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# hwsol2 - One-Dimensional Kinematics with Constant...

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Unformatted text preview: One-Dimensional Kinematics with Constant Acceleration :x ( t:) ::. ~; + Vi t .,..t :t a. cAt"oS 1'1 "",c. e."" e.IJi) ().. s 0.. In answering the following questions, assume that the acceleration is constant and nonzero: a =1= O. Part A: The quantity represented by xis a function of time (i.e., is not constant). x:: X{t) r e.p r t,S e,11'".r at j /j r ~tt Ce.., (D"" .functiol' of. time. Part B: The quantity represented by x1is a function of time (i.e., is not constant). ~i = initicd poSi+ion (no+ tl../-vncfiolt of- fime) CD" S-ra.."t Part C: The quantity represented by 1Jlisa function of time (i.e., is not constant). \I; ::. ,"t\ i t i 0\.1 V ~ I 0 C i ry -the- CO 11S ro..n -J- Part D: The quantity represented by vis a function of time (i.e., is not constant). ft..frt!.sents v~JoCJ'fy os o..fu'lc1-t'on.,f t \/~ vet) i. e. V (-t )::: Vi + ~;Part E: Which of the given equations is not an explicit function of tand is therefore useful when you don't know or don't need the time? Ih~ 1'(;"em().ti(~~ e~u~icn: [V (t;)] ~ v(t)= Vi Vi;l of. ~ 4. ( X (t) - X,' ) 0. Part F: A particle moves with constant acceleration velocity at what instant in time? The expression 'Vi + atrepresents the particle's +-o..t =) ref re,s foY' t. n 7-J ; Y1 S To.n to..n e Ou.J vt.lo c" t y t Z. 0 o..=;to.. Part G: What is the equation describing the position of particle B? For Pa.rtic.l e. 13: v: Y;a V,' t-~ "i ~1f X6 (t):; X; ,he. ve)o c.;ty ~ (;l./).) (t-t I):I.:=. Xi+- -k v, ( + -t,) t- D.. (+-- f,) -k v. I ~ Part H: At what time does the velocity of particle B equal that of particle A? ..5 v 0.. f ioY' rot": A: Vlt)~ v,' +o.:t ~ / 8: V(t)::: Yav,' or;). 0... Ct-t I ) vt,IOGI tl es se,+,. v, +-CA.t :: l .,.. (t: -t,) ~~ 20.. t I - e~Cltll ~ Vi = :J 0.-1;- a.. i ~.:: t-~t ~ 0.. '_---=--=-] [+~n + E.. l What Velocity vs. Time Graphs Can Tell You Part A: What is the initial velocity of the particle, t'O? vx(m/s) 2.0 1.5 V~loG; ry (X ....o..)("/'S) l 1"1 j 1- }'a..ll y /; I (). 5 n'I/s 1.0 I / i ( OAf o Part B: What is the total distance LlXtraveled by the particle? f{e.~i on ~: "'i::: 0, 5 rvsls a,...:::: ~~~ , \ I I (S) I0 20 30 40 50 . 1: ~ If (~o) ~ 2- 0.07 S M/S ~ X =- o. 5 JII1/s :t 5m riO. 075 "S'" (;). 0) ~X%:. ~ e ~;o" II V.':' ::l.O Mis V; (J..::::.u p.. e?/ 01-\ ur =- ~." ""'/s D >c:. '~.O,.ys (;"0)::::' ~ a..:: - o. ~ rttl s ':l. A X:= :2. 6 nt/s (I - C - ~ (I). ;). "s a) (I 0) 6. X= ~5 0.. :: ~ t l(o +'10 :::-fZs ~ m 1 6 X tIC ::. I () Part C: What is the average acceleration Ga-vofhe particle over the first 20.0 seconds? t cJ.t = C h<l.I19 (. (v) Ch().n~e.- (t) (J. 0 - 0.5) :lOs "'/s_ ::::. O. 0 75 Mls:l. . . . e P art:D Wh at IS the Instantaneous acce I . a0 f th e partIc I at t = 45.0 s? eratlOn . o.-m = (. - ~ .O),.,,?" .:::. lOs 0~ O. ~O m/! ~ ~ o Ve.,... ts 0.. D-- k <:'0 n.s+o-n + -r-= '-I5.s +A rrr .. , Is 0 -0. :l<' l'I1/s ~ Part E: Which of the graphs shown below is the correct acceleration vs. time plot for the motion described in the previous parts? p.. e.~'-o" T. 1< e. e i(') :rr /\ ~e~)"~Y) 0- "70 e,f't.t-ore. 13 1..5 . +A~ 0..=-0 rrr (1<0 C6r r eC t c.A~ " c.. 'e.., Motion of a Shadow Find an expression for V., the magnitude of the velocity tIme . . t At time t: v"of the top of the object's shadow, at Not e.. t' he a..&d; f..l I'~ -/-ion 1-1 S of e CA. J1 cA fA -e.. f l' fj ure Q:i_e ~~ _ L ~ ~ Loo'" 0.;1- the. -I ;~LJre.. - s ee, J. f:. vtl --- +r; {)..nfj Ie s 1-0. Y\ bLA- +- e::. so ..b- -X tiL h vt S e.,+ ~ u ().. ( ",-I t-~n e = +0 fincA- ..!L -vt ) We. H -C -/ how --, -vl: J, L _ H it! ~ Y'lvs-r f!.in<A H eho.."fl e.s +iM~ +. tiYlJ.. this, we. fo../{e. the. time. c!e.riV"'+"ve. ... @_.t../~L) rJt - olr l;t h L I ) GL"ot V GLY<!. ",'I cOlll,;f(J,.YI+S -1 ... d.nol. (c.o~ a.1"e. n4t o..H-ecfed. ut"' ~ _ b1. J (..L) ~ !J, . fJ.. Sf)... / + ya ott - v' txt c.o.- c 10 CJO I~) ' t V --- ~H-_-hL Jt - ~ V t'J..:;' V.s IJ! e 1Ilj-J. f-h e. l.f u,J. J VII h j c.J.. is the. MAS 1'1" +u oA ! o.f i(;: Iv s 1== U vt~ Overcoming a Head Start Cars A and B are racing each other along the same straight road in the following manner: "I" "I"' hea d start an d"IS a d'Istance DAb eyon d t he startIng Ine at t= 0. Th e startIng me IS at ;! = at a constant speed VA. Car B starts at the starting line but has a better engine than Car A, travels at a constant speed 'Vn, which is greater thantJ l. Part A: How long after Car B started the race will Car B catch up with Car A? Car A has a 0. C ar A trave I s and thus Car B I h .. el{, Va. -r; X (f- JA= 01'\ 0 f rn 0 "I-; 0 n fa," c o-r A: DA .... VA t" and for etA."" 13: {3 X (t) =- Va t e.n >e(..f)A:::' X (-1:)8 .I fhey m~et wA t-Aer~f-ore 0A +0A::; r VA t .:: s t V - VA ) (v 6 - t t=- OA. (1/8VA) Part B: How far from Car B's starting line will the cars be when Car B passes Car A? We.. f\no \AI' ~{tJ(3:::'''8 t 0..;1' timt.. t~e.d"Ch . ~ OA >' (1113 -1104) X B ::: _Ve OA__ (Va -VA) or Jin"t''Y \/6 - t~~ rcA Clear The Runway To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeoff velocity, depends on several factors, including the weight of the aircraft and the wind velocity. III Part A: A plane accelerates from rest at a constant rate of 5.00 along a runway that is 1800 In long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the time t'1'needed to take off? / s- ., re.call pJrJCjCji,.,~ x in = X .,..V. t" 1" ~ c>..t ~ h t,r e... Xi ::::vi=-- 0 HrtOlV n vo.lve..-.s ~ 4. ) ~0 0 "'" =: '1. \ 3: { s WJ/s ) t. t :::.~().s:1.. 7 [f~ Jf:,.3 3iJ Part B: What is the speed We... I~YlOW VTo of the plane as it takes off? V(-l:) = VI +o..t Vi::;O Vi 0 =. 5. 0 ~ ~ (:;.b . 8' '3 .s) V.,o:::' 13'-1 m/5 Part C: What is the distance j dll~t traveled by the plane in the first second of its run? ( d I rs -r := f (5 mj.5 d';j r.s '1. ) 1 s) '2. -r :::. 1. f, 6 ..., Part D: What is the distance db'lL traveled by the plane in the last second before taking off? CA.. Y 1-0 ().f'1' r-o...ch rJ, ;.5 is+J1 e, : J. is t-o.n C~ liU f- .= ti"'Oh:hl - 0<. ;.25.3"3 (p f ()..~6 -If 3 .5 Y'""un e-S.s +io..l\ y o..vo icA.i" 5 ~x ~ e..s.s I ve ca.l c () Ict+1 o~ J. ~ X { ~ ~ . 83)::: J,s. (5. 0 0 ~ ~ ) (1S . 2 3) :L =- /6 6 7. 9 7 m Lv' 0.. e,J'\ -r h e..c.a.sy ) boX (:J.l,. g 3) - If:> X( ;15. 8'3)::: I J 00 .. -I G 6 7. 91m == L I 3 :< ~ Part E: What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? ~;::; Cf DO rn ( m i 0.. pOin+ 0 f rlln liUo-y) X' ~ ~ 0 0 yY\ =- t ( .00 5 ""/s~) f ')., t .=. 18. 't 7 s o...+t':::./g.Cf7s 7-0 ...eo...c h ru Yl W(X Y M icA. po i n-r { po,' r V (, K. C[ 7) == CA.ll X . cr 7).:::; 9'(. 87 fYI/s 0- +(j rn i d. Vt 1'7' . 8 7 "'/S J 3l.f rn/S ~ TCJ ~() of fo.. ~e If s~ e eO< ! E1Jualty VI).:: Vj 'J.... va-IleA. o..flpl'"Oo.Ch: 6X : -r;to....6 x q 00"" v. -- D I 0 v; :l:::. :l (5. Vof ~/s"1)(lf60 1t1) =- 9lf. g 1 '0/5 -7 s a.m (2 r'"" e S fA If ~ Problem 2.23 Automobile Airbags. The human body can survive a negative acceleration trauma incident (sudden stop) 250 mN~. 25q if the magnitude of the acceleration is less than (approximately' ). If you are in an automobile accident with an initial speed of 120 km/h and are stopped by an airbag that inflates from the dashboard, over what distance must the airbag stop you for you to survive the crash? J - 7.0 I"\m hr - I:lOJ - OOOm ? 36005 3 3. ~ 3 mls r e..c.o..ll '). '. . Vt 'l.:::. Vi 'l.. -%. =-0 oJ- 2~ A X ~ ~ \I; ~ 33. 3 3~/s ~ ~ ;J. 50 rn/s"1 ?I u ~ j it'\ j ; 1\ V t).. ~l t>. b ( e 5 .' . (3 3. 3 3 "'/5) ~ ;l. (;2 50 '"/s"l) b X I L:i. X 2:. .::l. ;7.;;l ;y Problem 2.44 A hot-air balloonist, rising vertically with a constant velocity of magnitudelJ = 5.00 mis, releases a sandbag at an instant when the balloon is a height II. = 40.0 In above the ground ~j . After it is released, the sandbag is in free fall. For the questions that follow, take the origin of the coordinate system used for measuring displacements to be at the ground, and upward displacements to be positive. Part A: Compute the position of the sandbag at a time 0.335 S after its release. X (t CA.k 1~ \$I O. 0 m -I- 5 1"/5 t - t q. 8' - ~:1 t 1 -t :;:.O. 335.s J'Y\t X -:;,lO. D 5 ~ (0. 335s) -k 'i. 8' ~~(O. 335 s),). X=- 41.1~ -- . after its release. Part B: Compute the velocity of the sandbag at a time 0.335 S v(t );;.5 W'/.s - '[. 3 ~" ta..t- o. 335..> 11;:: 5";5 - ? l' 72. (0. 335 cS) v::::. /. --- 7 -:J.. ~!5 c: Compute the position of the sandbag at a time 1.30 S after its release. X ().30):: LlO -r 5 ~ (,. 3D 5) - .; 9. 8' ~ (t. 30.\$)'1.. Part x' ::::.3 g. ;).~ Part D: Compute the velocity of the sandbag at a time 1.30 S after its release. o...t t~ I. 36J V:::' 5 "'/05 - 1. 3' ?>- (f. 30S) it 11- ~ 1-_7.1'1 m/57 Part E: How many seconds after its release will the bag strike the ground? o iVI"cleoRe.~ion +he. +rAj e.c...f-ory into a r~~iQns I. 0:;; I{(.~ iO?\ v" - 5t ='1 t'" :5 {5Y '" 'fo,., . *~ 5 Lc D'/'I\ AX 1I I[ ,'..s A ~ ~x:;. y" ~ t ').. ') ..,vh""t ~ to LJ 7 11)c '" ,,~'" 40-6. )(::. 5 "'Is (3) - Y:. j I V 'l. Om -r /.1 ~ ::::. == 7,..1 ~ _/ i =- J ~'j('to) ~ ;;0=::;, ~ ~-a. TOt-lAl r-rrne,:/t ()J'f' V -ryv'.l~~Q.ll{b",J 1~f\(J Part F: With what magnitude of veloci~ does it strikt!? Pre, IY\ f E) we.. - M/ -rh e.. f' -t)me. e.(A/J.. rs; ~ W-t.. }t pec;..~ ~T\ow' \If'=' t ~f'ouJ'\c;L.l.= j l~(lf'o) +-~-a. ::::. :!> (j t- Vi - ~ ~ t - /~/:In J ~~(Y()}rv~ ~ ....... _ -- f ~('1(J)-rv~ Part G: What is the greatest height above the ground that the sandbag reaches? p ().. w~ c.. 0.. I c. u.lo..+eo\. bX r-r )to 2Cfe..CA.I, =we.. -r V t - ~~ t: k:::. ~ G 0.. l c.. r..d 0.. <.OL + .so A;" e. "",",.::;' '10,., 0.. IJ... + -, :;.. \/') I /;;l "3 j vQ. ,x. f e. 'i()rn +'...!... yI'7, ;)o- Problem 2.65 A ball starts from rest and rolls down a hill with uniform acceleration, traveling 170 4.10 S of its motion. How far did it roll during the first 4.10 S III during the second of motion? -r" t h c. ~x:::. / d" '-I. lOs : ~ fa. ('f./Os) 0. ('I. (I) V I" t-hf... = J '0) t!4 If.loJ.' I:l X.:: V ( 0/. I a) .,..::()... 'I. I()) ~::::I 70 YY\ ( ~/ ~ ~ II! 'l/o s M.>e. .k.cw: .. v.::. ()... l.(. I()) ( }Jq I 70 fY\ ::: a. ( '/. 10) J..}. t CA. (if. 10)~ ..=- f (~ ) ~ los ()): 0.. ::;, 6. 7'1;) ~ l plv~<jjnj L1 x':: -rAi::> b()...c.l~ l,,+o -k (&. 7lf 2. ~~) ('i. 10 is) ~ L1x= 5(,. 1m Problem 2.70 Two stunt drivers drive directly toward each other. At time t = 0 the two cars are a distance D apart, car 1 is at rest, and car 2 is moving to the left with speed Lb. Car 1 begins to move at t = 0, speeding up with a constant acceleration (lx. Car 2 continues to move with a constant velocity. Part A: At what time do the two cars collide? C~r ~ ?C=- D-..ln V\o)t I ,. t~ 'tV A J(,:::,x4 ~ 0::> en --rAe c.o.,....s c.o /I,'d.e : .::;'? C 0..(' ;;t 'X~=:.. Vb-t o us; n3 X.= - b..:t -~ fhe fJUo.d.ra.tic I D-.J.. /\ t ~-v 0 t a. V\o)< - F()~)'f1lJlo.: Jb 'J_ '1O-C - J.V<- ~ ... -/;""-t ~o.- - (/v,,' .y:l~)<b -vc) Part B: Find the speed of car 1 just before it collides with car 2. \I :: ~ O-j V:=.. o...x o..)C (J Vo ~~ ~c>.)( 0-) - Vo ..:: J va ~+),~ - ~0 -~ --- Problem 2.76 Egg Drop. You are on the roof of the physics building, 46.0 m above the ground [email protected] . Your physics professor, who is 1.80 tall, is walking alongside the building at a constant speed of 1.20 m/s. If you wish to drop an egg on your professor's head, how far from the building should the professor be when you release the egg? Assume that the egg is in free fall. 46.0 v = 1.20 m/s 1.80 m ",.~ III 1-10 w I"n~ 1-1) will if /a.}i(. +Ae (2,33 talt ~ / ('If, m -I. gM) == ;r B t- ~ 3.0D s !? T'" e.. .jhe q. a nrs:l. (n.:1- '",1 =. t =- pro I-e.ss or bviJoti"lj t\ e Q.cJ.s +0 h ~ 3S tllV(). Y t rc1'J"\ ;::n:~.S.5i n 3 01"' ... x= V t:::= I. ;;'0 5:60m) P 3s Problem 2.85 Juggling Act. A juggler performs in a room whose ceiling is a height 3.90 m above the level of his hands. He throws a ball upward so that it just reaches the ceiling. Part A: What is the initial velocity of the ball? ;\noVViY\:) (A.ISO Ax=-h::: Y:!lQ-t4 ~ =? ~::. j V~ ~!:J, ( I) (;2) V~.::. 2~(~)C) J:i;h U~in<j V.::: (J.) J;H" 8"'Y's ~ '3. , 0 n" -- v:::. g. 7lf mls ... Part B: What is the time required for the ball to reach the ceiling? USiY\3 0) t~ (Yi :) -=- - 2 (3. 'lOIt\) 't. 11'1/05 ~ t -:,0. 8 9 ;l s Part C: At the instant when the first ball is at the ceiling, he throws a second ball upward with two-thirds the initial velocity of the first. How long after the second ball is thrown did the two balls pass each other? rh~ e..~l)...~t-.I Vo-,=,O 2Go :~ ())I\ 0 (. h'\ 0+; 01\ for ball ~: X,(~)==h-~f)-c':l. for bo..l1)': fA .,tJ ~ V. o 3 = 2:.lP0 1\.9 X it ):;..J ;)..3 ~ 3" ..., - -& - ~ I 5t ~ ~o:::- s e. t+} rnes ~ e.f6 CA~ 1: x, (of)::. }(~ ( t ) h- y~ ~::: 3 J-~-5-;': t- ~J;),(bJ,'& . fi :l. Y~L, It; -t rfJ ;;: Part D: At what distance above the juggler's J..::. - 0.661.5_ hand do they pass each other? ~(f) = h3.90rn- Y.:t ~ t :l- ak t-f 0.. S5 ;x{t)= f(1.JOP~)(I).b69.s)~ YYl X{t)=- 1.766 --------------------- Problem 2.91 A student walks off the top of the eN Tower in Toronto, which has height 553 m and falls freely. His initial velocity is zero. The Rocketeer arrives at the scene a time of 5.15 G later and dives off the top of the tower to save the student. The Rocketeer leaves the roof with an initial downward speed 1,'0. In order both to catch the student and to prevent injury to him, the Rocketeer should catch the student at a sufficiently great height above ground so that the Rocketeer and the student slow down and arrive at the ground with zero velocity. The upward acceleration that accomplishes this is provided by the Rocketeer's jet pack, which he turns on just as he catches the student; before then the Rocketeer is in free fall. To prevent discomfort to the student, the magnitude of the acceleration of the Rocketeer and the student as they move downward together should be no more than five times g. Part A: What is the minimum height above the ground at which the Rocketeer should catch the student? ,Ae. s rvcJ.e,.~ Ve, 10 C I ~y when c.Q.tJf?n+; ~ ;, -e.. Q.jO I' V/'=?~o( ~ j,-;X t\,.s Tho(., ~f' tJ cl..U\ .,..'3 v e..l 0 ci +y r"OoC;' e.s ~ r~v",d.... h Vx ==;;l o e+f-; I\~ ~ (58 ) ~ Vx ':::..c 'J. V xLL 2 3 (J,-?c)':: 10 B" ~,B = 10Rf?<' -r ';1 fix. J, Jt ::;. :=[-7 J-.17-~ -t Part B: What must the Rocketeer's initial downward speed be so that he catches the student at the minimum height found in part (a)? fhe- stvcA-r'\t- is. -/-h e.. i1'\ Ire..e. f()..// 0 '1 for l s. .. I!lSO""l~ '160. Ern e.~ u AX'::' +-: c f h:..s m 6 f i h!) 6'J ='16 () .. Q.. 0 1') 81?'\ 4 fa.Xe. 1"0 C it i rn ".. Vo . ( t;:.. JI'1 r &> Cf s t h. e /J. e, -reo ~ us t- 6X~ \$o)l!i ?b9.s-[;./5s r V(J t-A (.s ~ o...p. CIS ve. r tJt e. sa.. m ~ d; oS -ra.n c:. -e. i n /e.s. s + ;h1e. .. ' ) l ~ -t-h.~(9.6fs-S.).ss) ::::."60.S'm f" f().;/! "j f(J v;, = 7'f. ~"7 yY& /s ...
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