Chapter 3 Motion in Two and Three Dimensions

Transcription

1 Chapte 3 Motion in Two Thee Dimensions Conceptual Poblems * Detemine the Concept The distance taeled alon a path can be epesented as a sequence of displacements. Suppose we take a tip alon some path conside the tip as a sequence of man e small displacements. The net displacement is the ecto sum of the e small displacements, the total distance taeled is the sum of the manitudes of the e small displacements. That is, total distance,,,3... N, N whee N is the numbe of e small displacements. (Fo this to be eactl tue we hae to take the limit as N oes to infinit each displacement manitude oes to zeo.) Now, usin the shotest distance between two points is a staiht line, we hae..., N,,,3 N, N, whee, N is the manitude of the net displacement. Hence, we hae shown that the manitude of the displacement of a paticle is less than o equal to the distance it taels alon its path. Detemine the Concept The displacement of an object is its final position ecto minus its initial position ecto ( f i ). The displacement can be less but nee moe than the distance taeled. Suppose the path is one complete tip aound the eath at the equato. Then, the displacement is but the distance taeled is πre. 3

2 4 Chapte 3 3 Detemine the Concept The impotant distinction hee is that aeae elocit is bein equested, as opposed to aeae speed. The aeae elocit is defined as the displacement diided b the elapsed time. the end of each complete cicuit. t t a The displacement fo an tip aound the tack is zeo. Thus matte how fast the ace ca taels, the aeae elocit is alwas zeo at What is the coect answe if we wee asked fo aeae speed? we see that no The aeae speed is defined as the distance taeled diided b the elapsed time. a total distance t Fo one complete cicuit of an tack, the total distance taeled will be eate than zeo the aeae is not zeo. 4 False. Vectos ae quantities with manitude diection that can be added subtacted like displacements. Conside two ectos that ae equal in manitude oppositel diected. Thei sum is zeo, showin b counteeample that the statement is false. 5 Detemine the Concept We can answe this question b epessin the elationship between the manitude of ecto A its component A S then usin popeties of the cosine function. Epess A S in tems of A θ : A S A cosθ Take the absolute alue of both sides of this epession: A S A cosθ Acosθ cosθ A A S

3 Motion in One Two Dimensions 5 Usin the fact that < cosθ, A substitute fo cosθ to obtain: < S o A to the manitude of the ecto. < A S No. The manitude of a component of a ecto must be less than o equal If the anleθ shown in the fiue is equal to o multiples of 8, then the manitude of the ecto its component ae equal. *6 Detemine the Concept The diaam shows a ecto A its components A A. We can elate the manitude of A is elated to the lenths of its components thouh the Pthaoean theoem. A Suppose that A is equal to zeo. Then A A A. But A A A A. No. If a ecto is equal to zeo, each of its components must be zeo too. 7 Detemine the Concept No. Conside the special case in which B A. If B A, then C the manitudes of the components of A lae than the components of C. B ae *8 Detemine the Concept The instantaneous acceleation is the limitin alue, as t appoaches zeo, of t. Thus, the acceleation ecto is in the same diection as. False. Conside a ball that has been thown upwad nea the suface of the eath is slowin down. The diection of its motion is upwad. The diaam shows the ball s elocit ectos at two instants of time the detemination of. Note that because is downwad so is the acceleation of the ball.

4 6 Chapte 3 9 Detemine the Concept The instantaneous acceleation is the limitin alue, as t appoaches zeo, of t is in the same diection as. Othe than thouh the definition of a, the instantaneous elocit acceleation ectos ae unelated. Knowin the diection of the elocit at one instant tells one nothin about how the elocit is chanin at that instant. (e) is coect. Detemine the Concept The chanin elocit of the olf ball duin its fliht can be undestood b econizin that it has both hoizontal etical components. The natue of its acceleation nea the hihest point of its fliht can be undestood b analzin the etical components of its elocit on eithe side of this point. At the hihest point of its fliht, the ball is still taelin hoizontall een thouh its etical elocit is momentail zeo. The fiue to the iht shows the etical components of the ball s elocit just befoe just afte it has eached its hihest point. The chane in elocit duin this shot inteal is a non-zeo, downwad-pointin ecto. Because the acceleation is popotional to the chane in elocit, it must also be nonzeo. (d) is coect. Remaks: Note that is nonzeo is zeo, while a is zeo a is nonzeo. Detemine the Concept The chane in the elocit is in the same diection as the acceleation. Choose an - coodinate sstem with east bein the positie diection noth the positie diection. Gien ou choice of coodinate sstem, the component of a is neatie so will decease. The component of a is positie so will incease towad the noth. (c) is coect. * Detemine the Concept The aeae elocit of a paticle, a, is the atio of the paticle s displacement to the time equied fo the displacement. (a) We can calculate fom the ien infomation t is known. (b) We do not hae enouh infomation to calculate (a) is coect. cannot compute the

5 Motion in One Two Dimensions 7 paticle s aeae acceleation. (c) We would need to know how the paticle s elocit aies with time in ode to compute its instantaneous elocit. (d) We would need to know how the paticle s elocit aies with time in ode to compute its instantaneous acceleation. 3 Detemine the Concept The elocit ecto is alwas in the diection of motion, thus, tanent to the path. (a) The elocit ecto, as a motion, is tanent to the path. consequence of alwas bein in the diection of (b) A sketch showin two elocit ectos fo a paticle moin alon a path is shown to the iht. 4 Detemine the Concept An object epeiences acceleation whenee eithe its speed chanes o it chanes diection. The acceleation of a ca moin in a staiht path at constant speed is zeo. In the othe eamples, eithe the manitude o the diection of the elocit ecto is chanin, hence, the ca is acceleated. (b) is coect. *5 Detemine the Concept The elocit ecto is defined b d / dt, while the acceleation ecto is defined b a d / dt. (a) A ca moin alon a staiht oad while bakin. (b) A ca moin alon a staiht oad while speedin up. (c) A paticle moin aound a cicula tack at constant speed. 6 Detemine the Concept A paticle epeiences acceleated motion when eithe its speed o diection of motion chanes. A paticle moin at constant speed in a cicula path is acceleatin because the

6 8 Chapte 3 diection of its elocit ecto is chanin. If a paticle is moin at constant elocit, it is not acceleatin. 7 Detemine the Concept The acceleation ecto is in the same diection as the chane in elocit ecto,. (a) The sketch fo the dat thown upwad is shown to the iht. The acceleation ecto is in the diection of the chane in the elocit ecto. (b) The sketch fo the fallin dat is shown to the iht. Aain, the acceleation ecto is in the diection of the chane in the elocit ecto. (c) The acceleation ecto is in the diection of the chane in the elocit ecto hence is downwad as shown the iht: *8 Detemine the Concept The acceleation ecto is in the same diection as the chane in elocit ecto,. The dawin is shown to the iht. 9 Detemine the Concept The acceleation ecto is in the same diection as the chane in elocit ecto,. The sketch is shown to the iht.

7 Motion in One Two Dimensions 9 Detemine the Concept We can decide what the pilot should do b considein the speeds of the boat of the cuent. Gie up. The speed of the steam is equal to the maimum speed of the boat in still wate. The best the boat can do is, while facin diectl upsteam, maintain its position elatie to the bank. (d) is coect. * Detemine the Concept Tue. In the absence of ai esistance, both pojectiles epeience the same downwad acceleation. Because both pojectiles hae initial etical elocities of zeo, thei etical motions must be identical. Detemine the Concept In the absence of ai esistance, the hoizontal component of the pojectile s elocit is constant fo the duation of its fliht. At the hihest point, the speed is the hoizontal component of the initial elocit. The etical component is zeo at the hihest point. (e) is coect. 3 Detemine the Concept In the absence of ai esistance, the acceleation of the ball depends onl on the chane in its elocit is independent of its elocit. As the ball moes alon its tajecto between points A C, the etical component of its elocit deceases the chane in its elocit is a downwad pointin ecto. Between points C E, the etical component of its elocit inceases the chane in its elocit is also a downwad pointin ecto. Thee is no chane in the hoizontal component of the elocit. (d) is coect. 4 Detemine the Concept In the absence of ai esistance, the hoizontal component of the elocit emains constant thouhout the fliht. The etical component has its maimum alues at launch impact. (a) The speed is eatest at A E. (b) The speed is least at point C. (c) The speed is the same at A E. The hoizontal components ae equal at these points but the etical components ae oppositel diected. 5 Detemine the Concept Speed is a scala quantit, wheeas acceleation, equal to the ate of chane of elocit, is a ecto quantit. (a) False. Conside a ball on the end of a stin. The ball can moe with constant speed

8 3 Chapte 3 (a scala) een thouh its acceleation (a ecto) is alwas chanin diection. (b) Tue. Fom its definition, if the acceleation is zeo, the elocit must be constant so, theefoe, must be the speed. 6 Detemine the Concept The aeae acceleation ecto is defined b a / t. The diection of a a is that of f i, as shown to the iht. a 7 Detemine the Concept The elocit of B elatie to A is. The diection of the iht. BA is shown to B A BA B A *8 (a) The ectos A () t A ( t t) ae of equal lenth but point in slihtl diffeent diections. A is shown in the diaam below. Note that A is neal pependicula to A () t. Fo e small time inteals, A A ( t) ae pependicula to one anothe. Theefoe, d A / dt is pependicula to A. (b) If A epesents the position of a paticle, the paticle must be undeoin cicula motion (i.e., it is at a constant distance fom some oiin). The elocit ecto is tanent to the paticle s tajecto; in the case of a cicle, it is pependicula to the cicle s adius. (c) Yes, it could in the case of unifom cicula motion. The speed of the paticle is constant, but its headin is chanin constantl. The acceleation ecto in this case is

9 alwas pependicula to the elocit ecto. Motion in One Two Dimensions 3 9 Detemine the Concept The elocit ecto is in the same diection as the chane in the position ecto while the acceleation ecto is in the same diection as the chane in the elocit ecto. Choose a coodinate sstem in which the diection is noth the diection is east. (a) (b) Path Diection of elocit ecto Path Diection of acceleation ecto AB noth AB noth BC notheast BC southeast CD east CD DE southeast DE southwest EF south EF noth (c) The manitudes ae compaable, but lae fo DE since the adius of path is smalle thee. the *3 Detemine the Concept We ll assume that the cannons ae identical use a constantacceleation equation to epess the displacement of each cannonball as a function of time. Hain done so, we can then establish the condition unde which the will hae the same etical position at a ien time, hence, collide. The modified diaam shown below shows the displacements of both cannonballs. Epess the displacement of the cannonball fom cannon A at an time t afte bein fied befoe an collision: Epess the displacement of the cannonball fom cannon A at an time t afte bein fied befoe an collision: t t t t

10 3 Chapte 3 If the uns ae fied simultaneousl, t t' the balls ae the same distance t below the line of siht at all times. Theefoe, the should fie the uns simultaneousl. Remaks: This is the monke hunte poblem in disuise. If ou imaine a monke in the position shown below, the two uns ae fied simultaneousl, the monke beins to fall when the uns ae fied, then the monke the two cannonballs will all each point P at the same time. 3 Detemine the Concept The doplet leain the bottle has the same hoizontal elocit as the ship. Duin the time the doplet is in the ai, it is also moin hoizontall with the same elocit as the est of the ship. Because of this, it falls into the essel, which has the same hoizontal elocit. Because ou hae the same hoizontal elocit as the ship does, ou see the same thin as if the ship wee stin still. 3 Detemine the Concept (a) Because A D epesent the elocit of the stone. ae tanent to the path of the stone, eithe of them could (b) Let the ectos A () t B( t t) be of equal lenth but point in slihtl diffeent diections as the stone moes aound the cicle.these two ectos A ae shown in the diaam aboe. Note that A is neal pependicula to A() t. Fo e small time inteals, A A() t ae pependicula to one anothe. Theefoe, da/ dt is pependicula to A onl the ecto E could epesent the acceleation of the stone.

11 Motion in One Two Dimensions Detemine the Concept Tue. An object acceleates when its elocit chanes; that is, when eithe its speed o its diection chanes. When an object moes in a cicle the diection of its motion is continuall chanin. 34 Pictue the Poblem In the diaam, (a) shows the pendulum just befoe it eeses diection (b) shows the pendulum just afte it has eesed its diection. The acceleation of the bob is in the diection of the chane in the elocit f i is tanent to the pendulum tajecto at the point of eesal of diection. This makes sense because, at an etemum of motion,, so thee is no centipetal acceleation. Howee, because the elocit is eesin diection, the tanential acceleation is nonzeo. 35 Detemine the Concept The pinciple eason is aeodnamic da. When moin thouh a fluid, such as the atmosphee, the ball's acceleation will depend stonl on its elocit. Estimation Appoimation *36 Pictue the Poblem Duin the fliht of the ball the acceleation is constant equal to 9.8 m/s diected downwad. We can find the fliht time fom the etical pat of the motion, then use the hoizontal pat of the motion to find the hoizontal distance. We ll assume that the elease point of the ball is m aboe ou feet. Make a sketch of the motion. Include coodinate aes, initial final positions, initial elocit components: Obiousl, how fa ou thow the ball will depend on how fast ou can thow it. A majo leaue baseball pitche can thow a fastball at 9 mi/h o so. Assume that ou can thow a ball at two-thids that speed to obtain:.447 m/s 6 mi/h mi/h 6.8 m/s

12 34 Chapte 3 Thee is no acceleation in the diection, so the hoizontal motion is one of constant elocit. Epess the hoizontal position of the ball as a function of time: Assumin that the elease point of the ball is a distance h aboe the ound, epess the etical position of the ball as a function of time: t () h () t at (a) Fo θ we hae: θ ( 6.8m/s) cos 6.8m/s sinθ cos ( 6.8m/s) sin Substitute in equations () () to obtain: Eliminate t between these equations to obtain: At impact, R: ( 6.8m/s)t m 9.8m/s t ( ) 4.9m/s m ( 6.8 m/s) 4.9m/s m R ( 6.8 m/s) Sole fo R to obtain: R 7.m (b) Usin tionomet, sole fo : Substitute in equations () () to obtain: Eliminate t between these equations to obtain: cosθ 9. m/s sinθ ( 6.8m/s) ( 6.8 m/s) cos 45 sin m/s ( 9.m/s)t m 9. m/s t 9.8m/s t ( ) ( ) 4.95m/s m ( 9. m/s)

13 Motion in One Two Dimensions 35 At impact, R. Hence: 4.95m/s m R R o R ( 9. m/s) ( 73.6 m) 47. m R Sole fo R (ou can use the sole o aph functions of ou calculato) to obtain: R 75.6m (c) Sole fo : Substitute in equations () () to obtain: Eliminate t between these equations to obtain: At impact, R: 6.8m/s ( 6.8m/s)t 4m 9.8m/s t ( ) 4.95m/s 4m ( 6.8m/s) 4.95m/s 4m R ( 6.8m/s) Sole fo R to obtain: R 45.3m (d) Usin tionomet, sole fo : Substitute in equations () () to obtain: Eliminate t between these equations to obtain: At impact, R: 9. m / s ( 9.m/s)t 4 m 9. m/s t 9.8m/s t ( ) ( ) 4.95m/s 4m ( 9. m/s) 4.95m/s 4m R R ( 9. m/s) Sole fo R (ou can use the sole o aph function of ou calculato) to obtain: R 85.6m

14 36 Chapte 3 37 Pictue the Poblem We ll inoe the heiht of Geoff s elease point aboe the ound assume that he launched the bick at an anle of 45. Because the elocit of the bick at the hihest point of its fliht is equal to the hoizontal component of its initial elocit, we can use constant-acceleation equations to elate this elocit to the bick s coodinates at impact. The diaam shows an appopiate coodinate sstem the bick when it is at point P with coodinates (, ). Usin a constant-acceleation equation, epess the coodinate of the bick as a function of time: Epess the coodinate of the bick as a function of time: Eliminate the paamete t to obtain: t at o, because a, t t at o, because a, t t ( tan ) θ Use the bick s coodinates when it tanθ R R whee R is the ane of the bick. R tanθ stikes the ound to obtain: ( ) Sole fo to obtain: Substitute numeical alues ealuate ( 9.8m/s )( 44.5m) : 4.8m/s tan 45 Note that, at the bick s hihest point,. Vectos, Vecto Addition, Coodinate Sstems 38 Pictue the Poblem Let the positie diection be staiht up, the positie diection be to the iht, A B be the position ectos fo the minute hou hs. The

15 Motion in One Two Dimensions 37 pictoial epesentation below shows the oientation of the hs of the clock fo pats (a) thouh (d). (a) The position ecto fo the minute h at: is: The position ecto fo the hou h at : is: A B : : (.5m)j ˆ (.5m)j ˆ (b) At 3:3, the minute h is positioned alon the ais, while the hou h is at an anle of (3.5 h)/ h 36 5, measued clockwise fom the top. The position ecto fo the minute h is: A 3:3 (.5m)j ˆ Find the -component of the ecto epesentin the hou h: B (.5m) sin5.4m Find the -component of the ecto epesentin the hou h: The position ecto fo the hou h is: B B (.5m) cos5.647m (.4m) iˆ (.647 m)j ˆ 3:3 (c) At 6:3, the minute h is positioned alon the ais, while the hou h is at an anle of (6.5 h)/ h 36 95, measued clockwise fom the top. The position ecto fo the minute h is: A 6:3 (.5m)j ˆ Find the -component of the ecto epesentin the hou h: B (.5m) sin95.647m Find the -component of the ecto epesentin the hou h: The position ecto fo the hou h is: B B (.5m) cos95.4m (.647 m) iˆ (.4m)j ˆ 6:3 (d) At 7:5, the minute h is positioned alon the ais, while the hou h is at an anle of (7.5 h)/ h 36 8, measued clockwise fom the top.

16 38 Chapte 3 The position ecto fo the minute h is: A 7 : 5 (.5m)iˆ Find the -component of the ecto epesentin the hou h: B (.5m) sin 8.54m Find the -component of the ecto epesentin the hou h: The position ecto fo the hou h is: B B (.5m) cos 8.97m (.54 m) iˆ (.97 m)j ˆ 7: 5 (e) Find A B at :: A B (.5m) ˆj (.5m) (.5m)j ˆ Find A B at 3:3: A B (.5m) ˆj (.4m) iˆ (.647 m) (.4m) iˆ (.435m)j ˆ ˆj [ ˆj ] Find A B at 6:3: A B (.5m) ˆj (.647 m) iˆ (.4m) ˆj (.647 m) iˆ (.59 m)j ˆ [ ] Find A B at 7:5: A B (.5m) ˆj (.5 m) iˆ (.97 m) (.5 m) iˆ (.697 m)j ˆ [ ˆj ] *39 Pictue the Poblem The esultant displacement is the ecto sum of the indiidual displacements. The two displacements of the bea its esultant displacement ae shown to the iht:

17 Motion in One Two Dimensions 39 Usin the law of cosines, sole fo the esultant displacement: Usin the law of sines, sole fo α: R R ( m) ( m) ( m)( m) cos35.m sin α sin35 m.m α.5 the anle with the hoizontal is Pictue the Poblem The esultant displacement is the ecto sum of the indiidual displacements. (a) Usin the endpoint coodinates fo he initial final positions, daw the student s initial final position ectos constuct he displacement ecto. Find the manitude of he displacement the anle this displacement makes with the positie -ais: He displacement is 5 (b) His initial final positions also ae the same as in ( a), so his displacement is *4 Pictue the Poblem Use the stad ules fo ecto addition. Remembe that chanin the sin of a ecto eeses its diection. (a) (b)

18 4 Chapte 3 (c) (d) (e) 4 Pictue the Poblem The fiue shows the paths walked b the Scout. The lenth of path A is.4 km; the lenth of path B is.4 km; the lenth of path C is.5 km: (a) Epess the distance fom the campsite to the end of path C:.4 km.5 km.9km (b) Detemine the anle θ subtended b the ac at the oiin (campsite): θ adians aclenth adius ad 57.3 is ad noth of.4km.4km His diection fom east. camp (c) Epess the total distance as the sum of the thee pats of his walk: d tot d east d ac d towad camp Substitute the ien distances to find the total: d tot.4 km.4 km.5 km 6.3 km

19 Motion in One Two Dimensions 4 Epess the atio of the manitude Manitude of his displacement.9 km of his displacement to the total Total distance walked 6.3km distance he walked substitute to obtain a numeical alue fo this atio: 7 43 Pictue the Poblem The diection of a ecto is detemined b its components. 3.5 m/s θ tan m/s The ecto is in the fouth quadant (b) is coect. 44 Pictue the Poblem The components of the esultant ecto can be obtained fom the components of the ectos bein added. The manitude of the esultant ecto can then be found b usin the Pthaoean Theoem. A table such as the one shown to the iht is useful in oanizin the infomation in this poblem. Let D be the sum of ectos A, B, C. Detemine the components of D b addin the components of A, B, C. Vecto -component -component A 6 3 B 3 4 C 5 D D 5 D 6 Use the Pthaoean Theoem to calculate the manitude of D : D D D (d) is coect. () 5 () Pictue the Poblem The components of the ien ecto can be detemined usin ihttianle tionomet. Use the tionometic elationships between the manitude of a ecto its components to calculate the - -components of each ecto. A θ A A (a) m m 5 m (b) 5 m m 3.54 m (c) 7 km km 6.6 km (d) 5 km 9 5 km

20 4 Chapte 3 (e) 5 km/s 5 3. km/s 7.5 km/s (f) m/s 4 5. m/s 8.66 m/s () 8 m/s 7 8. m/s *46 Pictue the Poblem Vectos can be added subtacted b addin subtactin thei components. Wite A in component fom: A (8 m) cos m A (8 m) sin m A ( 6.4m ) iˆ ( 4.8m )j ˆ (a), (b), (c) Add (o subtact) - D (.4m) iˆ ( 7.8m) -components: E F ˆj ( 3.4m) iˆ ( 9.8m) ˆj ( 7.6m) iˆ ( 3.8m)j ˆ (d) Sole fo G add components to obtain: G ( A B C ) (.3m) iˆ (.9m)j ˆ 47 Pictue the Poblem The manitude of each ecto can be found fom the Pthaoean theoem thei diections found usin the inese tanent function. (a) A 5 iˆ 3ˆj A A A 5.83 (b) B iˆ 7 ˆj, because A is in the st quadant, A tan θ 3. A B B B., because B is in the 4 th quadant, B tan θ 35. B (c) C iˆ 3 ˆj 4kˆ C C C C z 5.39 C z θ cos 4. C whee θ is the pola anle measued fom the positie z-ais

21 Motion in One Two Dimensions 43 φ cos C C cos 9 48 Pictue the Poblem The manitude diection of a two-dimensional ecto can be found b usin the Pthaoean Theoem the definition of the tanent function. (a) A 4iˆ 7 ˆj A A A 8.6 B 3iˆ ˆj, because A is in the 3 d quadant, A tan θ 4 A B B B 3.6 C A B ˆ i 9 ˆj, because B is in the 4 th quadant, B tan θ 33.7 B C C C 9.6, because C is in the 3 d quadant, C tan θ 64 C (b) Follow the same steps as in (a). A 4. ; θ 76. B 6.3 ; θ 7.6 C 3.6 ; θ Pictue the Poblem The components of these ectos ae elated to the manitude of each ecto thouh the Pthaoean Theoem tionometic functions. In pats (a) (b), calculate the ectanula components of each ecto then epess the ecto in ectanula fom.

22 44 Chapte 3 (a) Epess in ectanula fom: Ealuate : iˆ ˆj ( m/s) cos 6 5 m/s ( m/s) sin m/s Substitute to obtain: (b) Epess in ectanula fom: A ( 5m/s)ˆ i (8.66 m/s) ˆj A iˆ A ˆj Ealuate A A : A (5 m) cos m A (5 m) sin m Substitute to obtain: A ( 3.54m) iˆ ( 3.54m)j ˆ ( 4m) iˆ ( 6m)j ˆ (c) Thee is nothin to calculate as we ae ien the ectanula components: 5 Pictue the Poblem While thee ae infinitel man ectos B that can be constucted such that A B, the simplest ae those which lie alon the coodinate aes. Detemine the manitude of A : A A A Wite thee ectos of the same manitude as A : B 5iˆ, B 5ˆ i, B3 5 ˆj The ectos ae shown to the iht:

23 *5 Pictue the Poblem While thee ae seeal walkin outes the fl could take to et fom the oiin to point C, its displacement will be the same fo all of them. One possible oute is shown in the fiue. Motion in One Two Dimensions 45 Epess the fl s displacement D duin its tip fom the oiin to point C find its manitude: D A B C D ( 3 m) iˆ ( 3m) ˆj ( 3m)kˆ ( 3m) ( 3m) ( 3m) 5.m *5 Pictue the Poblem The diaam shows the locations of the tansmittes elatie to the ship defines the distances sepaatin the tansmittes fom each othe fom the ship. We can find the distance between the ship tansmitte B usin tionomet. Relate the distance between A B to the distance fom the ship to A the anle θ: tanθ D D AB SB Sole fo ealuate the distance fom the ship to tansmitte B: D DAB km tanθ tan 3 SB 73km

24 46 Chapte 3 Velocit Acceleation Vectos 53 Pictue the Poblem Fo constant speed diection, the instantaneous elocit is identical to the aeae elocit. Take the oiin to be the location of the stationa ada constuct a pictoial epesentation. Epess the aeae elocit: a t km ˆj Detemine the position ectos: ( ) Find the displacement ecto: Substitute fo t to find the aeae elocit. ( 4.km) iˆ ( 4.km)j ˆ a ( 4.km) iˆ ( 4.km)j ˆ ( 4.km) iˆ ( 4.km) h ˆj ( 4.km/h) iˆ ( 4.km/h)j ˆ 54 Pictue the Poblem The aeae elocit is the chane in position diided b the elapsed time. (a) The aeae elocit is: Find the position ectos the displacement ecto: Find the manitude of the displacement ecto fo the inteal between t t s: a t ( m) iˆ ( 3m)j ˆ ( 6m) iˆ ( 7m)j ˆ ( 4 m) iˆ ( 4 m)j ˆ ( 4m) ( 4m) 5.66m

25 Motion in One Two Dimensions 47 Substitute to detemine a : (b) Repeat (a), this time usin the displacement between t t 5 s to obtain: 5.66 m s a.83 m/s 4m θ tan 4m 45. measued fom the positie ais. ( 3 m) iˆ ( 4 m)j ˆ, ( m) iˆ ( m)j ˆ ( m) ( m) 5.6 m 5, 5.6 m a 5s 3.m/s, m θ tan m 45. measued fom the positie ais., *55 Pictue the Poblem The manitude of the elocit ecto at the end of the s of acceleation will ie us its speed at that instant. This is a constant-acceleation poblem. Find the final elocit ecto of the paticle: iˆ ˆj iˆ a tˆj ( ) iˆ 4. m/s ( 3. m/s )(.s) ( 4. m/s) iˆ ( 6. m/s)j ˆ ˆj Find the manitude of : ( ) ( ) 4. m/s 6. m/s 7. m/s (b) is coect. 56 Pictue the Poblem Choose a coodinate sstem in which noth coincides with the positie diection east with the positie diection. Epessin the west noth elocit ectos is the fist step in deteminin a a. (a) The manitudes of W N ae 4 m/s 3 m/s, espectiel. The chane in the manitude of the paticle s elocit duin this time is: N W m/s

26 48 Chapte 3 (b) The chane in the diection of the elocit is fom west to noth. The chane in diection is 9 (c) The chane in elocit is: ( ) ˆ N W 3m/s j ( 4m/s) ( 4m/s) iˆ ( 3m/s)j ˆ iˆ Calculate the manitude diection of : (d) Find the aeae acceleation duin this inteal: The manitude of this ecto is: ( 4 m/s) ( 3 m/s) 5 m/s 3 m/s θ tan ais m/s a a a a t ( 4m/s) iˆ ( 3m/s) 5s ( ) iˆ 8m/s ( 6m/s )j ˆ ˆj ( 8 m/s ) ( 6 m/s ) m/s its diection is 6m/s θ tan 8m/s 36.9 measued fom the positie ais. 57 Pictue the Poblem The initial final positions elocities of the paticle ae ien. We can find the aeae elocit aeae acceleation usin thei definitions b fist calculatin the ien displacement elocities usin unit ectos i ˆ ˆj. (a) The aeae elocit is: a t The displacement of the paticle duin this inteal of time is: ( m) iˆ ( 8m)j ˆ Substitute to find the aeae elocit: a ( m) iˆ ( 8m) 3s ˆj ( 33.3m/s) iˆ ( 6.7 m/s)j ˆ (b) The aeae acceleation is: a a t

27 Find,, Motion in One Two Dimensions 49 : ( 8.3m/s) iˆ ( 8.3m/s) ˆj ( 9.3m/s) iˆ ( 3. m/s) ˆj ( 9. m/s) iˆ ( 5.3 m/s)j ˆ Usin t 3 s, find the aeae aa ( 3. m/s ) i (.77 m/s )j acceleation: *58 Pictue the Poblem The acceleation is constant so we can use the constant-acceleation equations in ecto fom to find the elocit at t s the position ecto at t 4 s. (a) The elocit of the paticle, as a function of time, is ien b: Substitute to find the elocit at t s: at ( m/s) iˆ ( 9 m/s) ˆj [ iˆ (4 m/s ) (3 m/s ) ˆj ]( s) ( m/s) iˆ ( 3 m/s) ˆj ˆ ˆ (b) Epess the position ecto as a function of time: Substitute simplif: Find the manitude diection of at t 4 s: t at (4 m) iˆ (3 m) ˆj [( m/s) iˆ (-9 m/s) ˆj ]( 4 s) [(4 m/s ) iˆ (3 m/s ) ˆj ]( 4 s) (44 m) iˆ ( 9 m) ˆj (4 s) ( 44 m) ( 9 m) 44.9 m, because is in the 4 th quadant, 9m θ tan.6 44m 59 Pictue the Poblem The elocit ecto is the time-deiatie of the position ecto the acceleation ecto is the time-deiatie of the elocit ecto. Diffeentiate with espect to time: d d dt dt 3ˆ i ( 4 t)j ˆ [( ) iˆ 3t ( 4t 5t ) ˆj ]

28 5 Chapte 3 Diffeentiate with espect to time: whee has units of m/s if t is in seconds. d d a [ 3ˆ i ( 4 t) ˆj ] dt dt ( m/s )j ˆ 6 Pictue the Poblem We can use the constant-acceleation equations in ecto fom to sole the fist pat of the poblem. In the second pat, we can eliminate the paamete t fom the constant-acceleation equations epess as a function of. (a) Use at with to find : [ ˆj ]t ( ) ( ) 6m/s iˆ 4m/s Use t a with ( m)iˆ t (b) Obtain the components of the path fom the ecto equation in (a): Eliminate the paamete t fom these equations sole fo to obtain: to find : [( ) ( ) ] iˆ m 3m/s t ( m/s ) [ t ]j ˆ ( 3m/s ) m t ( ) m/s t m 3 3 Use this equation to plot the aph shown below. Note that the path in the plane is a staiht line (m) (m)

29 Motion in One Two Dimensions 5 6 Pictue the Poblem The displacements of the boat ae shown in the fiue. We need to detemine each of the displacements in ode to calculate the aeae elocit of the boat duin the 3- s tip. (a) Epess the aeae elocit of the boat: a t Epess its total displacement: a N N ( t ) ˆj t ( iˆ ) N W W W To calculate the displacement we fist hae to find the speed afte the fist s: Substitute to find the aeae elocit: (b) The aeae acceleation is ien b: (c) The displacement of the boat fom the dock at the end of the 3-s tip was one of the intemediate esults we obtained in pat (a). W N, f a N t N 6 m/s so a t ˆj 6 m/s t a a a N( N ) ( ) ( 6m) ˆj ( 6m)iˆ t ( 6m)( iˆ ˆj ) 3s ( m/s)( iˆ ˆj ) W iˆ f i t t 6m/s iˆ m/s 3s ( ) ( )iˆ ( 6m) ˆj ( 6m) ( 6m)( iˆ ˆj ) iˆ

30 5 Chapte 3 *6 Pictue the Poblem Choose a coodinate sstem with the oiin at Petoske, the positie diection to the east, the positie diection to the noth. Let t at 9: a.m. θ be the anle between Robet s elocit ecto the eastel diection let M R denote Ma Robet, espectiel. You can epess the positions of Ma Robet as functions of time then equate thei noth () east () coodinates at the time the endezous. Epess Ma s position as a function of time: Note that Robet s initial position coodinates ( i, i ) ae: ˆj ( 8 )j ˆ M Mt t whee M is in miles if t is in hous. ( i, i ) ( 3 mi,.5 mi) Epess Robet s position as a function of time: iˆ R [ i ( Rcosθ )( t )]) [ i ( Rsinθ )( t )]ˆj [ 3 {6( t )cosθ}] iˆ [.5 {6( t )sinθ}] ˆj whee R is in miles if t is in hous. When Ma Robet endezous, thei coodinates will be the same. Equatin thei noth east coodinates ields: Sole equation () fo cosθ : East: 3 6t cosθ 6 cosθ () Noth:.5 6t sinθ 6 sinθ 8t () 3 cosθ (3) 6 ( t ) Sole equation () fo sinθ : 8t.5 sinθ (4) 6( t ) Squae add equations (3) (4) to obtain: 8t.5 sin θ cos θ 6 3 ( t ) 6( t ) Simplif to obtain a quadatic equation in t: 8t 88t 639 Sole (ou could use ou calculato s sole function) this t 3.4h 3h 5min

31 Motion in One Two Dimensions 53 equation fo the smallest alue of t (both oots ae positie) to obtain: Now ou can find the distance taeled due noth b Ma: t ( 8mi/h)( 3.4h) 5.9mi M M Finall, solin equation (3) fo θ substitutin 3.4 h fo t ields: 3 θ cos cos 6 so Robet should head 4.7 noth of east. 3 ( t ) 6( 3.4 ) 4.7 Remaks: Anothe solution that does not depend on the components of the ectos utilizes the law of cosines to find the time t at which Ma Robet meet then uses the law of sines to find the diection that Robet must head in ode to endezous with Ma. Relatie Velocit 63 Pictue the Poblem Choose a coodinate sstem in which noth is the positie diection east is the positie diection. Let θ be the anle between noth the diection of the plane s headin. The elocit of the plane elatie to the ound, PG, is the sum of the elocit of the plane elatie to the ai, PA, the elocit of the ai elatie to the ound, AG. i.e., PG PA AG The pilot must head in such a diection that the east-west component of PG is zeo in ode to make the plane fl due noth. (a) Fom the diaam one can see that: Sole fo ealuate θ : AG cos 45 PA sinθ 56.6 km/h θ sin 5 km/h 3. west of noth (b) Because the plane is headed due noth, add the noth components of PG (5 km/h) cos 3. (8 km/h) sin 45

32 54 Chapte 3 PA AG to detemine the plane s ound speed: 3 km/h 64 Pictue the Poblem Let SB epesent the elocit of the swimme elatie to the bank; SW the elocit of the swimme elatie to the wate; WB the elocit of the wate elatie to the shoe; i.e., SB SW WB The cuent of the ie causes the swimme to dift downsteam. (a) The tianles shown in the fiue ae simila iht tianles. Set up a popotion between thei sides sole fo the speed of the wate elatie to the bank: WB SW 4m 8m (.6 m/s).8 m/s WB (b) Use the Pthaoean Theoem to sole fo the swimme s speed elatie to the shoe: ( ).6 m/s (.8m/s ) (c) The swimme should head in a diection such that the upsteam component of he elocit is equal to the speed of the wate elatie to the shoe: SB SW.79 m/s WS Use a tionometic function to.8 m/s ealuate θ: θ sin 3..6 m/s

33 *65 Pictue the Poblem Let the elocit of the plane elatie to the ound be epesented b PG; the elocit of the plane elatie to the ai b PA, the elocit of the ai elatie to the ound b AG. Then () PG PA AG Choose a coodinate sstem with the oiin at point A, the positie diection to the east, the positie diection to the noth. θ is the anle between noth the diection of the plane s headin. The pilot must head so that the east-west component of PG is zeo in ode to make the plane fl due noth. Motion in One Two Dimensions 55 Use the diaam to epess the condition elatin the eastwad component of AG the westwad component of PA. This must be satisfied if the plane is to sta on its nothel couse. [Note: this is equialent to equatin the - components of equation ().] (5 km/h) cos 45 (4 km/h) sinθ Now sole fo θ to obtain: ( 5 km/h) Add the noth components of PA AG to find the elocit of the plane elatie to the ound: cos45 θ sin 4 km/h PG AG sin45 PA cos8.47 PG (4 km/h)cos 8.47 (5 km/h)sin 45 km/h 8.47 Finall, find the time of fliht: t fliht distance taelled PG 5 km.57 h km/h

34 56 Chapte 3 66 Pictue the Poblem Let BS be the elocit of the boat elatie to the shoe; BW be the elocit of the boat elatie to the wate; WS epesent the elocit of the wate elatie to the shoe. Independentl of whethe the boat is oin upsteam o downsteam: BS BW WS Goin upsteam, the speed of the boat elatie to the shoe is educed b the speed of the wate elatie to the shoe. Goin downsteam, the speed of the boat elatie to the shoe is inceased b the same amount. Fo the upsteam le of the tip: Fo the downsteam le of the tip: Epess the total time fo the tip in tems of the times fo its upsteam downsteam les: Multipl both sides of the equation b ( BW WS )( BW WS ) (the poduct of the denominatos) eaane the tems to obtain: BS BW WS BS BW WS t total t BW upsteam L WS t downsteam BW L L BW BW WS ttotal WS Sole the quadatic equation fo BW. (Onl the positie oot is phsicall meaninful.) 67 Pictue the Poblem Let p be the elocit of the plane elatie to the ound; a be the elocit of the ai elatie to the ound; pa the elocit of the plane elatie to the ai. Then, p pa a. The wind will affect the fliht times diffeentl alon these two paths. BW 5.8km/h

35 Motion in One Two Dimensions 57 The elocit of the plane, elatie to the ound, on its eastbound le is equal to its elocit on its westbound le. Usin the diaam, find the elocit of the plane elatie to the ound fo both diections: p pa a ( 5m/s) ( 5m/s) 4.m/s Epess the time fo the east-west oundtip in tems of the distances elocities fo the two les: t oundtip,ew t eastbound t westbound adius of the cicle p,eastbound adius of the cicle p,westbound 3 m 4s 4.m/s Use the distances elocities fo the two les to epess ealuate the time fo the noth-south oundtip: t oundtip,ns t nothbound t southbound 3 m (5 m/s) (5 m/s) Because t < t NS, oundtip, EW oundtip, adius of the cicle p,nothbound 3 m (5 m/s) (5 m/s) adius of the cicle 5 s p,southbound ou should fl ou plane acoss the wind. 68 Pictue the Poblem This is a elatie elocit poblem. The ien quantities ae the diection of the elocit of the plane elatie to the ound the elocit (manitude diection) of the ai elatie to the ound. Asked fo is the diection of the elocit of the ai elatie to the ound. Usin PG PA AG, daw a ecto addition diaam sole fo the unknown quantit. Calculate the headin the pilot must 3 kts take: θ sin.5 5 kts Because this is also the anle of the plane's headin clockwise fom noth, it is also its azimuth o the equied tue headin: Az (.5 )

36 58 Chapte 3 *69 Pictue the Poblem The position of B elatie to A is the ecto fom A to B; i.e., AB The elocit of B elatie to A is AB dt B d AB the acceleation of B elatie to A is a AB dt d AB Choose a coodinate sstem with the oiin at the intesection, the positie diection to the east, the positie diection to the noth. A (a) Find,, B A AB : [ 4m ( m/s ) t ] B A AB [( m/s) t]iˆ B A [( m/s) t] iˆ ˆj [ 4m ( m/s ) t ]j ˆ Ealuate AB at t 6 s: (6s) ( m) iˆ (4 m) ˆ j AB (b) Find AB d AB dt : d AB d AB [{( m/s) t} i dt dt 4 m m/s t ˆj { ( ) } ] ( m/s) iˆ ( m/s ) t ˆj Ealuate AB at t 6 s: ( 6s ) ( m/s ) iˆ ( m/s )j ˆ AB (c) Find a AB d AB dt : d a [ iˆ ˆ AB ( m/s) ( m/s ) t j] dt ( m/s )j ˆ Note that a AB is independent of time. *7 Pictue the Poblem Let h h epesent the heihts fom which the ball is dopped to which it ebounds, espectiel. Let epesent the speeds with which the ball stikes the acket ebounds fom it. We can use a constant-acceleation equation to elate the pe- post-collision speeds of the ball to its dop ebound heihts.

37 Motion in One Two Dimensions 59 (a) Usin a constant-acceleation equation, elate the impact speed of the ball to the distance it has fallen: Relate the ebound speed of the ball to the heiht to which it ebounds: h o, because, h ' h' o because, ' h' Diide the second of these equations b the fist to obtain: ' h' h h' h Substitute fo h ealuate the '.64h atio of the speeds:. 8 h '. 8 (b) Call the speed of the acket V. In a efeence fame whee the acket is unmoin, the ball initiall has speed V, moin towad the acket. Afte it "bounces" fom the acket, it will hae speed.8 V, moin awa fom the acket. In the efeence fame whee the acket is moin the ball initiall unmoin, we need to add the speed of the acket to the speed of the ball in the acket's est fame. Theefoe, the ball's speed is: ' V.8V.8V 45m/s mi/h This speed is close to that of a tennis po s see. Note that this esult tells us that the ball is moin sinificantl faste than the acket. (c) Fom the esult in pat ( as the acket. b), the ball can nee moe moe than twice as fast Cicula Motion Centipetal Acceleation 7 Pictue the Poblem We can use the definition of centipetal acceleation to epess a c in tems of the speed of the tip of the minute h. We can find the tanential speed of the tip of the minute h b usin the distance it taels each eolution the time it takes to complete each eolution. Epess the acceleation of the tip of the minute h of the clock as a function of the lenth of the h the speed of its tip: Use the distance the minute h taels ee hou to epess its speed: a c R πr T

38 6 Chapte 3 Substitute to obtain: Substitute numeical alues ealuate a c : a a c c 4π R T (.5m) ( 36s) 4 6 π.5 m/s Epess the atio of a c to : 6.5 m/s 7.55 ac 9.8m/s 7 Pictue the Poblem The diaam shows the centipetal tanential acceleations epeienced b the test tube. The tanential acceleation will be zeo when the centifue eaches its maimum speed. The centipetal acceleation inceases as the tanential speed of the centifue inceases. We can use the definition of centipetal acceleation to epess a c in tems of the speed of the test tube. We can find the tanential speed of the test tube b usin the distance it taels each eolution the time it takes to complete each eolution. The tanential acceleation can be found fom the chane in the tanential speed as the centifue is spinnin up. (a) Epess the acceleation of the centifue am as a function of the lenth of its am the speed of the test tube: Use the distance the test tube taels ee eolution to epess its speed: a c R πr T Substitute to obtain: Substitute numeical alues ealuate a c : a c a c 4π R T 4π min 5 e 3.7 (.5m) 5 m/s 6s min

39 Motion in One Two Dimensions 6 (b) Epess the tanential acceleation in tems of the diffeence between the final initial tanential speeds: Substitute numeical alues ealuate a T : a a t t f i t 3.4 m/s πr T πr t T t (.5m) π min 6s 5 e min ( 75s) 73 Pictue the Poblem The diaam includes a pictoial epesentation of the eath in its obit about the sun a foce diaam showin the foce on an object at the equato that is due to the eath s otation, F R, the foce on the object due to the obital motion of the eath about the sun, F. o Because these ae centipetal foces, we can calculate the acceleations the equie fom the speeds adii associated with the two cicula motions. Epess the adial acceleation due to the otation of the eath: Epess the speed of the object on the equato in tems of the adius of the eath R the peiod of the eath s otation T R : a R R πr T R R R Substitute fo R in the epession 4π R ar fo a R to obtain: T R Substitute numeical alues ealuate a R : a R 4 π 3 ( 637 m) ( 4h) s h 3 m/s Note that this effect ies ise to the wellknown latitude coection fo.

40 6 Chapte 3 Epess the adial acceleation due to the obital motion of the eath: a o o Epess the speed of the object on the equato in tems of the eath-sun distance the peiod of the eath s motion about the sun T o : o π T o Substitute fo o in the epession 4π ao fo a o to obtain: T o Substitute numeical alues ealuate a o : a o 4π ( 365d) 5.95 (.5 m) 4h 36s d h 3 m/s Pictue the Poblem We can elate the acceleation of the moon towad the eath to its obital speed distance fom the eath. Its obital speed can be epessed in tems of its distance fom the eath its obital peiod. Fom tables of astonomical data, we find that the sideeal peiod of the moon is 7.3 d that its mean distance fom the eath is m. Epess the centipetal acceleation of the moon: Epess the obital speed of the moon: Substitute to obtain: Substitute numeical alues ealuate a c : ac π T a a c c 4π T 8 ( 3.84 m) 4π 4h 36s 7.3d d h 3.7 m/s 4.78

41 Motion in One Two Dimensions 63 a Remaks: Note that c adius of eath (a c is just the acceleation distance fom eath to moon due to the eath s ait ealuated at the moon s position). This is Newton s famous fallin apple obseation. 75 Pictue the Poblem We can find the numbe of eolutions the ball makes in a ien peiod of time fom its speed the adius of the cicle alon which it moes. Because the ball s centipetal acceleation is elated to its speed, we can use this elationship to epess its speed. Epess the numbe of eolutions pe minute made b the ball in tems of the cicumfeence c of the cicle the distance the ball taels in time t: Relate the centipetal acceleation of the ball to its speed the adius of its cicula path: n () c ac R Sole fo the speed of the ball: R Epess the distance taeled in time t at speed : t Substitute to obtain: Rt The distance taeled pe eolution is the cicumfeence c of the cicle: c π R Substitute in equation () to obtain: n Rt π R π t R Substitute numeical alues n π 9.8m/s 33.4 min ealuate n: ( ).8m Remaks: The ball will oscillate at the end of this stin as a simple pendulum with a peiod equal to /n. Pojectile Motion Pojectile Rane 76 Pictue the Poblem Nelectin ai esistance, the acceleations of the ball ae constant the hoizontal etical motions of the ball ae independent of each othe. We can use the hoizontal motion to detemine the time-of-fliht then use this infomation to detemine the distance the ball dops. Choose a coodinate sstem in which the oiin is at the point of elease of the ball, downwad is the positie diection, the hoizontal 6s

42 64 Chapte 3 diection is the positie diection. Epess the etical displacement of the ball: Find the time of fliht fom / t: ( ) t a t o, because a, t t ( ) ( 8.4m)( 36s/h) ( 4 km/h)( m/km).473s Substitute to find the etical displacement in.473 s: ( 9.8m/s )(.473s).m 77 Pictue the Poblem In the absence of ai esistance, the maimum heiht achieed b a pojectile depends on the etical component of its initial elocit. The etical component of the pojectile s initial elocit is: Use the constant-acceleation equation: sinθ a Set, a, h to obtain: h ( sin ) θ *78 Pictue the Poblem Choose the coodinate sstem shown to the iht. Because, in the absence of ai esistance, the hoizontal etical speeds ae independent of each othe, we can use constant-acceleation equations to elate the impact speed of the pojectile to its components. The hoizontal etical elocit components ae: Usin a constant-acceleation equation, elate the etical cosθ sinθ a o, because a h,

43 Motion in One Two Dimensions 65 component of the elocit to the etical displacement of the pojectile: ( sin ) h θ Epess the elationship between the manitude of a elocit ecto its components, substitute fo the components, simplif to obtain: ( cosθ ) ( sin θ cos θ ) h h Substitute fo : (. ) h Set., h 4 m sole fo : 4. m/s Remaks: Note that is independent of θ. This will be moe obious once conseation of ene has been studied. 79 Pictue the Poblem Eample 3- shows that the dat will hit the monke unless the dat hits the ound befoe eachin the monke s line of fall. What initial speed does the dat need in ode to just each the monke s line of fall? Fist, we will calculate the fall time of the monke, then we will calculate the hoizontal component of the dat s elocit. Usin a constant-acceleation equation, elate the monke s fall distance to the fall time: h t Sole fo the time fo the monke to fall to the ound: Substitute numeical alues ealuate t: t t h ( ).m 9.8m/s.5s Let θ be the anle the bael of the dat un makes with the hoizontal. Then: Use the fact that the hoizontal elocit is constant to detemine : m θ tan.3 5m ( 5m.5s) cosθ cos m/s

44 66 Chapte 3 8 Pictue the Poblem Choose the coodinate sstem shown in the fiue to the iht. In the absence of ai esistance, the pojectile epeiences constant acceleation in both the diections. We can use the constant-acceleation equations to epess the coodinates of the pojectile alon its tajecto as functions of time. The elimination of the paamete t will ield an epession fo as a function of that we can ealuate at (R, ) (R/, h). Solin these equations simultaneousl will ield an epession fo θ. Epess the position coodinates of the pojectile alon its fliht path in tems of the paamete t: ( cosθ )t ( ) sinθ t t Eliminate the paamete t to tanθ () cos θ obtain: ( ) Ealuate equation () at (R, ) to obtain: R sinθ cosθ Ealuate equation () at (R/, h) to obtain: Equate R h sole the esultin equation fo θ : h ( sin ) θ θ tan ( 4) 76. Remaks: Note that this esult is independent of. 8 Pictue the Poblem In the absence of ai esistance, the motion of the ball is unifoml acceleated its hoizontal etical motions ae independent of each othe. Choose the coodinate sstem shown in the fiue to the iht use constant-acceleation equations to elate the components of the ball s initial elocit. Use the components of to epess θ in tems of : θ tan ()

45 Motion in One Two Dimensions 67 Use the Pthaoean elationship between the elocit its components to epess : Usin a constant-acceleation equation, epess the etical speed of the pojectile as a function of its initial upwad speed time into the fliht: Because halfwa thouh the fliht (at maimum eleation): Detemine : () a t (9.8 m/s )(. s). m/s t 4m.44s 6.4 m/s Substitute in equation () ealuate : ( 6.4 m/s) (. m/s).3m/s Substitute in equation (). m/s ealuate θ : θ tan m/s *8 Pictue the Poblem In the absence of fiction, the acceleation of the ball is constant we can use the constantacceleation equations to descibe its motion. The fiue shows the launch conditions an appopiate coodinate sstem. The speeds,, ae elated thouh the Pthaoean Theoem. The squaes of the etical hoizontal components of the object s elocit ae: The elationship between these aiables is: Substitute simplif to obtain: sin θ h cos θ h Note that is independent of θ... as was to be shown.

46 68 Chapte 3 83 Pictue the Poblem In the absence of ai esistance, the pojectile epeiences constant acceleation duin its fliht we can use constant-acceleation equations to elate the speeds at half the maimum heiht at the maimum heiht to the launch anle θ of the pojectile. The anle the initial elocit makes with the hoizontal is elated to the initial elocit components. Wite the equation a, fo h : tan θ h h () Wite the equation h h () a, fo h/: We ae ien (3/4). Squae both sides epess this usin the components of the elocit. The component of the elocit emains constant. ( ) 3 (3) 4 whee we hae used. (Equations,, 3 constitute thee equations fou unknowns,,, h. To sole fo an of these unknowns, we fist need a fouth equation. Howee, to sole fo the atio ( / ) of two of the unknowns, the thee equations ae sufficient. That is because diidin both sides of each equation b ies thee equations thee unknowns /, /, h/. Sole equation fo h substitute in equation : ( ) h Substitute fo in equation 3: ( ) 3 4

47 Motion in One Two Dimensions 69 Diide both sides b sole fo / to obtain: Usin tan θ /, sole fo θ : θ tan tan ( 7 ) Pictue the Poblem The hoizontal speed of the cate, in the absence of ai esistance, is constant equal to the speed of the cao plane. Choose a coodinate sstem in which the diection the plane is moin is the positie diection downwad is the positie diection appl the constant-acceleation equations to descibe the cate s displacements at an time duin its fliht. (a) Usin a constant-acceleation equation, elate the etical displacement of the cate to the time of fall t: ( ) t t o, because, t ( ) Sole fo t: Substitute numeical alues ealuate t: t t 3 ( m) 49.5s 9.8m/s (b) The hoizontal distance taeled in 49.5 s is: R t ( 9 km/h) ( 49.5s).4 km h 36s (c) Because the elocit of the plane is constant, it will be diectl oe the cate when it hits the ound; i.e., the distance to the aicaft will be the eleation of the aicaft.. km

48 7 Chapte 3 *85 Pictue the Poblem In the absence of ai esistance, the acceleations of both Wile Coote the Roadunne ae constant we can use constant-acceleation equations to epess thei coodinates at an time duin thei leaps acoss the oe. B eliminatin the paamete t between these equations, we can obtain an epession that elates thei coodinates to thei coodinates that we can sole fo thei launch anles. (a) Usin constant-acceleation equations, epess the coodinate of the Roadunne while it is in fliht acoss the oe: Usin constant-acceleation equations, epess the coodinate of the Roadunne while it is in fliht acoss the oe: Eliminate the paamete t to obtain: Lettin R epesent the Roadunne s ane usin the tionometic identit sinθ sinθ cosθ, sole fo ealuate its launch speed: (b) Lettin R epesent Wile s ane, sole equation () fo his launch anle: Substitute numeical alues ealuate θ : t at o, because, a cosθ, ( cosθ )t t at o, because, a sinθ, ( sinθ ) t t ( tanθ ) () cos θ R sin θ 8. m/s θ sin θ sin 3. R ( 6.5m)( 9.8m/s ) sin 3 ( 4.5m)( 9.8m/s ) ( 8. m/s)

49 86 Pictue the Poblem Because, in the absence of ai esistance, the etical hoizontal acceleations of the cannonball ae constant, we can use constantacceleation equations to epess the ball s position elocit as functions of time acceleation. The maimum heiht of the ball its time-of-fliht ae elated to the components of its launch elocit. Motion in One Two Dimensions 7 (a) Usin a constant-acceleation equation, elate h to the initial final speeds of the cannonball: Find the etical component of the fiin speed: a o, because a, sinθ (3 m/s)sin 45 m/s Sole fo ealuate h: ( m/s) h 9.8m/s ( ).9 km (b) The total fliht time is: t t up t dn t up ( m/s) 9.8m/s 43.s (c) Epess the coodinate of the ball as a function of time: ( cos ) t t θ Ealuate ( R) when t 43. s: [( 3 m/s) cos45 ]( 43.s) 87 Pictue the Poblem Choose a coodinate sstem in which the oiin is at the base of the towe the - -aes ae as shown in the fiue to the iht. In the absence of ai esistance, the hoizontal speed of the stone will emain constant duin its fall a constant-acceleation equation can be used to detemine the time of fall. The final elocit of the stone will be the ecto sum of its components. 9.6 km

50 7 Chapte 3 (a) Usin a constant-acceleation equation, epess the etical displacement of the stone (the heiht of the towe) as a function of the fall time: ( ) t a t o, because a, t ( ) Sole fo ealuate the time of fall: t ( 4m) 9.8m/s.s Use the definition of aeae elocit to find the elocit with which the stone was thown fom the towe: t 8m.s 8.4m / s (b) Find the component of the stone s elocit afte. s: t (9.8 m/s)(.s).7 m/s Epess in tems of its components: Substitute numeical alues ealuate : ( 8.4 m/s) (.7 m/s) 3. m/s 88 Pictue the Poblem In the absence of ai esistance, the acceleation of the pojectile is constant its hoizontal etical motions ae independent of each othe. We can use constant-acceleation equations to epess the hoizontal etical displacements of the pojectile in tems of its time-of-fliht. Usin a constant-acceleation equation, epess the hoizontal displacement of the pojectile as a function of time: Usin a constant-acceleation equation, epess the etical displacement of the pojectile as a function of time: Substitute numeical alues to obtain the quadatic equation: Sole fo t: ( ) t a t o, because cosθ a, cosθ ( ) t ( ) t a t o, because sinθ a, ( cosθ ) t ( t ) m t 3.6 s ( 6m/s)( sin 6 ) t ( 9.8m/s )( t)

51 Motion in One Two Dimensions 73 Substitute fo t ealuate the hoizontal distance taeled b the pojectile: (6 m/s)(cos6 )(3.6 s) 48 m 89 Pictue the Poblem In the absence of ai esistance, the acceleation of the cannonball is constant its hoizontal etical motions ae independent of each othe. Choose the oiin of the coodinate sstem to be at the base of the cliff the aes diected as shown use constant- acceleation equations to descibe both the hoizontal etical displacements of the cannonball. Epess the diection of the elocit ecto when the pojectile stikes the ound: Epess the etical displacement usin a constant-acceleation equation: θ tan ( ) t a t o, because a, t ( ) Set (R h) to obtain: ( ) t t Sole fo : t t Find the component of the pojectile as it hits the ound: Substitute ealuate θ : θ tan a t t tan ( ) Pictue the Poblem In the absence of ai esistance, the etical hoizontal motions of the pojectile epeience constant acceleations ae independent of each othe. Use a coodinate sstem in which up is the positie diection hoizontal is the positie diection use constant-acceleation equations to descibe the hoizontal etical displacements of the pojectile as functions of the time into the fliht.

52 74 Chapte 3 (a) Use a constant-acceleation equation to epess the hoizontal displacement of the pojectile as a function of time: Ealuate this epession when t 6 s: (b) Use a constant-acceleation equation to epess the etical displacement of the pojectile as a function of time: t ( cosθ ) t ( 3 m/s)( cos6 )( 6s) 9m ( ) t ( ) sinθ t Ealuate this epession when t 6 s: ( 3 m/s)( sin6 )( 6s) ( 9.8m/s )( 6s).38km 9 Pictue the Poblem In the absence of ai esistance, the acceleation of the pojectile is constant the hoizontal etical motions ae independent of each othe. Choose the coodinate sstem shown in the fiue with the oiin at the base of the cliff the aes oiented as shown use constant-acceleation equations to find the ane of the cannonball. Usin a constant-acceleation equation, epess the hoizontal displacement of the cannonball as a function of time: Usin a constant-acceleation equation, epess the etical displacement of the cannonball as a function of time: ( ) t a t o, because cosθ a, cosθ ( ) t ( ) t a t o, because 4 m, a, sinθ, 4m ( 4. m/s)( sin 3 ) t 9.8m/s t ( )( ) Sole the quadatic equation fo t: t 5.73 s Calculate the ane: R ( 4. m/s)( cos3 )( 5.73s) 9m

53 *9 Pictue the Poblem Choose a coodinate sstem in which the oiin is at ound leel. Let the positie diection be to the iht the positie diection be upwad. We can appl constantacceleation equations to obtain paametic equations in time that elate the ane to the initial hoizontal speed the heiht h to the initial upwad speed. Eliminatin the paamete will leae us with a quadatic equation in R, the solution to which will ie us the ane of the aow. In (b), we ll find the launch speed anle as iewed b an obsee who is at est on the ound then use these esults to find the aow s ane when the hose is moin at m/s. Motion in One Two Dimensions 75 (a) Use constant-acceleation equations to epess the hoizontal etical coodinates of the aow s motion: Sole the -component equation fo time: Eliminate time fom the -component equation: R t h ( ) t t whee cosθ sinθ t R R cosθ R R h, at (R, ), h R cos θ ( tanθ ) R Sole fo the ane to obtain: R sin h sin θ θ Substitute numeical alues ealuate R: R ( 45m/s) ( ) 9.8m/s sin ( )(.5m) 9.8m/s ( 45m/s) ( sin ) 8.6m

54 76 Chapte 3 (b) Epess the speed of the aow in the hoizontal diection: aow ( 45m/s) 56.3m/s ache cos m/s Epess the etical speed of the aow: Epess the anle of eleation fom the pespectie of someone on the ound: θ tan ( 45 m/s) sin 7.8m/s tan 7.8m/s m/s Epess the aow s speed elatie to the ound: ( 56.3m/s) ( 7.8m/s) 56.8 m/s Substitute numeical alues ealuate R: R ( 56.8m/s) ( ) 9.8m/s sin5.8 ( )(.5m) 9.8m/s ( 56.8m/s) ( sin 7.9 ) 4m Remaks: An altenatie solution fo pat (b) is to sole fo the ane in the efeence fame of the ache then add to it the distance the fame taels, elatie to the eath, duin the time of fliht. 93 Pictue the Poblem In the absence of ai esistance, the hoizontal etical motions ae independent of each othe. Choose a coodinate sstem oiented as shown in the fiue to the iht appl constant-acceleation equations to find the time-of-fliht the ane of the spudplu. (a) Usin a constant-acceleation equation, epess the etical displacement of the plu: ( ) t a t o, because a, t ( ) Sole fo ealuate the fliht time t: t (.m) 9.8m/s.45s

55 Motion in One Two Dimensions 77 (b) Usin a constant-acceleation equation, epess the hoizontal displacement of the plu: Substitute numeical alues ealuate R: ( ) t a t o, because a, t R ( 5 m/s)(.45s).6m 94 Pictue the Poblem An eteme alue (i.e., a maimum o a minimum) of a function is detemined b settin the appopiate deiatie equal to zeo. Whethe the etemum is a maimum o a minimum can be detemined b ealuatin the second deiatie at the point detemined b the fist deiatie. Ealuate dr/dθ : dr dθ d dθ [ sin( θ )] cos( θ ) Set dr/dθ fo etema sole θ fo θ : cos( θ ) Detemine whethe 45 is a cos 45 d R maimum o a minimum: [ ( ) sin θ ] 95 Pictue the Poblem We can use constantacceleation equations to epess the coodinates of a bullet in fliht on the moon as a function of t. Eliminatin this paamete will ield an epession fo as a function of that we can use to find the ane of the bullet. The necessit that the centipetal acceleation of an object in obit at the suface of a bod equal the acceleation due to ait at the suface will allow us to detemine the equied muzzle elocit fo obital motion. dθ 4 θ 45 θ 45 < R is a maimum at θ 45 (a) Usin a constant-acceleation equation, epess the coodinate of a bullet in fliht on the moon: t at o, because, a cosθ, ( cosθ )t

56 78 Chapte 3 Usin a constant-acceleation equation, epess the coodinate of a bullet in fliht on the moon: Eliminate the paamete t to obtain: When R: Substitute numeical alues ealuate R: t at o, because, a moon sinθ, ( sinθ ) t moont moon ( tanθ ) cos θ R cos θ R sin θ moon ( tanθ ) R moon ( m/s) 485km 9 5 R sin m.67 m/s (b) Epess the condition that the centipetal acceleation must satisf fo an object in obit at the suface of the moon: This esult is pobabl not e accuate because it is about 8% of the moon s adius (74 km). This bein the case, we can no lone assume that the ound is flat because of the cuatue of the moon. a c Sole fo ealuate : 6 (.67 m/s )(.74 m) moon moon.7 km/s 96 Pictue the Poblem We can show that R/R / b diffeentiatin R with espect to then usin a diffeential appoimation. Diffeentiate the ane equation with espect to : dr d d d sin θ R sin θ

57 Motion in One Two Dimensions 79 Appoimate dr/d b R/ : R R Sepaate the aiables to obtain: R R i.e., fo small chanes in ait ( ± ), the factional chane in R is lineal opposite to the factional chane in. Remaks: This tells us that as ait inceases, the ane will decease, ice esa. This is as it must be because R is inesel popotional to. 97 Pictue the Poblem We can show that R/R / b diffeentiatin R with espect to then usin a diffeential appoimation. Diffeentiate the ane equation with espect to : Appoimate dr/d b R/ : Sepaate the aiables to obtain: dr d d sin sin d θ θ R R R R R i.e., fo small chanes in the launch elocit ( ± ), the factional chane in R is twice the factional chane in. Remaks: This tells us that as launch elocit inceases, the ane will incease twice as fast, ice esa. 98 Pictue the Poblem Choose a coodinate sstem in which the oiin is at the base of the suface fom which the pojectile is launched. Let the positie diection be to the iht the positie diection be upwad. We can appl constant-acceleation equations to obtain paametic equations in time that elate the ane to the initial hoizontal speed the heiht h to the initial upwad speed. Eliminatin the paamete will leae us with a quadatic equation in R, the solution to which is the esult we ae equied to establish. Wite the constant-acceleation equations fo the hoizontal etical pats of the pojectile s t

58 8 Chapte 3 motion: ( ) h t t whee cosθ sinθ Sole the -component equation fo time: Usin the -component equation, eliminate time fom the -component equation to obtain: t h cosθ cos θ ( tanθ ) When the pojectile stikes the ound its coodinates ae (R, ) ou equation becomes: h R cos θ ( tanθ ) R Usin the plus sin in the quadatic fomula to ensue a phsicall meaninful oot (one that is positie), sole fo the ane to obtain: R h sin θ sin θ *99 Pictue the Poblem We can use tionomet to elate the maimum heiht of the pojectile to its ane the sihtin anle at maimum eleation the ane equation to epess the ane as a function of the launch speed anle. We can use a constant-acceleation equation to epess the maimum heiht eached b the pojectile in tems of its launch anle speed. Combinin these elationships will allow us to conclude that tanφ tanθ. Refein to the fiue, elate the maimum heiht of the pojectile to its ane the sihtin anle φ: tanφ h R Epess the ane of the ocket use the tionometic identit sin θ sinθ cosθ to ewite the epession as: Usin a constant-acceleation equation, elate the maimum heiht of a pojectile to the etical component of its launch speed: Sole fo the maimum heiht h: R sin(θ ) sinθ cosθ h o, because sinθ, sin θ h h sin θ

59 Motion in One Two Dimensions 8 Substitute fo R h simplif to obtain: sin θ tanφ tanθ sinθ cosθ Pictue the Poblem In the absence of ai esistance, the hoizontal etical displacements of the pojectile ae independent of each othe descibable b constant-acceleation equations. Choose the oiin at the fiin location with the coodinate aes as shown in the fiue use constant-acceleation equations to elate the etical displacement to etical component of the initial elocit the hoizontal elocit to the hoizontal displacement the time of fliht. (a) Usin a constant-acceleation equation, epess the etical displacement of the pojectile as a function of its time of fliht: o, because a, t ( ) t a t ( ) t Sole fo : ( t) t Substitute numeical alues ealuate : 45m ( 9.8m/s )( s) s m/s (b) The hoizontal elocit emains constant, so: * Pictue the Poblem In the absence of ai esistance, the acceleation of the stone is constant the hoizontal etical motions ae independent of each othe. Choose a coodinate sstem with the oiin at the thowin location the aes oiented as shown in the fiue use constant- acceleation equations to epess the coodinates of the stone while it is in fliht. 3 m s t 5 m/s

60 8 Chapte 3 Usin a constant-acceleation equation, epess the coodinate of the stone in fliht: Usin a constant-acceleation equation, epess the coodinate of the stone in fliht: Refein to the diaam, epess the elationship between θ, at impact: t a t o, because, a, t t a t o, because, a, t tan θ Substitute fo sole fo the time to impact: Sole fo t to obtain: t tan θ t t tanθ t Refein to the diaam, epess the elationship between θ, L, at impact: L cosθ tanθ Substitute fo to obtain: Substitute fo t sole fo L to obtain: t L Lcosθ tanθ cosθ Pictue the Poblem The equation of a paticle s tajecto is deied in the tet so we ll use it as ou statin point in this deiation. We can elate the coodinates of the point of impact (, ) to the anle φ use this elationship to eliminate fom the equation fo the cannonball s tajecto. We can then sole the esultin equation fo elate the hoizontal component of the point of impact to the cannonball s ane. The equation of the cannonball s tajecto is ien in the tet: Relate the components of a point on the ound to the anle φ: ( ) (tanθ ) cos θ () (tan φ) Epess the condition that the tanφ (tanθ) cannonball hits the ound: ( ) cos θ

61 Motion in One Two Dimensions 83 Sole fo to obtain: Relate the ane of the cannonball s fliht R to the hoizontal distance : Substitute to obtain: cos θ(tanθ tanφ) R cosφ cos θ(tanθ tan R cosφ φ ) Sole fo R: 3 Pictue the Poblem In the absence of ai esistance, the acceleation of the ock is constant the hoizontal etical motions ae independent of each othe. Choose the coodinate sstem shown in the fiue with the oiin at the base of the buildin the aes oiented as shown appl constant-acceleation equations to elate the hoizontal etical displacements of the ock to its time of fliht. R cos θ(tanθ tanφ) cosφ Find the hoizontal etical components of : Usin a constant-acceleation equation, epess the hoizontal displacement of the pojectile: Usin a constant-acceleation equation, epess the etical displacement of the pojectile: Sole the -displacement equation fo t: Substitute t into the epession fo : Sole fo to obtain: Find t at impact: cos53.6 sin (. ) t m t 6 m t ( t) (.799 ) t ( t) m.6 t (.799 ) t ( 4.9 )( ) m m/s t t.8m/s (.8m/s) m 3.8s cos53

62 84 Chapte 3 Usin constant-acceleation equations, find at impact: 6.5m/s t m/s Epess the elocit at impact in ecto fom: (6.5 m/s) iˆ (.6 m/s) ˆj 4 Pictue the Poblem The ball epeiences constant acceleation, ecept duin its collision with the wall, so we can use the constant-acceleation equations in the analsis of its motion. Choose a coodinate sstem with the oiin at the point of elease, the positie ais to the iht, the positie ais upwad. Usin a constant-acceleation equation, epess the etical displacement of the ball as a function of t: When the ball hits the ound, m: Sole fo the time of fliht: Find the hoizontal distance taeled in this time: ( ) m t t ( m/s) t ( 9.8m/s )( t) t fliht t. s ( m/s) (. s). m The distance fom the wall is: 4 m 8. m Hittin Taets Related Poblems 5 Pictue the Poblem In the absence of ai esistance, the acceleation of the pebble is constant. Choose the coodinate sstem shown in the diaam use constantacceleation equations to epess the coodinates of the pebble in tems of the time into its fliht. We can eliminate the paamete t between these equations sole fo the launch elocit of the pebble. We can detemine the launch anle fom the sihtin infomation, once the ane is known, the time of fliht can be found usin the hoizontal component of the initial elocit.

63 Motion in One Two Dimensions 85 Refein to the diaam, epess θ 4.85 m in tems of the ien distances: θ tan m Use a constant-acceleation equation to epess the hoizontal position of the pebble as a function of time: Use a constant-acceleation equation to epess the etical position of the pebble as a function of time: Eliminate the paamete t to obtain: At impact, R: t a t o, because, cosθ, a, ( cosθ )t () t a t o, because, sinθ, a, ( sinθ ) t t ( tanθ ) cos θ R cos θ ( tanθ ) R Sole fo to obtain: R sin θ Substitute numeical alues ealuate : Substitute in equation () to elate R to t fliht : (4 m)(9.8m/s ) sin3.8 R ( cosθ ) t fliht 4.6m/s Sole fo ealuate the time of fliht: ( 4.6 m/s) t 4m cos6.9 fliht.99s *6 Pictue the Poblem The acceleation of the ball is constant (zeo hoizontall eticall) the etical hoizontal components ae independent of each othe. Choose the coodinate sstem shown in the fiue assume that t ae unchaned b thowin the ball slihtl downwad. Epess the hoizontal displacement of the ball as a function of time: ( ) t a t

64 86 Chapte 3 o, because a, t Sole fo the time of fliht if the ball wee thown hoizontall: t 8.4 m 37.5 m/s.49s Usin a constant-acceleation equation, epess the distance the ball would dop (etical displacement) if it wee thown hoizontall: Substitute numeical alues ealuate : The ball must dop an additional.6 m befoe it ets to home plate. ( ) t a t o, because a, t ( ) ( 9.8m/s )(.49s).8m (.5.8) m.3 m aboe ound Calculate the initial downwad speed the ball must hae to dop.6 m in.49 s:.6m.49s.6m Find the anle with hoizontal: θ tan.9 tan.6 m/s 37.5m/s Remaks: One can eadil show that 37.5 m/s to within %; so the assumption that t ae unchaned b thowin the ball downwad at an anle of.93 is justified. 7 Pictue the Poblem The acceleation of the puck is constant (zeo hoizontall eticall) the etical hoizontal components ae independent of each othe. Choose a coodinate sstem with the oiin at the point of contact with the puck the coodinate aes as shown in the fiue use constant-acceleation equations to elate the aiables, the time t to each the wall,,, θ. Usin a constant-acceleation equation fo the motion in the diection, epess as a function of the puck s displacement : a o, because a,

65 Motion in One Two Dimensions 87 Sole fo ealuate : (.8m)( 9.8m/s ) 7.4m/s Find t fom the initial elocit in the diection: t 7.4m/s 9.8m/s.756s Use the definition of aeae elocit to find : t.m.756s 5.9m/s Substitute numeical alues ealuate : Substitute numeical alues ealuate θ : ( 5.9 m/s) ( 7.4m/s) 7.5m/s θ tan 5. tan 7.4m/s 5.9 m/s 8 Pictue the Poblem In the absence of ai esistance, the acceleation of Calos his bike is constant we can use constant-acceleation equations to epess his coodinates as functions of time. Eliminatin the paamete t between these equations will ield as a function of an equation we can use to decide whethe he can jump the ceek bed as well as to find the minimum speed equied to make the jump. (a) Use a constant-acceleation equation to epess Calos hoizontal position as a function of time: Use a constant-acceleation equation to epess Calos etical position as a function of time: t a t o, because, cosθ, a, ( cosθ )t t a t o, because, sinθ, a, ( sinθ ) t t

66 88 Chapte 3 Eliminate the paamete t to obtain: Substitute R to obtain: Sole fo ealuate R: cos θ ( tanθ ) R cos θ ( tanθ ) R 4.3m (.m/s) R sin( θ ) 9.8m/s He should appl the bakes! sin (b) Sole the equation we used in the peious step fo,min :,min R sin ( θ ) Lettin R 7 m, ealuate,min : ( 7m)( 9.8m/s ) 9 Pictue the Poblem In the absence of ai esistance, the bullet epeiences constant acceleation alon its paabolic tajecto. Choose a coodinate sstem with the oiin at the end of the bael the coodinate aes oiented as shown in the fiue use constant-acceleation equations to epess the coodinates of the bullet as functions of time alon its fliht path.,min sin 4.m/s 5. km/h Use a constant-acceleation equation to epess the bullet s hoizontal position as a function of time: Use a constant-acceleation equation to epess the bullet s etical position as a function of time: Eliminate the paamete t to obtain: t a t o, because, cosθ, a, ( cosθ )t t a t o, because, sinθ, a, ( sinθ ) t t ( tanθ ) cos θ

67 Motion in One Two Dimensions 89 Let when R to obtain: Sole fo the anle aboe the hoizontal that the ifle must be fied to hit the taet: Substitute numeical alues ealuate θ : Refein to the diaam, elate h to θ sole fo ealuate h: R cos θ ( tanθ ) R θ sin R ( m)( 9.8m/s ) θ sin ( 5 m/s).45 Note: A second alue fo θ, 89.6 is phsicall uneasonable. h tanθ m h ( m) tan(.45 ).785m Geneal Poblems Pictue the Poblem The sum diffeence of two ectos can be found fom the components of the two ectos. The manitude diection of a ecto can be found fom its components. (a) The table to the iht summaizes the components of A B. Vecto component component (m) (m) A B (b) The table to the iht shows the components of S. Vecto component component (m) (m) A B S.57.7 Detemine the manitude diection of S fom its components: S S S.59m, because S is in the st S tan θ S 7.5 S

68 9 Chapte 3 (c) The table to the iht shows the components of D : Vecto component component (m) (m) A B D.59. Detemine the manitude diection of D fom its components: D D D.m, because D is in the nd quadant, D tan θ D 97.5 D * Pictue the Poblem A ecto quantit can be esoled into its components elatie to an coodinate sstem. In this eample, the aes ae othoonal the components of the ecto can be found usin tionometic functions. The components of ae elated to thouh the sine cosine functions: sin3 cos3 4.9m/s 8.5 m/s Pictue the Poblem The fiue shows two abita, co-plana ectos that (as dawn) do not satisf the condition that A/B A /B. Because A Acosθ cosθ A B B cosθ B, fo the cosθ B condition to be satisfied. A A/B A/B if onl if A B ae paallel (θa θb) o on opposite sides of the -ais (θa θb). 3 Pictue the Poblem We can plot the path of the paticle b substitutin alues fo t ealuatin coodinates of. The elocit ecto is the time deiatie of the position ecto. (a) We can assin alues to t in the paametic equations (5 m/s)t ( m/s)t to obtain odeed pais (, ) that lie on the path of the paticle. The path is shown in the followin aph:

69 Motion in One Two Dimensions 9 5 (m) (m) (b) Ealuate d dt : [( 5m/s) t iˆ ( m/s) ˆj ] d d t dt dt ( 5m/s) iˆ ( m/s)j ˆ Use its components to find the manitude of :. m/s 4 Pictue the Poblem In the absence of ai esistance, the hamme epeiences constant acceleation as it falls. Choose a coodinate sstem with the oiin coodinate aes as shown in the fiue use constant-acceleation equations to descibe the coodinates of the hamme alon its tajecto. We ll use the equation descibin the etical motion to find the time of fliht of the hamme the equation descibin the hoizontal motion to detemine its ane. Usin a constant-acceleation equation, epess the coodinate of the hamme as a function of time: Usin a constant-acceleation equation, epess the coodinate of the hamme as a function of time: t a t o, because, cosθ, a, ( cosθ )t t a t o, because h, sinθ, a, ( ) h sinθ t t

70 9 Chapte 3 Substitute numeical alues to obtain: Substitute the conditions that eist when the hamme hits the ound: Sole fo the time of fall to obtain: Use the -coodinate equation to find the hoizontal distance taeled b the hamme in.4 s: m ( 4 m/s)( sin 3 ) ( 9.8 ) t m/s m t.4 s R ( 4 m/s) ( 9.8 ) t m/s sin 3 t ( 4m/s)( cos3 )(.4s) 4.9m t 5 Pictue the Poblem We ll model Zacchini s fliht as thouh thee is no ai esistance, hence, the acceleation is constant. Then we can use constant- acceleation equations to epess the coodinates of Zacchini s motion as functions of time. Eliminatin the paamete t between these equations will leae us with an equation we can sole foθ. Because the maimum heiht alon a paabolic tajecto occus (assumin equal launch lin eleations) occus at half ane, we can use this same epession fo as a function of to find h. Use a constant-acceleation equation to epess Zacchini s hoizontal position as a function of time: Use a constant-acceleation equation to epess Zacchini s etical position as a function of time: Eliminate the paamete t to obtain: t a t o, because, cosθ, a, ( cosθ )t t a t o, because, sinθ, a, ( sinθ ) t t ( tanθ ) cos θ Use Zacchini s coodinates when he tanθ R R cos θ ls in a safet net to obtain: ( )

71 Motion in One Two Dimensions 93 Sole fo his launch anle θ : θ sin R Substitute numeical alues ealuate θ : θ ( 53m)( 9.8m/s ) sin ( 4. m/s) 3.3 Use the fact that his maimum heiht was attained when he was halfwa thouh his fliht to obtain: h ( tanθ ) R R cos θ Substitute numeical alues ealuate h: h ( tan 3.3 ) 53m 9.8m/s 53m cos 3.3 ( 4. m/s) 8.6m 6 Pictue the Poblem Because the acceleation is constant; we can use the constantacceleation equations in ecto fom the definitions of aeae elocit aeae (instantaneous) acceleation to sole this poblem. (a) The aeae elocit is ien b: a t t (3 m/s)ˆ i (.5 m/s) ˆj The aeae elocit can also be epessed as: a a Substitute numeical alues to obtain: (b) The acceleation of the paticle is ien b: ( m/s) iˆ ( m/s) ˆj a t t ( m/s ) iˆ ( 3.5 m/s ) ˆj (c) The elocit of the paticle as a function of time is: () a ] iˆ t t [(m/s) ( m/s ) t [(m/s) ( 3.5 m/s ) ] ˆj t (d) Epess the position ecto as a function of time: ( t) t at

72 94 Chapte 3 Substitute numeical alues ealuate ( t) ( t) [(4 m) (m/s) t (m/s ) t : ] iˆ [(3 m) ( m/s) t (.75 m/s ) t ] ˆj *7 Pictue the Poblem In the absence of ai esistance, the steel ball will epeience constant acceleation. Choose a coodinate sstem with its oiin at the initial position of the ball, the diection to the iht, the diection downwad. In this coodinate sstem a. Lettin (, ) be a point on the path of the ball, we can use constant-acceleation equations to epess both as functions of time, usin the eomet of the staicase, find an epession fo the time of fliht of the ball. Knowin its time of fliht, we can find its ane identif the step it stikes fist. The anle of the steps, with espect.8 m to the hoizontal, is: θ tan 3..3m Usin a constant-acceleation equation, epess the coodinate of the steel ball in its fliht: Usin a constant-acceleation equation, epess the coodinate of the steel ball in its fliht: The equation of the dashed line in the fiue is: t at o, because a, t t at o, because,, a, t t tanθ Sole fo the fliht time: t tanθ Find the coodinate of the lin position: tanθ tanθ Substitute the anle detemined in the fist step: ( 3m/s) 9.8m/s tan3. m The fist step with >. m is the 4th step.

73 8 Pictue the Poblem Inoin the influence of ai esistance, the acceleation of the ball is constant once it has left ou h we can use constant-acceleation equations to epess the coodinates of the ball. Elimination of the paamete t will ield an equation fom which we can detemine. We can then use the equation to epess the time of fliht of the ball the equation to epess its ane in tems of,,θ the time of fliht. Motion in One Two Dimensions 95 Use a constant-acceleation equation to epess the ball s hoizontal position as a function of time: Use a constant-acceleation equation to epess the ball s etical position as a function of time: Eliminate the paamete t to obtain: Fo the thow while stin on leel ound we hae: Sole fo : t a t o, because, cosθ, a, ( cosθ )t () t a t o, because, sinθ, a, ( ) θ () sin t t ( tanθ ) cos θ ( tanθ ) cos θ sin θ sin ( 45 ) At impact equation () becomes: ( ) θ t t sin fliht fliht Sole fo the time of fliht: t ( sinθ sin ) θ fliht

74 96 Chapte 3 Substitute in equation () to epess the ane of the ball when thown fom an eleation at an anle θ with the hoizontal: R ( cosθ ) t fliht ( cosθ ) ( sinθ sin θ ) cosθ ( sinθ sin θ ) Substitute θ, 3, 45 : ( ).4 ( 3 ).73 ( 45 ).6 9 Pictue the Poblem Choose a coodinate sstem with its oiin at the point whee the motoccle becomes aibone with the positie diection to the iht the positie diection upwad. With this choice of coodinate sstem we can elate the coodinates of the motoccle (which we e teatin as a paticle) usin Equation 3-. (a) The path of the motoccle is ien b: Fo the jump to be successful, h < (). Solin fo, we find: () (tanθ) cos θ min > cosθ ( tanθ h) (b) Use the alues ien to obtain: min > 6. m/s o 58.mph (c) In ode fo ou epession fo min to be eal alued; i.e., to pedict alues fo min that ae phsicall meaninful, tanθ h >. h ma < tanθ The intepetation is that the bike "falls awa" fom taelin on a staiht-line path due to the fee-fall acceleation downwads. No matte what the initial speed of the bike, it must fall a little bit befoe eachin the othe side of the pit.

75 Motion in One Two Dimensions 97 Pictue the Poblem Let the oiin be at the position of the boat when it was enulfed b the fo. Take the diections to be east noth, espectiel. Let BW be the elocit of the boat elatie to the wate, BS be the elocit of the boat elatie to the shoe, WS be the elocit of the wate with espect to the shoe. Then BS BW WS. θ is the anle of WS with espect to the (east) diection. (a) Find the position ecto fo the boat at t 3 h: boat {( 3 km)( cos35 ) t} iˆ {( 3 km)( sin35 ) t 4 km} {(.6 km) t} iˆ {(.6 km) t 4 km}j ˆ ˆj Find the coodinates of the boat at t 3 h: Simplif the epessions inolin equate these simplified epessions to the components of the position ecto of the boat: Diide the second of these equations b the fist to obtain: [( km/h) cos35 WS cosθ ]( 3h) km/h sin35 WS sinθ [( ) ]( 3h) 3 WS cosθ.4 km/h 3 WS sinθ.586 km/h.586 km tanθ.4km o.586 km θ tan 6.4 o 4.4.4km Because the boat has difted south, use θ 4.4 to obtain: WS.4km/h 3 cosθ cos ( 4.4 ).98 km/h atθ 4.4

76 98 Chapte 3 (b) Lettin φ be the anle between east the pope headin fo the boat, epess the components of the elocit of the boat with espect to the shoe: BS, ( km/h) cosφ (.98 km/h) cos(4.3 ) BS, ( km/h) sinφ (.98 km/h) sin(4.3 ) Fo the boat to tael nothwest: Substitute the elocit components, squae both sides of the equation, simplif the epession to obtain the equations: BS, BS, sinφ cosφ.33, sin φ cos φ sinφ cosφ.77, sin(φ).77 Sole fo φ: φ 9.6 o 4.4 Because the cuent pushes south, the boat must head moe nothel than 35 : (c) Find BS : To find the time to tael 3 km, diide the distance b the boat s actual speed: Usin 9.6, the coect headin is 39.6 west of noth. BS, 6.84 km/h BS B /cos km/h t (3 km)/(9.68 km/h) 3.3h 3h 8min * Pictue the Poblem In the absence of ai esistance, the acceleation of the pojectile is constant the equation of a pojectile fo equal initial final eleations, which was deied fom the constant-acceleation equations, is applicable. We can use the equation iin the ane of a pojectile fo equal initial final eleations to ealuate the anes of launches that eceed o fall shot of 45 b the same amount. Epess the ane of the pojectile as a function of its initial speed anle of launch: Let θ 45 ± θ: Because cos( θ) cos(θ) (the cosine function is an een function): R sin θ R sin cos ( 9 ± θ ) ( ± θ ) R (45 θ ) R(45 θ )

77 Pictue the Poblem In the absence of ai esistance, the acceleation of both balls is that due to ait the hoizontal etical motions ae independent of each othe. Choose a coodinate sstem with the oiin at the base of the cliff the coodinate aes oiented as shown use constant-acceleation equations to elate the components of the ball s speed. Motion in One Two Dimensions 99 Independentl of whethe a ball is thown upwad at the anle α o downwad at β, the etical motion is descibed b: a h The hoizontal component of the motion is ien b: Find at impact fom its components: h h