul is v p g(y y ) The average acceleraion is q (9:8 m/s )( : m + 4: m) 7:67 m/s : a v v 7:67 m/s + 8:85 m/s : 3 s :65 3 m/s :

37 CHAPTER MOTION ALONG A STRAIGHT LINE 47 79P Le he ime of igh of he nh drop be n (n ; 3; 4), hen 3 3 and 3. Since y n g n, we have y 3 y ; which gives y 3 y 9 cm. Also, y y which gives y 4y 9 89 cm ; y fauce y 3 y y 8P (a) Le he heigh of he diving board be h, he deph of he lake be D, and he oal ime for he ball o descend be T. The speed of he ball as i reaches he surface of he lake is hen v p gh. The ime he ball spends in he air is p hg, while he ime i spends descending in he lake is Dv D p gh. Thus which gives T + s h g + D p gh ; D p ght h p ()(9:8 m/s )(5: m)(4:8 s) ()(5: m) 38: m : (b) The average speed is v D + h T 38: m + 5: m 4:8 s 9: m/s : (c) In his case, he ball has o be hrown upward. Following he soluion o par (c) in 74P, we have gt p g + v + v, which yields v gt h + D T (9:8 m/s )(4:8 s) 38:6 m + 5: m 4:8 s 4:5 m/s : 8P Le h be he heigh of he fall and be he ime of he fall. Take he y axis o be upward and place he origin a he poin of release. Then h g : Since i falls half he disance down in he las second, i falls he rs half in ime, where is in seconds. Thus h g( ). Solve hese wo equaions simulaneously for and h. Firs

38 48 CHAPTER MOTION ALONG A STRAIGHT LINE subsiue g for h in h g( ) o obain ( ). A lile manipulaion gives he quadraic equaion 4 +. I has he soluions p +3:4 s and :586 s. You wan he posiive soluion. To nd he heigh subsiue his value for ino h g o obain h (9:8 m/s )(3:4 s) 57 m. The negaive soluion for corresponds o a ball hrown upward, during he ime before i reaches is highes poin. 8P The speed of he woman jus before she lands in he venilaor box is given by v p gh p ()(3 f/s )(44 f) 96 f/s. The acceleraion she experiences as she crashes ono he box is hen a v x (96 f/s) ()(8 in.) 37 f/s ; which is equivalen o [(37 f/s )(3 f/s )]g 96g. 83P (a) The ime of igh for he rs sone is given by y (f) r y g s (44 f) 3: f/s 3: s : Thus he ime of igh for he second sone mus be iniial speed be v, hen or which gives v 4: f/s. (b) y v + g ; 3: f/s 44 f v (: s) + (: s) ; : s : s. Le is 44 nd sone s sone.. 3.

39 CHAPTER MOTION ALONG A STRAIGHT LINE 49 84P (a) The ime during which he parachuis is in free fall is given by h 5 m g (9:8 m/s ), which yields 3: s. The speed of he parachuis jus before he opens he parachue is given by v gh, or v p gh p ()(9:8 m/s )(5 m) 3 m/s : If he nal 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 v a 3 m/s 3: m/s m/s 4 s : The oal ime of igh T is herefore T + 7 s. (b) The disance hrough which he parachuis falls afer he parachue is opened is given by h v v (3 m/s) (3: m/s) 4 m a ()(: m/s : ) Thus he fall begins a a heigh of H 5 m + h 9 m. 85P Take he y axis o be upward and place he origin a he poin from which he objecs are dropped. Suppose he rs objec is dropped a ime. Then is coordinae is given by y g and, since he second objec sars : s laer, is coordinae is given by y g( ), where is in seconds. You wan he ime for which y y m. Solve g( ) + g, wih g 9:8 m/s. Afer wriing ( ) as + and canceling he erms in you should ge (g) + :5 s. 86P Le he verical disances beween Jim's and Clara's fee and he jump-o level be H J and H C, respecively. A he insan his phoo was aken, Clara has fallen for a ime T C, while Jim has fallen for T J. Thus H J gtj and H C gtc. Measuring direcly from he phoo, we ge H J 3:6 m and H C 6:3 m, which yields T J :85 s and T J : s. Thus Jim has waied for T T C T J :3 s: 87P (a) Take he y axis o be upward and place he origin on he ground, under he balloon. Since he package is dropped, is iniial velociy is he same as he velociy of he balloon, + m/s. The iniial coordinae of he package is y 8 m; when i his he ground is coordinae is. Solve y y + v s v g v g + y g g for : s ( m/s) (8 m) + (9:8 m/s ) 9:8 m/s 5:4 s ; m/s 9:8 m/s + where he posiive soluion was used. A negaive value for corresponds o a ime before he package was dropped.

40 5 CHAPTER MOTION ALONG A STRAIGHT LINE (b) Use v v g m/s (9:8 m/s )(5:4 s) 4 m/s. Is speed is 4 m/s. 88P (a) Take he y axis o be upward and place he origin a he poin where he ball is hrown. Take he ime o be a he insan he ball is hrown. Then he velociy of he ball is given by v v g and a he highes poin v. Solve v g for and subsiue he resul ino y y + v g o obain y v g. The iniial velociy is m/s + m/s 3 m/s. The resul is y (3 m/s) (9:8 m/s ) 46 m : I is hen 46 m + 3 m 76 m above he ground. (b) The coordinae of he ball is given by y v g. When he ball is hrown he oor of he elevaor has coordinae y f : m and he elevaor is moving upward wih consan velociy v f. Thus he coordinae of he oor a ime is given by y f y f + v f. You wan o solve for he ime when he coordinae of he ball and elevaor oor are he same: y y f. This means v g y f + v f. The soluion o his quadraic equaion is s (v v f ) (v v f ) g g (3 m/s m/s) 9:8 m/s + y f g s (3 m/s ms) ( : m) (9:8 m/s ) (9:8 m/s ) 4: s ; where he posiive roo was used. The negaive roo corresponds o a ime before he ball is hrown. 89P From he graph, we may wrie v v + g( ) y y v ( ) + g( ) : Subsiue :5 s, y y : m, and g 9:8 m/s ino he equaions above and solve for v and v : v 8:99 m/s, v : m/s. Thus (roof) (op of window) (boom window) and y v g 4: m y y + : m 5:3 m : y 3 (ground)

41 CHAPTER MOTION ALONG A STRAIGHT LINE 5 Also where 3 y 3 y v ( 3 ) + g( 3 ) ; : s : s: Thus he heigh of he building is y 3 5:3 m + (: m/s)(: s) + (9:8 m/s )(: s) :4 m : 9P The ime he po spends going up and going down pas he window of lengh L is :5 s apiece. Assume ha he po passed he boom of he window wih a speed v and i passed he op of he window wih a speed v. Then v v g and Solve for v : v L is herefore v v + v L : g. The disance H he po goes above he op of he window H v g (L g g) [: m:5 s (9:8 m/s )(:5 s)] ()(9:8 m/s ) :34 m : 9 (a) d v i a + T R v i ; (b) 9: m/s ; (c) :66 s 9 (a) d a ; (c) 7: m/s 93 (a) v j a d (j ) + v ; (c) 7: m/s ; (d) 4 m

42 5 CHAPTER MOTION ALONG A STRAIGHT LINE 94 Various mahemaics packages on a personal compuer can be used o solve he problems in he Elecronic Compuaion secion. As an example his problem is solved using he "Symbolic Calculaion" feaure on Mahcad. Load he symbolic processor afer you open up he Mahcad program. The following is an acual compuer prinou of a Mahcad documen desinged o solve his problem. (a) To obain an expression for he velociy v as a fuincion of ime, wrie down he expression for x() firs: x( ) ( e ) Use Mahcad o evaluae symbolically he derivaive of x() wih respec o velociy: v() d d ( e ) wih he resul v( ) exp( 3... ).7.. exp( 3... ) To obain he acceleraion as a funcion of ime, ake he derivaive of v( a() d d which yields ( exp( 3... ).7.. exp( 3... ) ) a( ) 48.. exp( 3... ).88.. exp( 3... ).6... exp( 3... ) (b) Now plo x, v and a as funcions of ime :, x( )

43 CHAPTER MOTION ALONG A STRAIGHT LINE 53, v( ) 4 6 8, a( ) (c) To find he ime when x(), use he "roo" feaure in Mahcad. Le he ime be when x reaches zero. To find, firs give i an iniial "guess" value, say. (as you can see clearly from he x vs curve ploed above). Then he value can be of found as follows:

44 54 CHAPTER MOTION ALONG A STRAIGHT LINE roo e,.75 So he ime is.75 sec when x. Once his is known, he velociy v and acc a ha momen are easily obained: v(.75 ) a(.75 ) (d) Similar o par (c) above, we firs find when he ime v. From he graph of v vs w give an inial guess of 65.Then 65 roo exp exp 3..., The acual ime is hen 66.7 sec. The corresponding values for x and a are x( ).44 4 a( ) (As anoher useful feaure, Mahcad does allow he quaniies o carry heir un calculaion. I can keep rack and figure ou he appropirae unis. In his proble are suppressed for convenience. The suppressed unis for x, v, a and are m, m sec, respecively.)