College physics 10th edition serway solutions manual by qiqi111

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Instructor Solutions Manual

College P hys ics !

TENTH EDITIO N

! ! ! !

RAY MO ND A . S ERWAY Emeritus , J ames M adis on U nivers ity

CH RIS V UIL L E Embry-Riddle A eronautical U nivers ity

! ! ! ! ! ! ! !

P repared by

! Vahe Pe roomian & J ohn G ordon ! 1 !

! 1 In troduction ANSWERS T O WARM- UP EX ERCISE S 1. ( a) T h e n u mb er g i v en , 5 6 8 017 , h a s s i x s ign if i c ant f i g u re s , wh i ch we wi l l r et ain i n co nv ert i n g th e nu mbe r t o s c i e nti f i c n o t a5

tion. Moving th e decimal five spaces to th e l eft gi ves us the an swer, 5.680 17 × 10 . ( b ) T h e n u m b er g i v en , 0 .0 0 0 3 09 , h a s t h r ee s ig ni f i c an t f ig u re s , wh i c h we wi ll r et a in i n co n v er tin g th e nu mb e r t o s c i ent ifi c –4

notation. Moving th e deci mal four spaces to th e righ t gives us th e answer, 3.09 × 10 . 2. W e f i rst c o ll e ct t er m s , t h en s i mp l i fy : [ M ][L [T ] [ M ][ L ]2 [T ]2 [ M ][L ] . [T ] = = ]2 [T ]3

[L ]

[T ]3 [L ]

[T ]

As we wi l l s e e i n C h apt e r 6 , t h ese a r e t h e u ni ts f o r mo m en t u m. 3. E x amin in g t h e ex p r e s sio n s hows t h a t th e un it s o f me t e rs a n d s e c o nd s s qu ar e d ( s2 ) appear in both the nu merat o r and the de- nominato r, and therefore cancel out. We comb ine th e numbers and uni ts separat el y , squari ng the last term before doing so: §

m · § 1.0 0 km · § 6 0.0 s · 7.00 2 ¨ s © 1. 00 × 1 0 3 ¨© 1.00 mi n m § 1.00 · § 3600 · § m · § km · § s2 · = (7 .00 ) ¨© 1.00 ¨ 2 ¸ m ¨ mi n 2 ¸ × 10 3 1. s 00 ©¹©¹ km = 25 .2 mi n 2

4. T h e r eq u i re d co nv e r s ion c an b e c ar r i e d o ut in o n e st ep: § 1. 00 cu bi t us· h = (2. 00 m ) ¨ ¸ = 4. 49 cu bi t i © 0.445 m ¹

5 T h e a re a o f t h e h o u se i n s q u are f e et ( 1 4 2 0 f t 2 ) contains 3 signif icant figu res. Ou r answer will th eref ore al so cont ai n three 2 signi fi can t fi gures. Al so not e th at th e conversi on from feet to met ers is squa re d t o accou nt un it s i n wh i ch th e a for th e ft rea is o rig in ally giv e n .

()

2 A = 1 4 20 ¨ ¸ ft

§ 1. 00 m · = 13 1. 909 m 2 = 132 m 2 © 3.281 ft ¹

6. Usi ng a c al c ula t o r to mu l ti p ly t h e l eng th by t h e wi dt h gi v es a r a w a n s wer o f 6 7 8 3 m 2. This answer mu st be ro unded to contain th e same number of signif icant figures as the least accurate factor in the product. The least accurate fact or is th e l eng t h, wh i ch c o nt ain s 2 s i gn if i c ant f ig u re s , s i n ce t he t r a i li ng z er o is n ot s i gni f i c ant ( s e e S ec t i on 1.6 ). Th e co r r ec t an s wer f o r t he 3

are a o f t h e ai rst r ip i s 6 . 8 m2 . 0 × 10 7. Ad di ng th e t h re e n u m b er s wi th a c al c ul a to r g i v es 2 1 . 4 + 1 5 + 17 . 1 7 + 4 .0 03 = 57 .5 7 3 . Ho wev e r , t h i s a n swer mu s t b e rounded to cont ain the same number of signifi cant figures as t h e least accurate number i n t h e su m, wh i ch is 15, with two sig ni fi can t fi g ures. Th e co rrec t an s wer i s t h ere fo re 5 8 . 8. T h e gi v en C a rt esi an c o o rdi n at es a re x = – 5 .0 0 a n d y = 1 2 .00 . T h e l ea s t a cc u r at e o f th e s e c o o rd in at e s c on t ain s 3 s ig ni f i ca n t figu res, so we will express our answer in t h ree si gn if icant figu res. The speci fi ed poin t , (– 5.00, 12.00 ), is in the seco nd quad - rant since x < 0 a n d y > 0 . T o f ind t h e po l ar co o rd in at e s ( r, θ ) of t his po int, we use r = x 2+ y 2=

(5 . 00 ) 2 + (1 2.0 0) 2 = 1 3.0

and

θ = tan − 1

§y· § 12 .00 · = t an −1 = –67.3° x –5. 00

Si n ce t h e poin t i s i n th e s ec o nd q u ad r a nt , we a d d 1 80 ° t o th i s a ng l e to ob t ai n θ = − 67 .3° + 18 0° = 1 13 ° . T h e p o l a r c o o rd i - nat es of the poin t are therefore (13.0, 113°) . 9. R e f er t o ANS . FI G 9 . Th e h ei g ht of t h e t re e is d e s c ri b ed by t h e t ang e nt of th e 2 6° a n gl e , o r tan 2 6° =

h 45 m

f r o m wh i ch we o b t ain h = ( 45 m ) t an 26 ° = 22 m

ANS . FIG 9

ANSWERS T O EVEN NUMB E RE D CONCEPTUAL Q UE ST IONS 2. At o mi c cl o ck s a r e b as ed o n t he e l ec t r o ma gn et i c wa v e s t h at a t o ms e mi t . Al s o , p ul s ars are h i gh ly re g ul a r a s t r on o mi cal cl o cks . 4. (a)

(b)

(c) 6. Let us assume the atoms are solid sphere s of diameter 10 − 10 m. T h e n , t h e vo lu me o f e ac h a t o m i s o f t h e o r d er o , the numb er of at oms in th e 1 cm 3 s o l id i s o f 10 − 30 m3 . (More precisely, volume = .) Therefore, si nce n the order of at o ms . A mo r e p rec i s e c a l c ul at io n wou ld re q uire k n o wl edg e o f th e d en s it y o f th e s ol id a nd t h e ma s s o f e ac h a t o m . Ho wev e r , o u r e s ti m at e ag re es wi t h t h e mo re prec ise ca lc ulation to wi th in a factor of 10. 8. Realisti cal ly , the only lengths you mi ght be ab l e to verify are the length of a football fiel d and the lengt h of a housefly. Th e only time in terv als subject to veri fi cation would be th e length of a day and the ti me between normal heart b eats. 10. I n t h e me t r i c s y st em, u n i t s dif f er b y po w e r s o f t en , s o i t ’ s v ery easy and accurate to conv ert from one unit to anot her. 12. Bo t h a n s we rs ( d ) a n d ( e ) c o u ld b e p hy si c al ly me an i n gfu l . An s we rs ( a ) , ( b ) , an d ( c) mu s t b e m ea n i n gl e ss s in c e q u ant i ti es c a n be added or subtracted on ly if they have the same dimensions. ANSWERS T O EVEN NUMB E RE D PROBLEMS 2 . (a)

(b ) L

4 . Al l th r e e eq u at i on s are d i m ens io n all y in co r re c t .

6 . (a)

(b ) Ft = p

8 . (a) 22 .6 (b ) 22 .7 (c) 2 2 . 6 is mo r e r el i ab l e 10. (a) 12. (a)

(b )

(c)

(b )

14. (a) 797 (b ) 1 . 1 (c) 1 7 . 66 16.

18. (a) (b)

(b)

(c)

34. (a) ~

(b)

(c) The very l arge mass of prokaryotes implies they are important to th e b i o sph e re . T h ey a r e r es p o n si bl e f o r f ix i ng c a r - bon, produ cing oxygen, and breaking up po llutants, among many other biol ogical roles. Humans depend on them! 36. 2 .2 m 38. 8 .1 c m 40. 42. 2 .3 3 m 44. (a) 1 .5 0 m (b) 2 . 6 0 m 46. 8 .6 0 m 48. (a) a n d (b)

(c)

(d)

50. 52. (a)

(b )

(c)

54. As s u me s p o pula t io n o f 3 00 m i l li o n , av e r ag e o f 1 ca n/ we ek p e r p e rso n , an d 0 . 5 o z p e r c a n . (a)~

(b ) ~

56. (a)

(b )

58. (a)

(b )

60. (a) 500 y r (b ) 6 .6 Ă&#x2014; 1 0 4 t i me s

62. ~ . Assu mes 1 l o st b a ll p er h itte r, 1 0 h itt e rs p e r i n ni ng , 9 i nni n g s p e r g a me , a n d 81 g a me s p e r y e a r.

PRO BL E M S O L UT IO N S

a n d re c o g ni z ing t h at 2Ď&#x20AC; i s a d i me n s io n l

1.1 Su b s titu t in g dime n sio n s i nt o t h e g iv en eq u ati o n

es s c o n st an t , we h av e

Thus, t h e .

1.2 (a) From x = Bt 2 , we fi nd th at

(b) If

. Thus, B h a s u nits o f

, then But the sine of an angl e i s a d imensio nl ess rat io. Th e refo re,

1.3 ( a) T h e un it s of v ol u m e, are a , a n d h ei g ht a r e: ,

, and

We t h en ob se rv e t h at

Th u s , t h e e q u ati on is (b)

wh ere wh ere

1.4 (a) In the equation ,

wh il e

(b)In,

. Thus, the equati on is .

b ut

. H en c e , t h i s e qu at io n i s .

, wh il e

. T h e r ef o r e, t h i s equ at i o n

1.5 Fro m t h e u niv ers al g r av i t at ion l a w , t h e c o nsta n t G i s

. Its units are then

1.6 (a) So l vin g f o r t h e m o me n t u m, p , gives wh e re t h e n u me r al 2 i s a d i me n si onle s s co n st an t . Di - mensional analysis gives th e un its of mo mentum as:

T h e r efo r e , i n t h e SI s y s t e m, t h e u nit s o f mome n t u m a r e (b ) No t e th at t h e un it s of f o rc e are

. Then, observe that

F ro m t h i s , i t f ol lo w s th a t f o rc e mu l t ipl i ed by t i m e i s p r o po rt i on al t o m o me n tu m: . ( Se e t he i mp u ls eâ&#x20AC;&#x201C; m ome n t u m th eo rem i n C hap t er 6 , , wh ich says th at a constant force F m u l ti p li ed by a d u r a ti on o f ti me â&#x2C6;&#x2020;t e q u al s t h e change in mo mentum, â&#x2C6;&#x2020;p .)

1.7 1.8 (a) Comp uti ng

w i t hou t rou nd i ng t h e i nt e r me d i at e r e s

u lt y ie ld s

t o t h r ee s ig nif ic a n t f ig u r e s .

( b ) R o und ing t h e i n t er m edi a t e re s u lt to t h r e e s ign i fi ca nt f i gu r es y i e ld s

T h en , we o b t ain (c) 1.9 (a) h a s

t o t h re e sig ni fi can t fi g u res. b e c au s e ro un d i ng i n p a r t (b ) wa s ca r ri ed o ut t o o so on . wi th t h e un c ert ai nt y i n th e t e n th s po si ti on . (

b)has (c) h as (d)has

. The two ze ros we re origin al ly in cl ud ed only to position th e deci mal.

1.10 (a) Ro und e d to 3 si gni fi c an t fig u re s: (b ) Ro und e d to 5 si gni fi c an t fig u re s: (c) Ro und e d to 7 si gni fi c an t fig u re s: 1.11 O b s e rv e t h at t he l e n gth t h e w i dt h , an d t h e he i ght a l l con t ai n 3 si gni fi c an t f i gu re s . Thus, any p rodu ct of these quan tities should contain 3 signif icant figures .

(a) (b) (c)

( d ) I n t h e r o un ding p r o ce s s , s ma l l amo u nt s are e i th e r a d d ed t o or subt rac t ed from a n a nswe r to s at i s f y t h e r u l es o f s i gni f y c an t f ig u r e s . F o r a g i v en ro u n d ing , d i f f ere nt s ma l l ad j u st m en t s a r e ma d e, i n t ro du ci ng a ce r t a in a mo u n t o f ra n do m n es s i n th e l as t s ign if i c an t d i gi t of th e f i n al a n s w er. 1.12 (a) R eco gni z e t h at t h e l ast t e r m i n t h e b r ack e t s is i n si g ni f i can t i n co m p ari son t o th e ot h e r t wo . T h us,wehave

(b) 1.13 T h e l e ast ac cu r a t e d i m en sion o f th e b o x h a s t wo s i gn if i c ant f i gu r e s . T h u s , t h e vol u m e ( p ro du ct o f th e th r ee d i me n si on s ) wi l l co n t ai n onl y t wo si gni fi c ant f igu res .

1.14 (a) The sum is rounded to (b)

b e c aus e 7 56 i n t h e t erms t o b e ad d ed h as no p o sit i o n s b eyon d th e d ec i ma l . mu s t b e ro u n ded t o

b e c au s e h as on ly t wo s ig nifi c

ant f igu re s. (c) m u s t b e r o u nde d t o

b e c au s e 5 .6 20 h a s on l y f ou r s i gni f i c a nt f i gu r e s .

1.15 T h e a n s w er i s l i mi t ed to on e s i gni f i c an t f ig u re b e c au s e o f the accuracy to wh ich the conv ersion from fathoms to feet is given.

1.16 g ivi ng

1.17

1.18 (a)

)

(c)

)

1 0

I n ( a ) , t h e an s wer i s l i mi t ed to t h r ee s ig nif i ca n t fi g u re s b ecause of the accu racy of th e orig inal data value, 348 miles. In ( b), (c), and (d), the answers a re l i mi ted to four si gni fican t figures b e c au s e of th e a c cu r a cy to w h i ch t h e kilom et er s - t o - f ee t co n version factor is given.

1 1

1.19

. 1.20

1.21 (a)

(b) (c)

1.22

T hi s me an s t h at t h e p ro t ein s a r e as s em b l e d a t a ra t e o f ma n y l ay e r s o f at oms ea c h s e c o n d ! 1.23

1.24

1.25

1.26

1.27

(W h e re L = l en g th o f on e s i de o f t h e cub e .)

Thus, an d 1.28 W e est i ma t e t hat t h e len gt h o f a s t e p f o r an a v era g e p er s on is a b ou t 18 i n che s , o r ro u gh l y 0 . 5 m. T h en , an e s t i m at e f o r t h e nu mb er o f s t ep s re q u i r ed t o t r a vel a d i st a n ce equ al t o th e c i r cu m f er e n ce o f t he E a r t h wo ul d b e

1 2

or 1.29 W e ass u me a n a v e ra g e r e s p i r at ion ra t e o f abo u t 10 b r ea th s /mi n u te and a typi cal li fe span of 70 years. Then, an esti mate of the nu mber of breath s an average person would take i n a lifet ime i s

or 1.30 W e ass u me t h at t h e a v e rag e pe r s o n ca t ch es a c o l d t wi c e a y e ar an d i s s i c k an av e r ag e o f 7 d ay s ( o r 1 wee k ) e ac h t i me . T h u s , on averag e, each person is sick for 2 weeks out of each year ( 52 weeks). The prob ability th at a particul ar person will be sick at a n y giv e n t i me e q u a l s th e p e rc e nt a g e of ti me t h at p e rso n i s s i ck , o r

T h e po pul ati o n o f t h e E a rt h i s a p p ro xi m at el y 7 b ill i on . T h e n u mb er o f p eo p l e e xp ect ed t o h a v e a co l d on an y g iv en d ay i s then

1.31 ( a) A s s u m e t h a t a t ypi c a l i nt e st in al t r a ct h a s a l en g th o f ab out 7 m and average diameter of 4 cm. The estimat ed total intestinal volume is then

T h e a pp r o xi m at e v olu me o cc up i ed by a s i ng le b a c t e ri u m i s

I f i t i s as s u me d t h at b a ct eri a o cc u py o n e h und r edt h of t h e t ot al int es t in al vo lu me , t h e est ima t e o f t h e numb er o f mi c r o o r g an - isms i n th e human i n testin al tract is

( b ) T h e l ar g e v alue o f t h e nu m b er o f b a ct eri a e st i ma ted to e xi st i n th e i nt es tina l t r ac t me an s th a t th ey a r e p r o b abl y not dangerous. Intestinal bacteria help dig est food and provide i mportant nutrients. Hu mans and bact eria enjoy a mu tually bene- fici al symbioti c relationship.

1.32 (a) ( b ) C o n sid e r y ou r b ody t o b e a cyl ind e r h a vi ng a r ad i u s of a b out 6 i n ch es ( o r 0 . 1 5 m ) a n d a h e i ght o f abou t 1 . 5 m et er s . Then , its volume is

( c) T h e e s ti ma t e of t h e nu mb e r o f ce l l s in th e b o dy i s th en

1.33 A r e aso n ab l e gu ess f o r t h e d i ame t e r o f a t i r e mi g ht b e 3 f t , wi th a c i r cu m f er e n ce ( tan ce t ra v el s p er re v o -

=dis

luti on ) of about 9 ft . Thus, the total number of revolutions the tire might ma ke is

1.34 A n s w ers t o t his p r o bl e m w i l l v a ry , d e p en d ent o n th e ass u mp ti o ns o n e ma kes . T h i s s o lut i on a s s u m es t h at b ac t e ri a a n d ot h er prok aryotes occu py approx i mately one ten-millionth (10 â&#x2C6;&#x2019;7) of the Earthâ&#x20AC;&#x2122;s volume, and that th e densi ty of a prokary ot e, l i ke the density of the human body, i s approximat ely equ al to that of wat er (103 kg/m3).

(a)

) ( c) T h e v e r y l ar g e m as s o f p r o k a ry ot e s i mpli es th e y a r e i mportant to th e bi osph ere. They are respon si bl e for fi xing ca r- bon, producing oxygen, and breaking up po llu t ant s, among many other biol ogi cal rol es. Hu mans depend on th em! 1.35 T h e x coo r d in at e i s f oun d a s an d th e y c oo r d in at e 1.36 T h e x di s t an c e o ut to t h e f ly i s 2 .0 m a n d t h e y d i s t an ce u p to t h e f ly i s 1 . 0 m . T h u s , we c a n u s e t h e Py t h ag o r e an t h eo r em t o find the di st ance from the origin to the fly as

1.37 T h e di s t an c e f ro m t h e o r ig i n to t h e f ly i s r i n p ol ar c o o r din ate s , a n d t hi s was f o und t o b e 2 .2 m i n Pro b l em 3 6 . T h e a n g le Î¸ is the angl e between r and the ho rizo nt al reference line (the x axis in this case). Thus, th e an gl e can be found as

a nd T h e po l a r c o o rd in at e s a r e 1.38 T h e x di s t an c e b e t w een t h e t w o p oin t s i s a n d t h e y d i s t an ce b et we en t h e m is . T h e d i s t an c e b e t ween t h e m is f o und f r o m t h e Py t h ag o re a n t h eo r em :

1.39 R e f er t o t h e Fi g u re gi v en i n P ro bl em 1 . 4 0 b el o w. T h e C a rt es i an c o o rd in at es f o r t h e t wo gi v en poi n t s a r e:

Th e di st an c e be t we en th e t wo p o i nt s i s th en :

1.40 Co n s id er t h e Fi gu r e s h o wn at t h e ri g ht . T h e C ar t e si an co o rd i n at e s f o r th e t wo po in t s a r e:

Th e di st an c e be t we en th e t wo p o i nt s i s th e l en gth o f t h e hy po t enu se o f th e sh ad e d t ri angl e an d i s gi v en by

Ap pl yin g th e id e nti ti es

and

, this red u ces to

1.41 (a) W ith a n d b b ei ng t wo si d es of t hi s ri ght t r i an g l e h av ing h yp ot en us e agorean theorem g i v es th e u n kno wn s i d e a s

(b)

(c)

1.42 Fro m t h e d i ag r a m, Thus,

, t h e Py th

1.43 Th e c i rcu mf eren ce o f th e f ount a in i s

Thus,

1.44 (a) )

wh ich giv es

so, ( b so,

1.45 ( a) T h e s id e o ppo s i t e θ =

(c)

, so the radius is

( b ) T h e s i de a d j ac en t to φ =

(d)

(e)

1.46 U s i ng th e d i agr a m at t h e r i gh t, t h e P y t h a go r ea n t h eo re m y i el ds 1.47 Fro m t h e d i ag r a m g i v e n i n Pro b l em 1 . 4 6 abo v e , i t is s e e n th

at and 1.48 ( a) a n d (b ) Se e t h e Fi g ur e g i v en at t h e ri gh t . (c) Ap pl y in g t h e de fi ni ti o n of t h e t a n g ent fun ct io n to th e l arg e ri gh t t ri angl e co nt ai nin g th e 12 .0 ° a ng l e giv es:

[1] Al s o , ap pl y ing t h e d ef i nit i on of t h e t ang e nt fu n cti o n t o t h e s ma l l e r r i g ht t r ia n gl e con t ain

i ng t h e 14 . 0 ° a ngl e g

i v es : [ 2 ]

( d ) Fro m E q u a ti o n [ 1 ] a b o v e , o bse r v e t h at Su b s ti tu tin g thi s r e s u l t int o E qu a tio n [ 2 ] g iv es

T h en , s o l vin g f o r t h e h ei ght o f t h e m o u nt ain , y , y i el d s

1.49 Using th e sketch at the right: 1.50 T h e f igu r e a t t h e r i gh t s ho ws t h e si tu at ion d es cr i b ed i n t h e probl e m st at ement . Ap pl yin g th e def ini tio n of th e t a n g ent f un ct io n to th e l a rg e ri gh t t ri angl e c o nt ai ni n g th e an gl e Î¸ i n th e Fi g u r e , o n e ob t ai n s

[1] Al so , a p ply i ng t h e d efi n it ion of t h e t ang en t fun c tio n to th e s ma l l ri gh t t ri ang l e co nt ai nin g t h e a ngl e Ď&#x2020; g i ve s

[2 ] So l vin g Eq u at io n [1 ] f o r x a n d s u bsti t uti ng th e re s u lt i n to E q u ati on [ 2 ] y i el ds

T h e l a st r es u l t s i mpl if i es t o So l vin g fo r y :

1.51 ( a) Gi v en t h at

, we h a v e . Th e re f o re, t h e u nit s o f f o rc e are t h o se o f ma,

(b)

1.52 (a) (

(c)

1.53 (a) Since , t h en

, g i ving

As a ro u g h c al cu l ati on , t r e at e ac h o f th e f oll o wi n g ob j ec ts a s i f t h ey

we r e 1 00 % wa t e r. ( b ) c e l l:

(d) f ly:

1.54 Assu me an av era g e o f 1 ca n p er p erso n ea c h we ek a n d a p op ul at ion o f 300 mi l l ion .

(a)

(b)

As s u m es a n a v e ra g e we i g ht of 0 . 5 o z o f a lumi n u m p er ca n .

1.55 The term s h a s d i me nsion s o f L , a h a s d i me n s io n s of LT â&#x2C6;&#x2019;2, and t h a s di me n si o n s o f T . T here f o r e , t h e e q u a tio n , with k b e i ng dime n s io n l es s , ha s d i me n si on s o f or T h e po we rs o f L a n d T mu s t be t h e s a me o n ea c h s i d e o f t he eq u a ti o n . T he re f o re , L 1 = L m a n d

L ik e wi s e , eq u at i ng p o we r s o f T , we s e e t h a t , or

, a d i me n s i on l es s c o n st an t .

1.56 ( a) T h e r a t e o f f i l lin g in g al lon s pe r s e co n d i s

(b) Note that

Thus, (c)

1.57 T h e vo lu m e o f p a int u s ed i s g iv e n by V = A h, wh ere A i s t h e are a c o v er e d a n d h is t h e thi ck n es s o f th e l ay e r . T h u s ,

1.58 (a) For a sphere, .

. In t hi s c a s e, t h e r ad i u s of t h e s e c on d s ph er e i s t w i c e th at o f t h e f i rst , o r

He n c e, (b) For a sphere, the volume is

T hu s,

1.59 T h e e s ti m at e of t h e tot a l di st an c e ca r s a r e d r i v e n e ac h y ea r i s

At a ra t e o f 2 0 m i / g al , t h e f u el u s e d p e r y e a r wo ul d b e

I f t h e r a t e i n c r ea s e d t o

, the annual fuel consumpt ion wo uld be

an d th e fu e l s av ing s ea c h y e ar wo ul d b e

1.60 (a) The t i me i nt e rva l requi re d t o re pa y t he de bt wi ll be ca l c ul a ted b y di vi di ng t he t ot al de bt b y t he ra t e a t w hi c h i t i s re - pa i d.

(b) The numbe r of ti mes $17 tril lion in bills enci rcles the Earth i s gi ven by 17 t rillion times th e lengt h of on e dol lar bill di vi de d by t he c i rc umfere nce o f t he Ea rt Eh (C = 2Ę&#x152; R ).

(17 × 1 0 ) ( 0.15 5 m ) 12

N = = = 6.6 × 10 4 times 2π RE 2π 6.3 78 × 10 6 m

(

)

1.61 (a) (b) Consid er a segmen t of the surface of th e Moon which has an area of 1 m2 and a depth of 1 m. When filled with meteorit es, each having a diameter 10 − 6 m, t h e n u mb e r o f m et e o r it e s al ong e a ch ed g e of th i s b ox i s

The total number of meteorites in the filled box is t h en

At t h e ra t e o f 1 met eo ri t e p er sec on d , t h e ti me t o fi ll t h e box i s

1.62 W e wi ll a s s u me t h at , o n a ve ra g e , 1 b a l l wil l b e l os t p er h itt er, t h a t th e re wi ll b e a bo ut 10 h i tt e r s p e r i n n in g , a g a me h a s 9 inni ngs, and th e team plays 81 ho me games per season. Ou r esti mate of the number of game balls need ed per season i s t h en

1.63 Th e vo lu me o f t h e M il k y W ay g a l a xy i s ro ugh l y

I f , wi t hin t h e Mi lky W a y g ala x y , t h ere i s typ i call y on e n eut r on s t a r i n a s p h e ri c al vol u me o f r a d i us , t h en the gal acti c vo lu me per neut ron star i s

T h e o r d er o f ma g nit ud e of t h e n u mb e r o f n eut r on s t a r s i n t he M il k y W ay is t h en

2 Motion in One Dimension QUICK QUIZZES 1. (a) 2 00 yd (b ) 0 (c) 0 2. ( a ) F a l s e . T h e c a r m a y b e s l o w i n g d o wn , s o t h at t h e d i r e ct i o n o f i t s a c c el e r a t i o n i s opposite the direct ion of its velocity. (b) True. If the velocity is in th e direct ion chosen as negative, a posi tive acceleration causes a decrease in speed. (c) True. For an ac celerating particle to st op at all, the velocity and acceleration m u s t h av e o p p o si t e signs, so that the speed is decreasing. If this is the case, t h e p a r t i cl e w i l l e v e n t u al ly come to rest. If the acceleration remains constant, howev er, the particle mu st begin to move again, oppo site to the direct ion of its origin al velocity. If the particle comes to rest and then stays at re st, the acceleration has b ecome zero at the moment the motion st ops. Th is is the case for a braking car—the acceleration is negative and goes to zero as the car comes to rest. 3. The velocity-vs.-time graph (a) has a constant sl ope, indicating a constant acceleration, which is re presented by the acceleration -v s.-time grap h (e). Gra ph (b) re pre se nt s a n obj e ct whose spe ed a l wa ys i nc re a se s, and does so at an ever incr easing rate. Thus, the accelerat ion must be increasing, and the acceler ation-vs.-time graph that best indicates th is behavior is (d). G r a p h ( c ) d e p i c t s a n o b j e ct w h i ch f i r s t h as a veloci ty that increases at a co nstant rate, which means that the object ’s acceleration is constant. The motion then changes to one at constant sp eed, i ndicating that the acceleration of the object beco mes zero. Thus, the best match to th is si tu at ion i s g raph (f). 4. Ch oi ce (b ). Ac co rdi ng to graph b , there are some inst ants in time when the object is si mu lt an eou sly at two d ifferent x- coordinates. This is physical ly impossibl e. 5. (a) Th e bl u e g raph o f Fi gu re 2 .1 4 b b es t s ho ws th e p u ck ’s p os iti on a s a f u n cti on o f ti me. As s e e n i n Fi gu re 2.1 4 a , t h e d is - tance the puck has travel ed grows at an increasing rate for approximatel y th re e t i me i nt e rv a ls , g ro ws at a steady rate for about four time interv als, and then grows at a dimi ni shing rate for the last two in terv als. (b ) Th e red grap h o f Fi gu re 2 .1 4 c b e s t il lu st rat e s t h e s p e ed (d i stan c e t ra v el e d per t i me i n t e rv al) o f t h e p u ck a s a f u n c tio n of time. It shows the pu ck gaining speed fo r approx imately three time intervals, moving at cons tant sp eed for about four time interv al s, then slowing to rest during the last two interv als. (c) Th e green grap h of Figure 2.14d best shows th e puck’s acceleration as a function of time. The puck gains velocity (positive acceleration) for approximately three time intervals, moves at constant ve loci ty (zero acceler ation) for about four time intervals, a nd then loses velocity (negative acceleration) f o r r o u g h l y t h e l as t t wo t i m e i n t e r v als . 6. Choice (e). The acceleration of the ball remains constant while it is in the air. The magnitude of its acceleration is the free- fall acceleration, g = 9 .8 0 m/ s2. 7. Choice (c). As it trav els upward, its speed decreases by 9.80 m/s during each s econd of its motion. When it reaches the peak of its motion, it s speed becomes zero. As the ball moves downwar d, its speed increases by 9.80 m/s each second. 8. Ch oi ces (a) an d (f ). Th e f i rst j u mp er wi ll a lway s b e mo v ing wi th a h i gh e r v e l o cit y t h an th e s e c on d . T h us , i n a g i v en t ime interv al, th e first ju mp er co v ers mo re d istan ce th an th e s e c on d , a n d th e s e parat io n di st an ce b e t we e n th em increases. At an y given instant of time, the velocities of the jumpers are definitely different, because one

had a head start. In a time interv al after this instant, however, each jumper increases his or her ve locity by the same a mount, b ecause they have the same acceleration. Thus, the diffe ren ce in v elo ci ti es st a y s t h e sa me .

ANSW ERS TO WARM-UP EXERCISES 1. Fo r a q u a d ra ti c e q u a tio n in the f o rm o f ax 2 + bx + c = t h e q u ad ra ti c f o rmu l a gives the answer for x as 0,

−b ± b 2 − 4 ac 2a

Th e qu ad rati c eq u ati on giv en h as a = 2 .0 0 , b = –6 .0 0 , a n d c = – 9 .00 . Su b s ti tut ing i nto t he q u a d rat i c fo rmu l a g iv e s t wo s o lu - tion s for t: 6.00 ±

( −6.00 ) 9.00 )

− 4 ( 2.00 ) ( –

t== 2 ( 2.00

6.00 ± 108 4.00

)

wh i ch g iv e s t = −1.10 or t = 4 . 1 0 . 2. (a) So l vin g th e fi rst equ atio n fo r t h e ti me t (as s u mi n g SI u n it s ): −9.8t + 4 9 = 0

→ − 9.8t = − 4 9

→ t = 5.0 s

(b ) Su b stitu tin g the v al u e of t f rom a b o v e i nto th e e q u ati on fo r x: 2

x = −4.9t 2 + 49t + 1 6 = − 4.9 ( 5.00 s ) + 4 9 ( 5.00 s ) + 16 = 13 8 . 5 m = 14 0 m 3. (a) Set t i ng th e t wo e qu a tio ns f o r x e q u a l to on e a n oth e r, we o bt ai n 3.00t 2 = 24.0t + 72.0 R e a rran gin g giv e s u s a q u ad ra t i c e qu ati on , 3.00t 2 − 2 4.0t − 7 2.0 = 0 Di vi din g out a f a c to r o f 3 .00 , t 2 − 8.00t − 24 .0 = 0 Su b stitu tin g into t h e qu ad rati c fo rmu l a g iv es 8.00 ±

( −8.00 ) 2 − 4 (1.00 ) ( – 24.00 ) t== 2 (1 .0 0

)

8.00 ± 160 2 .0 0

= 10 .3 s o r − 2.32 s W he re we ha ve c hosen t h e posi tive root for the time t . (b ) Su b stitu tin g the v al u e of t f rom a b o v e i nto th e f i rst e q u ati on f o r x:

x = 3.00t 2 = 3.00 (10.3 s ) = 3.20 Ă&#x2014; 10 3 m W he re t he a nswe r ha s be e n e xpre ssed in three significant figures.

4. (a) The football player covers a t o t al of 1 50 y a rd s in 1 8 .0 s . Hi s a v e ra g e s p e ed i s pat h leng t h 1 50 yd averag e s peed = = = 8. 33 yd / s elaps ed ti me 18 .0 s (b) The football pl ayer’s average veloci ty is h i s t o t al d i sp l a c e me n t d i v i d ed b y t h e elap sed time. His di sp l a c em e nt a t t he e n d of th e ru n is 5 0 .0 y a rd s , s in c e h e h a s ret urn e d to t h e fi fty -y a rd li n e .

υ=

∆x 50.0 yd = = 2.78 yd/ s ∆t 18.0 s

5. At g ro und l e v el , t h e d is pl a c emen t o f th e ro ck f ro m i t s l au nc h p oin t is ∆y = − h, where h i s t h e h e ig ht of t h e to wer a n d u p - ward h a s b e e n c h ose n a s t h e po si tiv e d i rect ion . F ro m E q u a t io n 2 .10 ,

υ 2 = υ 02 + 2 a ∆y we o b t ai n (wi t h a = –g ),

υ = ± υ 2 +0 2 ( − g ) ( − h ) 2

(

= ± (12.0 m/ s ) + 2 −9 .8 0 m/ s 2

) ( −4

= 3 0.5 m/ s

0.0 m )

6. Once the arrow has left th e bow, it has a constant do wnward acceleration equal to the free-fall acceleration, g . T a k i ng u p ward as the positive direction, t h e e l a p s ed t i me req u i red for t he velocity to change from an initial value of 15.0 m/ s upward ( υ 0 = +15.0 m/ s ) to a value of 8.00 m/ s do wnward (υ f = −8. 00 m/ s ) i s g i v e n b y ∆t =

∆υ a

υf − υ 0 == −g

− 8.00 m/ s − ( + 15.0 = 2.35 s m/s ) −9.80 m/ s 2

7. We set the i nit ial positi on of the blue ba ll , at a h ei ght o f 10 .0 m, as t h e o ri g in and take upward as the positive di rection. The initial posi tion of the bl ue ball is then 0, and the init ial po si t i o n of t h e red b a l l i s 1 0 .0 m – 6 .0 0 m = 4 .0 0 m b e l o w t h e bl ue ball, or at y = – 4 .0 0 m. At t h e i ns t ant t h at th e b lu e b al l c at c hes u p to t h e red b a l l , t h e y coordinate of both balls will be equal. The displacement of the red ball is given by ∆y red = y − y 0 , red = υ 0 , red t +

12 at 2

Si n c e t h e init i al v e lo cit y of t he red b a l l i s z e ro , y=y− red 2

gt 2 red 0 ,

The displacement of the blue ball is given by 1 ∆y blue = y blue − y 0 , blue = υ 0 , blue t + 2 a t 2 or

y=υt− blue 2

gt 2 blue 0 ,

Set t ing t h e t wo eq u atio ns eq ual an d su b sti tuti ng a = –g gi v es

12 1 y0, −red gt 2 = υ 0, blue t − 2 g t 2 Si mp l ify ing ,

υ 0, blue t = y 0, red or

y −4.00 m = = 1.00 s −4.00 m/s

0, red

υ 0, blue

ANSW ERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2. Yes. The particle may stop at some in stant, but still have an acceleration, as when a ball thrown straight up reaches its max i mu m h eigh t . 4. (a) No. They can be used only when the acceleratio n is constant. (b ) Yes. Zero i s a co nst ant .

6. (a) In Figure (c), the images are farther apart for each successive time interval. The object is moving to ward the right and speeding up. This means that the accele ration is positive in Figure (c).

(b ) In Fi g u re (a) , t h e f i rst f o ur images show an increasing distan ce travel ed each time interval and therefore a positive acceleration. However, after the fourth imag e, the spacing is decreasing, showing that the object is no w slowing do wn (o r has negative acceleration). (c) In Figure (b), the images are equally spaced, showing that the object mo ved the same distance in each time interval . Hence, the velocity is co nstant in Fi gu re (b ). 8. (a) At the maximum height, the ball is mo me ntarily at rest (i.e., has zero veloci ty ). The acceleration remains constant, with magnitude eq ual to the free-fall acceleration g an d di rect ed d o wn ward . Th u s, ev en th ough t h e v elo ci ty is mo men t ari ly zero, i t cont inues t o change, a nd the bal l will begin to gai n sp eed in the downward di rection. (b) The acceleration of the ball remains constant in magnitude and direction througho ut the ball’s free fl ight, from the instan t it leav es the hand until the instant just before it strikes the ground. The acceleration is direct ed downward and has a magnitude equal to the freefall acceleration g . 10. (a) Su c c e s s iv e i ma g e s on t h e fi l m wi l l b e s e p a rated b y a constant di st an ce if th e ball has constant velo city . (b) Starting at the ri ght-mo st image, the images wi ll be getting closer together as one moves toward the left . (c) Starting at the right-most image, the images will be getting fart her apart as one moves toward the left .

(d) As one moves from left to right, th e balls will first get farther apart in each successive image, then cl oser together whe n the ball begins to sl ow do wn . 12. On ce the ball has left the throwerâ&#x20AC;&#x2122;s hand, it is a freel y falling body wi th a cons tant, nonzero, acceleration of a = Ăg. Since the acceleration of the ball is not zero at any point on its traj ectory, c h oices (a) through (d) are all false and the co r- rect response is (e).

14. The initi al veloci ty of the car is υ 0 = 0 and the velocity at time t i s υ . The constant acceleration is therefore given by a=

∆υ υ − υ 0 υ − 0 υ = = = ∆tttt

and the average velocity of the car is

( υ+υ ) ( υ + 0 ) υυ =

222

The distan ce trav el ed in ti me t i s ∆x = υ t = υ t/2. In the special case where a = 0 (an d h enc e υ = υ 0 = 0 ), we see t h at st at emen t s (a), (b ), (c ), an d (d ) are all correct. However, in the general case ( a 0, and hence

υ ≠ 0 ) only statemen ts (b ) and (c) are true. St at ement (e) is not true in either case. ANSW ERS TO EVEN NUMBERED PROBLEMS 2. (a )

(b)

4. (a )

(b)

6. (a )

(b) (d)

8. (a )

(b)

(c)

(b)

(d) 0 10. (a ) 2.3 mi n (b) 64 mi 12. (a ) (d) 14. (a )

(b) 13 m

16. (a ) The trailing runner’s speed mu st be greater than that of the leader, and the l eader’s distance from the fi ni sh li n e mu st b e great enough to give the trailing runner time to make up the defici en t di st an ce. (b)

(c)

18. (a ) So me data point s that can be used to plot the graph are as given below: x ( m)

5 .7 5

1 6 .0

3 5 .3

6 8 .0

1 19

1 92

t (s)

1 .0 0

2 .0 0

3 .0 0

4 .0 0

5 .0 0

6 .0 0

(b) (c) 20. (a )

(b) 263 m

22. 0 .3 91 s

24. (i ) (a ) 0 (b)

(c)

(i i ) (a) 0 (b)

26. The curves intersect at

28.

30. (a ) (b)

(e) 32.

(d)

(a ) 13.5 m

(b)

13.5 m

No , it cannot land safely on the 0.800 km runway.

22.5 m

(d) 34.

(a ) 20.0 s

(b)

36.

(a ) 5.51 km

(b)

38.

(a)

40.

(a)

107 m

(b) (b)

95m

44.

2 9 .1 s

46.

1 .7 9 s

48.

(a) Yes. (d)

50. (a )

(b)

(c)

. The two rocks have the same acceleration, but the ro ck thrown downward has a higher average speed between the two levels, and is accelerated over a smaller time interval. (b) 19.6 m (c)

52.

(a )

(b)

(a )

(b) 44.1 m

56.

(a )

(b) 198 m

58.

(a ) 4.53 s

(b)

60.

(a )

(b)

62. See So l ut ion s Sect i on fo r Mot ion Di ag rams. 64. Yes. The mi nimu m acceleration need ed to co mp lete the 1 mile distan ce in the allotted time is bly less than wh at sh e is capab le of producing. 66. (a )

, considera-

(b) (c)

(d) as l o ng as b ot h b all s are still in the air.

68. 70. (a) 3.00 s (b)

(c)

72. (a) 2.2 s (b) 74. (a) only if acceleration = 0 (b) Yes , for all initial veloci ties and accelerations. PROB LEM SOLUTION S 2.1 We assume that you are approx imately 2 m tall and that the nerve impulse travel s at uniform sp eed. The elapsed time is then

2.2 (a) At constant speed,

, t h e d i st an c e l igh t t ra v e ls i n 0 .1 s i s

(b ) Co mp ari ng th e resu l t of p art (a) t o t h e di amet er o f t h e Earth, DE , we f i n d

2.3 Distan ces travel ed between pairs of cities are

Thus, the total distan ce traveled is , and the elapsed time is .

(a)

) (see above)

2.4 (a)

(b ) 2.5 (a) Bo a t A req u i res 1 .0 h t o c ro s s t h e l a k e an d 1 .0 h t o ret u rn , t o t al t i me 2 .0 h . B o at B req ui re s 2 .0 h t o c ro s s t h e l a k e at wh i ch t i me th e ra c e i s o v e r.

when the race ends.

(b) Average velocity is the net displacemen t of the boat divided by the total el apsed time. T h e winning boat is back where it started, its displacement thus being zero, yielding an average velocity of 2.6 The average velocity ov er any time interval is

(a)

) (c)

(d ) (e)

2.7 (a)

(b ) T h e to t al e l ap sed t i me i s

so, 2.8 The average velocity over a n y t i me i nt e rv a l is

(a)

(b ) (c)

(d )

2.9 The plane starts from rest and maintains a constant acceleration of . Thus, we find the distan ce it will trav el b efo re re ach ing th e requ ired tak eo ff sp eed , from , as

Si nce this di st an ce is less th an the lengt h of th e runway,

2.10 (a) The time for a car to make the trip is t rip i s

. Thus, the difference in the times for the two cars to comp lete th e same 10 mi l e

(b ) When the faster car has a 15.0 mi n l ead, it is ahead by a distance equal to that trav eled by the slower car in a time of 15.0 mi n. This distan ce is given by . The faster car pulls ahead of the slower car a t a rate of

Thus, the time required for it to get di st an ce

ahead is

Fi n ally , t h e di st ance the faster car has trav eled during th is time is

2.11 (a) From

, wi th

, and â&#x2C6;&#x2020;x = 45 m, the acceleration of the cheetah is found to be

(b) The cheetahâ&#x20AC;&#x2122;s displacement 3.5 s after starting from rest is

2.12 (a)

(b )

(c)

(d ) 2.13 (a) T h e to t al ti me f o r t h e t ri p i s , where t 1 i s t h e t i me s p en t t rave l ing a t Thus, the distan ce travel ed is , wh ich gives

o r, From wh ich, f o r a t o t al t i me o f (b) The distan ce travel ed du rin g th e t rip i s

, g i vin g

2.14 (a) At the end of the race, the to rt oise has been moving for time t and the hare for a time . The speed of the tort oi se is , and the speed of the hare is . The tortoise tr avels distance xt, wh ich is 0.20 m larg er than th e distan ce xh traveled by the hare. Hence,

wh i ch b e c o mes or This gives the time of the race as (b )

2.15 Th e max i mu m al lo wed ti me to co mp l et e th e t rip is

The time spent in the fi rst half of the t rip i s

Thus, the maximu m time that can be spen t on the seco nd half of the trip is

and the required average speed on the second half is

2.16 (a) In o rd e r f o r t h e t ra i l i ng a t h l et e t o b e a b l e t o cat c h t h e l e ad e r, h i s s p e ed (Ď&#x2026; 1) mu st be greater than that of the leading at h- lete (Ď&#x2026; 2 ), and the distan ce between the leading athlete and the finish lin e mu st be great en ough to give the trailing athlete sufficient ti me to make up the defi cient distance, d . (b ) Du ri ng a t i me t the leading athlete will travel a distance and the trailing athlet e will travel a distance . On ly wh en (wh ere d is the initial di stan ce the trai ling athlete was behind the l eader) will the trai ling athlete havecaughttheleader.Requiringthatthis condi tion be satisfied gives the elapsed time required for the second athlete to overtake the first:

g ivi ng (c) In order for the trailing athl et e to be able to at least tie for first place, the initial distance D between the leader and the fini sh line mu st be greater than or equal to the distance the leader can t ra v el i n th e ti me t calculated above (i.e ., the time requi re d t o ov ert a ke th e l e a der). That is, we mu st require that

or 2.17 The instantaneous veloci ty at any time is the slope of the x vs. t g ra p h a t t h at t i me . W e c o mp u t e t hi s s l o pe b y u si ng t wo points on a straight segment of the curv e, one point on each side of the po int of interest .

(a)

(b ) (c) (d )

2.18 (a) A fe w t yp i c al va l ue s a re t (s) 1 .0 0 2 .0 0 3 .0 0 4 .0 0 5 .0 0 6 .0 0

x (m) 5 .7 5 1 6 .0 3 5 .3 6 8 .0 1 19 1 92

(b ) We wi ll use a 0.400 s interv al centered at t = 400 s. We find at t = 3 .8 0 s, x = 6 0 .2 m a n d a t t = 4 .20 s , x = 7 6 .6 m. Therefore,

Usi ng a t i me i nt e rv al o f 0 .20 0 s , we f i n d t h e c o rres p ond i ng val u es t o b e: a t t = 3 .9 0 s , x = 6 4 .0 m a n d a t t = 4 .10 s , x = 72.2 m. Thus,

Fo r a t i me i n t erv a l o f 0 .100 s, the values are: at t = 3 .9 5 s , x = 6 6 .0 m, a n d a t t = 4 .05 s , x = 70.1 m. Therefore,

This value is than the instanta neous velocity at t = 4 .00 s .

2.19 Ch oo se a coordinate axis wi th the origin at the flagpole and east as the positive direction. Then, using th a = 0 for each runner, the x-coordinate of each runner at time t i s

Wh en th e ru nners meet , g ivi ng

This gives the elapsed time when they meet as

. At this ti me,

. Thus, they meet

2.20 From the figure at the right, ob se rv e th a t th e mo t ion of thi s pa rti c l e c a n be brok e n int o th re e di st in ct t i me i nt e rv a l s, during each of which the particle has a constant acceleration. These interv als an d the associ ated accelerations are

and (a) Ap pl yin g

to each of the three time intervals gives

for,

(b ) Ap pl yin g to each of the time interv als gives

for

Thus, the total distan ce traveled in th e fi rst 20 .0 s is

2.21

W e c h oo s e th e p o siti v e di re c tio n to po int a wa y f ro m t h e wal l . T h en , t h e i nit i al v elo c veloci ity o f ty t hise .bIfallt hi i ss c h a nge i n v e lo cit y o cc u rs o v e r a t i me i n t e rv a l of final ing which the ball is in contact with the wall), the av erage acceleration is

2.22 Fro m

2.23 Fro m

and the (i .e. , t h e i nt e rv al d u r-

, t h e req u i red t i me i s

, we have

2.24 (i ) (a) Fro m t = 0 t o t = 5 .0 s ,

(b ) Fro m t = 5 .0 s t o t = 1 5 s ,

(ii ) At any instant, the instan taneous acceleration equals the slope of the line tangent to the v v s. t graph at that poi nt in time. (a) At t = 2.0 s, the slop e of the tangent line to the curv e is .

(b ) At t = 1 0 s , t h e s l op e o f t h e t an g ent l i n e i s . (c) At t = 1 8 s , t h e s l op e o f t h e t an g ent l ineis

2.25 (a)

or Al t e rn ati v ely , (b) If the acceleration is constant,

or 2.26 As i n t h e a lg ebrai c s o lu tion t o E x a mp l e 2 .5 , we l e t t represent the time the trooper has been moving .

We graph

and

The curves intersect at 2.27 Ap pl y t o t h e 2.0 0 -secon d ti me i n t erv al du ring wh i ch t h e obj ect mo ves f rom With , this yields an acceleration of

2.28 Fro m

, we have

2.29 (a)

so t h at

becomes

wh i ch y i eld s

(b )

2.30 (a) (b) The known quantities are initial veloci ty, final velocity, and displacement. The kinematics equation that relates these quantities to acceleration is

(c)

(d )

(e) Usi ng , we f i nd t h at 2.31 (a) With

yields

(b) The required time is 2.32 (a) From

, wi th

, we f i nd

(b ) In t hi s c a se , t he obj e ct move s i n t he same direction for the entire time interval and the total distan ce travel ed is simply the magnitude or absolute valu e of the displacement. That is,

, and we find

[the same as in part (a)] (d ) In t hi s c a s e , t h e o b j e ct ini ti al ly s lo ws d o wn a s i t t ra v el s in th e n e g ati ve x-d i re c t i on, st op s mo ment a ri l y, a ndthengains speedasitbeginstravelinginthepo sitive x-direction. We find the total distan ce travel edbyfirstfindingthedisplace- mentduringeach phase of this mo tion. While coming to rest

Af t er rev ersi ng d i recti on

Note that the net displacement is er, t h e total distan ce traveled in this case is

, as f o und in p art (c) ab o ve. Ho wev

2.33 (a)

(b ) Fro m , t h e req u i red t i me i s

(c)

For uniform acceleration, the change in velocity ∆υ generated in time ∆t i s g i v en by . Fro m t h i s, i t is seen th at dou bling th e length of th e time in terv al ∆t wil l a l way s dou bl e t h e c h ang e in vel o cit y ∆υ . A mo re p reci se way of stat ing this is: “When acceleration is constant, veloci ty is a linear function of time.”

2.34 (a) T he t i me requi red t o st op t he pl an e i s (b ) Th e mi ni mu m d ist an ce n eed ed t o st op i s

Thus, the plan e requires a mi nimu m runway length of 1.00 km. 2.35 We choose at the location of Sue’ s car when she first spots the van and applies the brakes. Then, the initial conditions for Sue’s car a re a n d . Her constant acceleration for . T h e i n it i al conditions for the van are , and its constant acceleration is . We then use to write an equation fo r the x-coordinate of each vehicl e for . This gives Su e’s C ar: Van :

In order for a collisi on to occu r, the two vehi cles mu st be at the same location . Thus, we test for a collision by equating the two equation s fo r the x-coordinates and see if the resulting equation has any real solutions.

or Using the quadratic formula yi elds

The solutions are real , not imaginary, so . T h e s ma l l e r o f t he t wo s o lut ion s i s t h e c ol li sio n t i me. (Th e larger solution tells when the van woul d pull ahead of the car again if the vehi cl es could pass harmlessly thro ugh each other.) The x -coordinate wh ere the collisi on occurs is given by

2.36 T h e v el o cit y a t t h e e n d of th e f i rst i nt e rv al i s

This is also the constant veloci ty du ring th e second in terv al and the initi al veloci ty fo r t h e t h i rd i nt e rv al . Al s o , n ot e th a t the durat ion of the second interv al is .

(a) From

, the total displacement is

(b )

, and the average velocity for the t ot al t rip i s 2.37 Using the uniforml y accelerat ed mo tion eq uation

f o r t h e fu l l 40 s i nt e rv al y i eld s , wh ich is obviously wrong. The source of the erro r is found by comp uting the ti me required for the trai n to come to rest. This ti me is

Thus, the train is slowing do wn for the fi rst 20 s and is at rest for the last 20 s of the 40 s interv al . The acceleration is not constant during the full 40 s. It is, however, constant during the first 20 s as the train slows to re st. Ap pl i c ati on of to this interv al gives the stopp ing distan ce as

2.38 (a) To f ind t h e di stance travel ed, we use

(b) The constant acceleration is 2.39 At the end of the acceleration period, the velocity is

This is also the initial veloci ty for the braking period. (a) Af ter braking,

(b )

Th e to t al di st an ce travel ed is

2.40 For the acceleration period, the parameters f o r the car are: , , and . Fo r the braking period, the parameters are: ,

, and

(a) To determine the velocity of the car ju st before the brakes are engaged, we apply period and find

. to the acceleration

or (b ) We may u se to determine the distance tr aveled during the acceleration period (i.e., before the driver begins to brake). This gives

or (c) The displacement occu rring during the braking peri od is

Thus, the total displacement of the car during the two intervals combined is

2.41 Th e ti me t h e Th und erbi rd sp en d s sl o wi ng down i s

The time required to regain speed after the pit stop is

Thus, the total elapsed time before th e Th und erbi rd i s b ack u p t ospeedis

Duringthistime,the Mercedes has traveled (at constantspeed)adistance and

the Thun derb ird has fallen behind a distance

2.42 The car is distan ce d from the dog and has initial velocity Ď&#x2026; 0 when the brakes are applied, giving it a constant acceleration a . Ap pl y o o btai n

t o t h e e n ti re t ri p (f o r wh i ch

[1 ]

lyi

the

enti

trip

elds

Substitut e for Ď&#x2026; 0 from Equation [1 ] to fi nd that

[2 ] Fi nally, appl y

t o t h e fi rst 8 .0 s o f t h e t ri p (f or wh i ch

This gives

[3 ]

Substitut e Equat ions [1] and [2] i nt o Equati on [3] t o obt ain

wh i ch y i eld s 2.43 (a) Take t = 0 at the time when the player st arts to chase his opponent . At t h i s ti me, t h e o pp onen t is d i st an ce i n f ro nt of t h e p l ay e r. At t i me t > 0, the displacements of the players from thei r initial posi tions are [1 ] and [2 ]

When the players are side-by-side,

[3 ]

Substituting Equati ons [1] and [2] i nto Equat ion [3] gives

or Ap pl yin g th e qu ad rat i c fo rmu l a t o thi s resu lt g i v es

wh i ch h a s s ol ut i on s o f t = â&#x2C6;&#x2019;2.2 s and t = + 8 .2 s. Since the time mu st be greater than zero, we mu st choose as the proper answer. (b )

2.44 The initi al veloci ty of t h e t rai n i s a n d t h e fi nal v e lo cit y i s to p ass th e cro ssin g is fou nd fro m as

2.45 (a) From

wi th Ď&#x2026; = 0 , we h a v e

(b) The time to reach the highest point is

The time required for th e 400 m train

(d ) T h e v el o cit y of t h e b a ll wh en i t retu rn s to t h e o ri gi n al l ev e l (2 .5 5 s a f t e r it s t a rt s to f a ll f rom res t ) is

2.46 We t ak e up ward as t h e po siti ve y-d i re c t i on a nd y = 0 at the point where the ball is released. Then, and Fro m

Then ,

when the ball reaches the ground.

, the velocity of the ball just before it hits the ground is

gives the elapsed time as

2.47 (a) T he vel ocit y of t he obj e ct wh en i t was 30.0 m a bove t he grou nd c a n be det e rmi ne d by a ppl yi ng t o the last 1.50 s of the fall. This gives

(b) The displacement the object mu st have unde rgone, starting from rest , to achieve this velocity at a point 30.0 m above the ground is gi ven by

The to tal di st an ce the obj ect drop s during the fall is th en

2.48 (a) Co n si d e r t h e ro c k â&#x20AC;&#x2122;s e nt i re upwa rd f l i g ht , f o r wh i c h (t aking y = 0 a t g rou nd l ev e l ), a n d altitude reached b. Then applying to this upward fl ight gives

So lving fo r the maximu m altitude of the ro ck gives

Since (h ei ght o f th e wal l ),

(b) To find the velocity of the rock when it reaches the top of the wall, we use and solve for wh en (st art in g wi t h This yields

(c) A ro ck th ro wn downward at a speed of 7.40 m/ s from the top of the wall undergoes a displacemen t o f before reaching the level of the attacker. Its velocity when it reaches the attacker is

so the change in speed of th is rock as it goes between the 2 points located at the top of the wall and the attack er is given by

(d ) Ob serv e that the change in speed of the ball thro wn up ward a s i t went from t he a t ta c ke r t o t he t op of t he wal l was

The rocks have the same acc eleration, but the rock thro wn do wnward has a higher average speed between the two levels, and is accelerated over a smaller time interv al . 2.49 The velocity of the childâ&#x20AC;&#x2122;s head just be fore impact (after falling a distan ce of 0.40 m, starting from rest) is gi ven by as

If, upon impact, the childâ&#x20AC;&#x2122;s head un dergoes an additional displacement before comi ng to rest, the acceleration duri ng the i mpact can be found from t o b e . Th e du rat ion of t h e imp act is f ound f ro m Ap pl yin g th ese resu l t s to the two cases yields

Hard wo o d Floo r

as .

and

C arp et ed Fl oo r

and

2.50 (a) Af t e r 2.00 s, t he ve l oci t y of t he mai l bag i s

The negative sign tells us that the bag is mo ving do wn ward and the magnitude of the velocity gives the speed as

(b) The displacement of the mailbag after 2.00 s is

Du ring this time, the helicopter, moving down ward wi th constant velocity, undergoes a displacement of

The distan ce separating the package and the helicopter a t t his t i me i s th e n

and

wh i l e

In this case, the displacement of the helicopter during the 2.00 s interval is

Mean while, the mailbag has a displacement of

Af ter 2.00 s, the velocity

The distan ce separating the package and the helicopter a t t his t i me i s th e n

2.51 (a) From the instan t the ball leaves th e playerâ&#x20AC;&#x2122;s hand until it is caught, the ball is a fr eely falling body with an acceleration of

(b ) At i t s max i mum h ei g ht , t h e bal l co mes to rest mo mentarily and then begi ns to fall back do wn ward . Thus,

wi t h

When the ball is at the throwerâ&#x20AC;&#x2122;s hand , the

g i vin g

This equation has two solutions, t = 0, wh ich correspond s to wh en the ball was thro wn , and co rresp o nd- ing to when the ball is caught. Therefore, if the ball is caught at t = 2.00 s, the initial velocity mu st have been

wi t h Ď&#x2026; = 0 at t h e max imu m h ei g ht ,

(d ) Fro m

2.52 (a) Let t = 0 be the instant the pack age leav es the helicopter, so the package and the helicopt er have a common initial velocity of

(choosing up ward as positive).

At t i me s t > 0, the veloci ty of the pack ag e (i n free-fall with constant acceleration

i s g i v en b y

as and (b) After an elapsed time t , the downward disp lacement of th e packag e from it s po int of release will be

and the downward displacemen t of the helicopter (moving with constant ve locity, or acceleration a h = 0 ) f ro m t h e re - lease point at this time is

The distan ce separating the package and the helicopter a t t his t i me i s th e n

(c) If the helicopter and packag e are moving upward at th e instant of release, then th e common initial veloci ty is The accelerations of the heli co pter (moving wi th constant velocity) and the p ackage (a freely falling object) remai n uncha ng e d from t he previ ous c a se

In this case, the package speed at time t > 0 is At this time, the disp lacements from the release point of the package and the helico pter are given by

and The distan ce separa ting the package and helicopter at time t i s no w g i v e n b y

(t h e same as earl i er!) 2.53 (a) Af t er i t s engi nes st op , t h e ro ck et is a f reel y fal lin g bod y . It continues up ward , slowing under the influence of gravity unti l it comes to rest momentari ly at i ts maxi mum al ti tude. Then it falls back to Earth, gaining speed as it falls. (b) When it reaches a height of 150 m, the speed of the rocket is

Af t er t h e eng ines st op , t h e rocket cont inues moving upward wi th an initial velocity of

and

When the rocket r eaches maximu m height, Ď&#x2026; = 0. The displacement of the acceleration rocket abov e the point wh ere the engi nes stopp ed (that is, above the 150 m level) is

The maximu m height above ground that the rocket reaches is then given by (c) Th e to t al ti me o f th e up ward mo t ion o f the rocket is the sum of two intervals. Th e first is the ti me for t he rocket t o go from given by

at the ground to a veloci ty of

at an altitude of 150 m. This time is

The second interval is the ti me t o ri s e 1 58 m s t a rt i n g wit h and endi ng wi th Ď&#x2026; = 0 . T hi s t ime i s

The tot al ti me of the upward fl ight is t h en

(d) The time for the ro cket to fall 308 m back to the ground , with

and acceleration i s found from

s o th e t ot al t i me o f t h e fl ight is 2.54 (a) Fo r t he upwa rd fl i gh t of t he bal l , we ha ve Thus,

and

g i v es t h e ini ti al v e lo cit y a s

(b) The vertical displacement of the ball during this 3.00-s upward flight is

2.55 Du ri ng t h e 0 .60 0 s req ui re d fo r t h e ri g t o p as s c o mp l et el y o nt o t h e b ri dg e, t h e f ro n t bu mp e r o f t h e t ra ct o r mo v e s a d i s t a n c e eq u al to t he l eng th of t h e ri g at co nst ant v eloci ty of length of the rig is

While some part of the rig is on the bridge, the front bu mp er moves a dist an ce With a constant velocity of

t he t i me f o r t hi s to o c c u r i s

2.56 (a) The acceleration experi enced as he came to rest is given by as

Therefore the

(b ) Th e di st an ce t rav el ed wh il e sto ppi ng i s fo und f ro m

2.57 (a) The acceleration of the bullet is

(b) The time of contact with the board is

2.58 (a) From

we have

This reduces to

and the quadratic fo rmula gives

The desired time is the smaller solution of The larger solution of t = 12.6 s is the time when the boat would pass the buoy mo ving back wards, assumi ng it maintain ed a constant acceleration. (b) The velocity of the boat wh en it first reaches the buoy is

2.59 (a) The keys have acceleration

from the release point unt il they are caught 1.50 s later. Thus,

g i v es

(b ) Th e v el o cit y of t h e k eys just before the catch was

or 2.60 (a) T h e k ey s , mo v i ng f re e l y u nd er t h e i n flu e nc e o f g ra vi t y i n time t . We u se

undergo a vertical displacement of

t o f ind t h e ini ti a l v elo ci ty a s

(b ) T h e v el o cit y of t h e k e ys j u st bef o re t h ey were c a u g ht (at t ime t ) is given by

2.61 (a) Fro m

t h e insect â&#x20AC;&#x2122;s velocity aft er st rai ght ening i ts l egs is

(b) The time to reach this velocity is

(c) The upward displacement of the insect between when its feet leave the ground and it co mes to rest mo mentari ly at max i mu m al ti tu d e is

2.62 (a)

(b ) (c)

(d )

(e)

(f ) If t h e s p e ed d i d n ot c h an g e at a c o n s t an t rat e, t h e d rawings wo ul d have less regularity than those given ab ov e. 2.63 The falling ball moves a dist an ce of before they meet, where h is the height above the ground where they meet. Apply wi th t o o bt ain

[1]

Ap pl yin g

t o t he ri si ng b all giv es

[2] Co mb ining Equations [1] and [2] gives

or 2.64 The constant speed the stud en t has maintained for the fi rst 10 mi nut es, and hence her initial speed for the final 500 yard dash , is

With an i niti al speed of , the mi nimu m constant acceleratio n that would be needed to comp lete the last 500 yards i n t h e remai nin g 2 .0 mi n (120 s) o f h er al l ott ed ti me i s f oun d f ro m as

Since this acceleration is consider ab ly smaller than the acceleration of cing , sh e should be able to

t h at sh e i s cap ab l e o f p rodu

o f ru n ni ng 1 .0 mi l e i n 1 2 mi n ut e s .

2.65 Once the gymnastâ&#x20AC;&#x2122;s feet leav e th e ground, she is a freely falling body with cons tant acceleration wi th an initial upward velocity of , the vertical displacement of the gymnastâ&#x20AC;&#x2122;s center of mass from its starting po int is gi ven as a function of time by . (a) At

. St art- ing

(b ) At ,

(d ) At 2.66 (a)

While in the air, both balls have acceleration (where upward is taken as po si tive). Ball 1 (t hrown down. Taking at g rou ward) has i n it ial velocit y , while ball 2 (t hrown up ward ) has initial velocity to nd each ball gives their ylevel, the initial y-coordinate of each ball is . Applying coordinates at time t as: B al l 1: or

B al l 2: or (b ) At g ro und l ev el , . Thus, we equate each of the equations found abov e to zero and use the qu adratic fo rmula to solve for the times when each ball reaches the ground. This gives:

B al l 1:

Using only the positive soluti on gives

B al l 2:

and

Ag ain, using on ly the positi ve solution Thus, the difference in the times of flight of the two balls is

(c) Realizing that the balls are going down wa rd as they near the ground, we use find t h e velocity of each ball ju st before i t st ri kes t h e ground: B al l 1: B al l 2:

wi th

(d) Whil e both ball s are st i ll in t he air, the distance se parat ing them i s

2.67 (a) T h e f i rst b a l l i s d ro pp e d f ro m res t f ro m t h e hei ght h of the wi ndow. Thus, speed of this ball as it reaches the ground (and hence the initial veloci ty of the seco nd ball) as .

gives the

When ball 2 is thrown upward at the same time that ball 1 is dropped, their y -co o rdi n at es at t i me t during the flights are given by as: B al l 1:

B al l 2:

When the two balls pass, , o r

g ivi ng (b) When the balls meet,

and Thus, the distan ce below the win dow where this event occu rs is

2.68 We do not know either the initial velo city or the final velo city (tha t i s, vel oc i t y j ust be fo re i mpa c t ) for t he t ruc k. W hat we do know is that the truck skids 62.4 m in 4.20 s while accelerat ing at . We have

and

. Applied to the mo tion of the truck, these yi el d or

[1 ]

and

or [2 ] Ad di ng Equations [1 ] and [2 ] gives the velocity just befo re impact as

2.69 Wh en rel eased f ro m rest , the bill falls freely with a dow nward acceleration due to grav ity Thus, the magnitude of its down ward displacement during Davidâ&#x20AC;&#x2122;s 0.2 s reaction time will be

This is ov er twice the di st an ce from the center of the bill to its top edge

,so.

2.70 (a) T he vel ocit y wi th wh i ch th e first st on e hi t s the wat e r i s

The ti me for this stone to hit th e water is

(b ) Si nce they hit simu lt an eously , the second ston e, wh ich is released 1.00 s later, will hit the water after an flight ti me of 2.00 s. Thus,

The final veloci ty of the seco nd stone is

2.71 (a) The sled â&#x20AC;&#x2122;s displacement,

, after accelerating at

f o r t i me t 1, i s

[1 ]

At the end of time t 1, the sled had achieved a velocity of or

[2 ]

The displacement of the sled while moving at constant velocity Ď&#x2026; fo r t i me t 2 i s or It i s k no wn t h at

[3 ] , and substitution s from Equation s [1] and [3] give or

[4 ]

Al so , i t i s known t h at

[5 ]

So lving Equations [4 ] and [5 ] simu ltaneously yields or The quadratic fo rmula then gives

wi th sol uti ons

Since it is necessary that , the valid solutions are

. (b )

From Equation [2] above, (c) The displacement

of the sled as it comes to rest (with acceleration ) is

Thus, the total displacement for the trip (me asured from the starting point) is

(d ) The time required to come to rest from veloci ty υ (with acceleration a 3) i s

so the duration of t h e enti re t ri p is

2.72 (a) From

wi t h , we have

(b ) The final veloci ty is (c) The time it takes for the sound of the impact to reach the spectator is

s o th e t ot al e l ap s ed t i me i s . 2.73 (a) Si nc e t he so und ha s c on st ant veloci ty, the distan ce it trav eled is

(b) The plane travels th is di st an ce in a time of

, so its velocity mu st be

2.74 The distan ce the glider moves during the time i s g i v en b y , where υ 0 i s t h e g l id e r’s v e lo c it y wh en th e f l ag fi rst ent ers t h e ph oto g at e and a is the glider’s a cceleration. Thus, the average velocity is

(a) The gliderâ&#x20AC;&#x2122;s velocity when it is halfway thro ugh the photogate in space m as

Not e that thi s is

i s f oun d f ro

, in which case .

(b) The speed Ď&#x2026; 2wh en the gl ider is halfway through the phot ogate i n ti me (i .e. , wh en t h e el apsed ti me i s ) i s given by as

wh i ch

f o r a l l po s si bl e v al u es o f .

2.75 The time required for the stuntman to fall 3.00 m, starting from rest, is found from

so (a) With the ho rse moving wi th constant veloci ty of

, the horizontal distance is

(b) The required time is as calculated above.

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