Prof. Dr. I. Nasser T /16/2017

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1 Pro. Dr. I. Nasser T /16/017 Chapter Part 1 Moton n one dmenson Sectons -,, 3, 4, 5 - Moton n 1 dmenson We le n a 3-dmensonal world, so why bother analyzng 1-dmensonal stuatons? Bascally, because any translatonal (straght-lne, as opposed to rotatonal) moton problem can be separated nto one or more 1-dmensonal problems. Problems are oten analyzed ths way n physcs; a complex problem can oten be reduced to a seres o smpler problems. The rst step n solng a problem s to set up a coordnate system. Ths denes an orgn (a startng pont) as well as poste and negate drectons. We'll also need to dstngush between scalars and ectors. A scalar s somethng that has only a magntude, lke area or temperature, whle a ector has both a magntude and a drecton, lke dsplacement or elocty. In analyzng the moton o objects, there are our basc parameters to keep track o. These are tme, dsplacement, elocty, and acceleraton. Tme s a scalar, whle the other three are ectors. In 1 dmenson, howeer, t's dcult to see the derence between a scalar and a ector! The derence wll be more obous n dmensons. -3 Dsplacement The dsplacement, x, s the dstance moed n a gen drecton. Dsplacement: 1- s a ector quantty. - s poste the partcle moes n the poste drecton, and negate the partcle has moed n the negate drecton and 3- s measured n meters. Mathematcally, n one dmenson, the dsplacement s dened as: x x x where, x s the nal poston and x s the ntal poston o the partcle. x s a ector pontng rom x to x. Example: x 5 m and x 1 m, then x 1 m 5 m +7 m Comment: the poste result ndcates that the moton s n the poste drecton. Example: x 5 m and x 1 m, then x 1 m 5 m 4 m Comment: the negate result ndcates that the moton s n the negate drecton. H.W. Do Checkpont 1. - x Δx + x x 1

2 Pro. Dr. I. Nasser T /16/ Speed and elocty Example: Imagne that on your way to class one mornng, you leae home on tme, and you walk at 3 m/s east towards campus. Ater exactly one mnute you realze that you'e let your physcs assgnment at home, so you turn around and run, at 6 m/s, back to get t. You're runnng twce as ast as you walked, so t takes hal as long (30 seconds) to get home agan. There are seeral ways to analyze those 90 seconds between the tme you let home and the tme you arred back agan. One number to calculate s your aerage speed, whch s dened as the total dstance coered dded by the tme. I you walked or 60 seconds at 3 m/s, you coered 180 m. You coered the same dstance on the way back, so you went 360 m n 90 seconds. m m total dstance 3 60 s 6 30 s 360 m, s s m m dsplacement 3 60 s 6 30 s 0 m, s s Total elapsed tme 60 s 30 s 90 s total dstance traeled 360 m Aerage speed s ag 4, total tme taken 90 s dsplacement 0 m Aerage elocty ag 0, total tme taken 90 s In ths case, your aerage elocty or the round trp s zero, because you're back where you started so the dsplacement s zero -5 Instantaneous elocty and speed We usually thnk about speed and elocty n terms o ther nstantaneous alues, whch tell us how ast, and, or elocty, n what drecton an object s traelng at a partcular nstant. The nstantaneous elocty s dened as the rate o change o poston wth tme, or a ery small tme nteral. In a partcular tme nteral t, the dsplacement s x, the elocty durng that tme nteral s: x dx lm t 0 t dt The nstantaneous speed s smply the magntude o the nstantaneous elocty.

3 Pro. Dr. I. Nasser T /16/017 Acceleraton a : t s the rate o change o elocty. An object accelerates wheneer ts elocty changes. d a lm t 0 t dt 3

4 Pro. Dr. I. Nasser T /16/017 Aerage Speed: Usually, we are nterested to measure the aerage speed n two cases: a- For the whole journey, where total dstance traeled sag total tme taken b- For one part o a trp, where s the ntal speed, s the nal speed reached n the gen part, thus 1 s ag ( ) Instantaneous elocty We usually thnk about speed and elocty n terms o ther nstantaneous alues, whch tell us how ast, and, or elocty, n what drecton an object s traelng at a partcular nstant. The nstantaneous elocty s dened as the rate o change o poston wth tme, or a ery small tme nteral. In a partcular tme nteral t, the dsplacement s x, the elocty durng that tme nteral s: The nstantaneous speed s smply the magntude o the nstantaneous elocty. Example: see text sample problem.3 4

5 Pro. Dr. I. Nasser T /16/017 Velocty-Tme Graphs: The elocty-tme graph usually descrbes a certan trp. elocty acceleraton Unorm elocty deceleraton tme A journey usually starts rom rest. The body accelerates and the elocty ncreases to a certan alue, then t traels at a constant speed (unorm elocty), and nally t slows down (decelerates) untl t comes to a complete stop at a certan destnaton. (*) the slope o the elocty-tme graph at a certan moment ges the alue o the acceleraton at that moment. Notce that: 1- The slope s poste when the body s acceleratng (the elocty s ncreasng). - The slope s zero when the elocty s unorm (the elocty remans constant). 3- The slope s negate when the body s deceleratng (the elocty s decreasng). The dstance traeled durng a trp can be calculated by: Or Dstance traeled = aerage speed total tme Dstance traeled equals the area under the (,t) graph. dx Notes that: x dt area under the cure dt Sectons -6 Acceleraton: a s the rate o change o elocty. change n elocty Acceleraton, total tme a t - Acceleraton s a ector quantty measured n (m/s²). - Acceleraton occurs only durng the applcaton o unbalanced orce. - Acceleraton may reman constant, called "unorm acceleraton", or t may ncrease by tme "acceleraton", or decrease "deceleraton". Ths can be obsered rom the shapes o -t graphs shown: 5

6 Pro. Dr. I. Nasser T /16/017 elocty elocty elocty tme Constant acceleraton tme Non-unorm acceleraton (ncreasng acceleraton) tme Non-unorm acceleraton (decreasng acceleraton) at rest dstance dstance dstance deceleraton acceleraton Constant elocty tme tme Non-unorm elocty (acceleraton) tme Example: In the elocty-tme graph gen, nd: (a) acceleraton, (b) the deceleraton, (c) the total dstance traeled, (d) the aerage elocty durng the trp. elocty 1 b c d a 10 tme b a 1 0 m Acceleraton (ab) 1. t 10 s d -c 01 m Decelaraton (cd) 0.48 t 5 s Dstance traeled area under the trapezum = (65 30) m total dstance 570 m Aerage elocty 8.8 total tme 65 s 1 6

7 Pro. Dr. I. Nasser T /16/017 Instantaneous acceleraton a : t s the rate o change o elocty. An object accelerates wheneer ts elocty changes. d a lm t 0 t dt Example: see text sample problem.4 Equatons o Moton: For moton under constant (unorm) acceleraton, most o the problems can be soled by the equatons: 1 at, d x x x ( + ) t where s the ntal elocty, s the nal elocty, d s the dstance traeled, a s the acceleraton, and t s the tme taken. Secton.7 Knematcs equatons when the acceleraton s constant When the acceleraton o an object s constant, calculatons o the dstance traeled by an object, the elocty t's traelng at a partcular tme, and/or the tme t takes to reach a partcular elocty or go a partcular dstance, are smpled. There are our equatons that can be used to relate the derent arables, so that knowng some o the arables allows the others to be determned. Note that the equatons apply under these condtons: 1. the acceleraton s constant. the moton s measured rom t 0 3. the equatons are ector equatons, but the arables are not normally wrtten n bold letters. The act that they are ectors comes n, howeer, wth poste and negate sgns. The requred equatons are: 0 d x x t (to be used s constant,.e. a 0) 1 at, (to be used the x s not known) d x x t at, (to b ax, (to be used the t s not known) e used the s not known) 3 d x x ( + ) t, (to be used the a s not known) where s the nstantaneous elocty, s the ntal elocty, s the nal elocty, x s the poston, x s the ntal poston, x s the nal poston, and a s the acceleraton

8 Pro. Dr. I. Nasser T /16/017 Example: An object traelng at 8 m/s accelerates at m/s² or a perod o 10 seconds. a- What s the nal elocty? b- How ar does t trael durng ths perod? Inputs: m m 8 ; a ; s s t 10s a m at s b s 1 1 ag t ( + ) t = (8 8) m For part b, we can also use the relaton: 1 1 d x x t at m Example: A body, mong wth unorm acceleraton, has a elocty o m 0.1 s when ts x coordnate s 0.03 m. I ts x coordnate.00 s later s 0.05 m, what s the magntude o ts acceleraton? m Inputs: x 0.03 m; 0.1 ; x 0.03 m; t s s Usng the equaton: 1 1 d x x t at at x x t Substtute nputs: 1 m at x x t 0.05 m 0.03 m 0.1 s s 0.3 m Sole or a: 0.3 m 0.3 m a 0.16 m/s² t s The x acceleraton o the object s 0.16 m/s². (The magntude o the acceleraton s 0.16 m/s² ) Example: A jet plane lands wth a elocty o 100 m/s and can accelerate at a maxmum rate o a 5.0 m/s² as t comes to rest. a- From the nstant t touches the runway, what s the mnmum tme needed beore t stops? b- Can ths plane land at a small arport where the runway s 0.80 km long? 8

9 Pro. Dr. I. Nasser T /16/017 Plane touches down on runway at 100 m/s and comes to stop. a- The data gen n the problem s llustrated n the aboe Fgure. The mnus sgn n the acceleraton ndcates that the sense o the acceleraton s opposte that o the moton, that s, the plane s deceleratng. The plane wll stop as quckly as possble the acceleraton does hae the alue a 5.0 m/s², so we use ths alue n ndng the tme t n whch the elocty changes rom 100 m/s to 0. Use the equaton: t a Substtutng, we nd: m t s 0 s m 5.0 s The plane needs 0 s to come to stop. b- The plane also traels the shortest dstance n stoppng ts acceleraton s a 5.0 m/s². Wth x 0, we can nd the plane s nal x coordnate usng the equaton, 1 x x t at Usng t = 0 s whch we got rom part (a): 1 m 1 m x (0 s) s x t at s s 1000 m Or, we can use: 1 1 m x x ( ) t (0 s) s 1000 m So, the plane must hae at least 1.0 km o runway n order to come to rest saely km s not sucent. 9

10 Pro. Dr. I. Nasser T /16/017 Extra Problems (Chapter _I): Q: A car aerages 55.0 mph or the rst 4.0 hours o a trp and aerages 70.0 mph or each addtonal hour. The aerage speed or the entre trp was 60.0 mph. How long s the trp? x t 11t t 11t t ae 60.0 t t t1t 4.0 t t t 10.0t 0 t.0 hours Total Tme = t1t hours Q: A partcle s mong back and orth along the x-axs. A graph o ts poston ersus tme s gen n Fgure 1. What s the aerage speed o the partcle between t = 0 and t = 10 s? A) 9 m/s B) 8 m/s C) 7 m/s D) 6 m/s E) 5 m/s Fgure 1 What s the aerage elocty o the partcle between t = 0 and t = 10 s? Ans: 1 m/s Q: A rocket accelerates ertcally up rom ground leel rom rest at 30 m/s or 30 s; then runs out o uel. What s the rocket s nal alttude? (Ignore ar resstance) A) 55 km B) 45 km C) 35 km D) 5 km E) 15 km,, 10

11 Pro. Dr. I. Nasser T /16/017 3 Q: The poston o a partcle mong along the x-axs s gen by xt () 3t 9t 18, where x s n meters and t s n seconds. What s the alue o x when the partcle s acceleraton s zero? A) 1 m B) 10 m C) 19 m D) 0 m E) 15 m Q: A tme t = 0 an object s red ertcally up wth an ntal speed o. It takes the object 10 s to return back to ts startng pont. What s ts ntal speed o? (Ignore ar resstance) A) 49 m/s B) 98 m/s C) 5 m/s D) 10 m/s E) 55 m/s Q: A car traels up a hll at a constant speed o 40 km/h and returns down the hll at a constant speed o 60 km/h. Calculate the aerage speed or the round trp. Aerage speed, as opposed to aerage elocty, relates to the total dstance, as opposed to the net dsplacement. The dstance D up the hll s, o course, the same as the dstance down the hll, and snce the speed s constant (durng each stage o the moton) we hae speed = D/t. Thus, the aerage speed s ae Dup Ddown D t D D up tdown whch, ater cancelng D and pluggng n up = 40 km/h and down = 60 km/h, yelds 48 km/h or the aerage speed. Q: Compute your aerage elocty n the ollowng two cases: up down 11

12 Pro. Dr. I. Nasser T /16/017 (a) You walk 73. m at a speed o 1. m/s and then run 13. m at a speed o 3.05 m/s along a straght track. (b) You walk or 1.00 mn at a speed o 1. m/s and then run or 1.00 mn at3.05 m/s along a straght track. (a) Usng the act that tme = dstance/elocty whle the elocty s constant, we nd 73. m 73. m 1.74 m/s. ag 73. m 73. m 1. m/s 3.05 m (b) Usng the act that dstance x = t whle the elocty s constant, we nd (. 1 m / s)(60 s) (. 305 m / s)(60 s) ag 14. m/s. 10 s 1

13 Pro. Dr. I. Nasser Term171 October 16, 017 Chapter, Part Moton n one Dmenson Knematcs equatons when the acceleraton s constant When the acceleraton o an object s constant, calculatons o the dstance traeled by an object, the elocty t's traelng at a partcular tme, and/or the tme t takes to reach a partcular elocty or go a partcular dstance, are smpled. There are our equatons that can be used to relate the derent arables, so that knowng some o the arables allows the others to be determned. Note that the equatons apply under these condtons: 1. the acceleraton s constant. the moton s measured rom t 0 3. the equatons are ector equatons, but the arables are not normally wrtten n bold letters. The act that they are ectors comes n, howeer, wth poste and negate sgns. The requred equatons are: 0 d x x t (to be used s constant,.e. a 0) 1 at, (to be used the x s not known) d x x t at, (to b 1 1 e used the s not known) 3 d x x ( + ) t, (to be used the a s not known) 4 ax, (to be used the t s not known) where s the nstantaneous elocty, s the ntal elocty, s the nal elocty, x s the poston, x s the ntal poston, x s the nal poston, and a s the acceleraton. Secton.9 Free Fall: I a body s allng reely wthout ar rcton, ts acceleraton, a g 9.8 (m/s²), s constant and ts speed ncreases unormly, but the dstance moed ncreases non-unormly. The equatons o moton are gen by: 1 gt, gs, d t gt where the arables are dened earler. 1

14 Pro. Dr. I. Nasser Term171 October 16, 017 Example 1: A ball s thrown drectly downward wth an ntal speed o 8.00 m/s rom a heght o 30.0 m. When does the ball strke the ground? We dagram the problem as n the gure. We hae to choose a coordnate system, and here I wll let the orgn o the y axs be at the place where the ball starts ts moton (at the top o the 30m heght). Wth ths choce, the ball starts ts moton at y 0 and strkes the ground when y 30 m. We can now see that the problem s askng us: At what tme does y 30 m? We hae = 8.00 m/s (mnus because the ball s thrown downward!) and the acceleraton o the ball s a g 9.8 (m/s²), so at any tme t the y coordnate s gen by 1 1 y 0 8 m/s 9.8 m/s y t gt t t But at the tme o the mpact we hae y 30 m 8 m/st 4.9 m/s t an equaton or whch we can sole or t. We rewrte t as: 4.9t 8t 30 0 whch s just a quadratc equaton n t and ts solutons are: 3.4 s t 1.78 s Our answer s one o these... whch one? Obously the ball had to strke the ground at some poste alue o t, so the answer s t 1.78 s. The ball strkes the ground 1.78 s ater beng thrown. Example : A student throws a set o keys ertcally upward to hs brother n a wndow 4.00m aboe. The keys are caught 1.50 s later by the brother s outstretched hand. (a) Wth what ntal elocty were the keys thrown? (b) What was the elocty o the keys just beore they were caught? (a)we draw a smple pcture o the problem; such a smple pcture s gen n the gure. Hang a pcture s mportant, but we should be careul not to put too much nto the pcture; the problem dd not say that the keys were caught whle they were gong up or gong down. For all we know at the moment, t could be ether one! We wll put the orgn o the y axs at the pont where the keys were thrown. Ths smples thngs n that the ntal y coordnate o the keys s y = 0. O course, snce ths s a problem about ree all, we know the acceleraton: a g 9.8 (m/s²). What mathematcal normaton does the problem ge us? We are told that when t = 1.50 s, the y coordnate o the keys s y = 4.00 m. Is ths enough normaton to sole the problem? We wrte the equaton or y(t): 1 1 y y t gt t gt where ` s presently unknown. At t = 1.50 s, y = 4.00 m, so:

15 Pro. Dr. I. Nasser Term171 October 16, Now we can sole or. Rearrange ths equaton to get: 4 t gt m 10 s (b) We want to nd the elocty o the keys at the tme they were caught, that s, at t = 1.50 s. We know ; the elocty o the keys at all tmes ollows rom the equaton: m m m at s 4.68 s s s So the elocty o the keys when they were caught was 4.68 m/s. Note that the keys had a negate elocty; ths tells us that the keys were mong downward at the tme they were caught! Example 3: A ball s thrown ertcally upward rom the ground wth an ntal speed o 15.0 m/s. (a) How long does t take the ball to reach ts maxmum alttude? (b) What s ts maxmum alttude? (c) Determne the elocty and acceleraton o the ball at t =.00 s. (a) An llustraton o the data gen n ths problem s gen n the gure. We measure the coordnate y upward rom the place where the ball s thrown so that y = 0. The ball s acceleraton whle n lght s a g 9.8 (m/s²). We are gen that = m/s. The ball s at maxmum alttude when ts (nstantaneous) elocty s zero (t s nether gong up nor gong down) and we can use the expresson or to sole or t: at t a Plug n the alues or the top o the ball s lght and get: 015 t 1.53 s 9.8 The ball takes 1.53 s to reach maxmum heght. (b) Now that we hae the alue o t when the ball s at maxmum heght we can plug t nto 1 y y t gt and nd the alue o y y y at ths tme and that wll be the alue o the maxmum heght. But we can also use ay snce we know all the alues except or y. Solng or y we nd: ay y a Pluggng n the numbers, we get 0 15 y 11.5 m ( 9.8) 3

16 Pro. Dr. I. Nasser Term171 October 16, 017 The ball reaches a maxmum heght o 11.5 m. (c)at t =.00 s (that s,.0 seconds ater the ball was thrown), usng the equaton at to nd: m 15.0 ( 9.80)(.00) 4.60 s so at t =.00 s the ball s on ts way back down wth a speed o 4.60 (m/s). Example 4: A baseball s ht such that t traels straght upward ater beng struck by the bat. A an obseres that t requres 3.00 s or the ball to reach ts maxmum heght. Fnd (a) ts ntal elocty and (b) ts maxmum heght. Ignore the eects o ar resstance. (a) An llustraton o the data gen n the problem s gen n aboe gure. For the perod rom when the ball s ht to the tme t reaches maxmum heght, we know the tme nteral, the acceleraton (a = g) and also the nal elocty, snce at maxmum heght the elocty o the ball s zero. Then at ges us : m 0 ( 9.80)(3.00) 9.4 s The ntal elocty o the ball was +9.4 m/s. (b) To nd the alue o the maxmum heght, we need to nd the alue o the y coordnate at tme t = 3.00 s. We can use: ay y y y a Pluggng n the numbers we nd that the change n y coordnate or the trp up was: y y 44.1 m a ( 9.8) The ball reached a maxmum heght o 44.1m. Challengng problem: A allng object requres 1.50 s to trael the last 30.0m beore httng the ground. From what heght aboe the ground dd t all? Ths s an ntrgung sort o problem... ery easy to state, but not so clear as to where to begn n settng t up! The rst thng to do s draw a dagram. We draw the mportant ponts o the object s moton, as n the gure. The object has zero elocty at A; at B t s at a heght o 30.0m aboe the ground wth an unknown elocty. At C t s at ground leel, the tme s 1.50 s later 4

17 Pro. Dr. I. Nasser Term171 October 16, 017 than at B and we also don t know the elocty here. O course, we know the acceleraton: a = 9.80 m/s!! We are gen all the normaton about the trp rom B to C, so why not try to ll n our knowledge about ths part? We know the nal and ntal coordnates, the acceleraton and the tme so we can nd the ntal elocty (that s, the elocty at B). Let s put the orgn at ground leel; then, y0 = 30.0m, y = 0 and t = 1.50 s, and usng we nd: so that Ths s the elocty at pont B; we can also nd the elocty at C easly, snce that s the nal elocty, : Now we can consder the trp rom the startng pont, A to the pont o mpact, C. We don t know the ntal y coordnate, but we do know the nal and ntal eloctes: The ntal elocty s 0 = 0 and the nal elocty s = 7.3 m/s, as we just ound. Wth the orgn set at ground leel, the nal y coordnate s y = 0. We don t know the tme or the trp, but we use: we nd: and we can rearrange ths to get: and the so the object started allng rom a heght o 38.m. There are probably cleerer ways to do ths problem, but here I wanted to ge you the slow, patent approach! 5

18 Pro. Dr. I. Nasser Term171 October 16, 017 Phys101 Chapter Instructor: Dr. Al M. Al Shukr Selected problems rom the textbook 18. The poston o a partcle s gen by x = 0 t 5 t 3, where x s n meters and t s n seconds, (a) When, eer, s the partcle s elocty zero? (b) When s ts acceleraton a zero? (c) For what tme range s a negate? (d) For what tme range s a poste? (e) Graph x(t), (t), and a(t). We can get (t) and a(t) by derentatng x(t) = 0 t 5 t 3. (t) = dx/dt = 0 15 t and a(t) = d/dt = d x/dt = 30 t where x s n m, n m/s, a n m/s, and t s n s. (a) From 0 = 0 15 t, the only poste alue o t or whch the partcle s (momentarly) stopped s t = 0/15 = 1. s. (b) From 0 = 30t a(0) = 0 (t anshes at t = 0). (c) The acceleraton a(t) = 30t s negate or t > 0 (d) The acceleraton a(t) = 30t s poste or t < 0. (e) See the graphs Up and to the let 6

19 Pro. Dr. I. Nasser Term171 October 16, A car traelng 56.0 km/h s 4.0 m rom a barrer when the drer slams on the brakes. The car hts the barrer.00 s later. (a) What s the magntude o the car s constant acceleraton beore mpact? (b) How ast s the car traelng at mpact? (a) Snce a s constant, the ollowng equaton can be used: x = x0 + 0t + 1 at a = (x x0 0t)/t I we take x0 = 0 then a = (x 0t)/t Substtutng x = 4.0 m, 0 = 56.0 km/h = m/s and t =.00 s, we nd.00s 4.0m 15.55m/s.00s a 3.56m/s, The negate sgn s used to ndcate that the acceleraton s opposte to the drecton o moton o the car (The car s slowng down). (b) Snce a s constant, the ollowng equaton can be used = 0 + at Substtutng a = 3.56 m/s, 0 = 56.0 km/h = m/s, and t =.00 s, we nd = ( 3.56).00 = = 8.43 m/s = 8.43 (m/s) (1 km/1000 m) (3600 s/1 h) = = 30.3 km/h 67. How ar does the runner whose elocty tme graph s shown n the Fgure trael n 16 s? The gure s ertcal scalng s set by s = 8.0 m/s. s (m/s) Velocty = change n poston oer tme nteral = dx/dt dx = dt x x0 = Δx = dt In case o a graph: Δx = area under (t) cure Then dstance traeled n tme nteral t = 0 to t = 16 s t t (s) 7

20 Pro. Dr. I. Nasser Term171 October 16, 017 Δx = Area (t = 0 ) + Area (t = 10) + Area (t = 10 1) + Area (t = 1 16) Δx = area o a trangle + area o a rectangle + (areas o a trangle + and a rectangle) + area o a rectangle = ½ ( 0) (8 0) + (10 ) (8 0) + [½ (1 10) (8 4) + (1 10) (4 0)] + (16 1) (4 0) = [4 + 8] + 16 = 100 m 76. Two damond begn a ree all rom rest rom the same heght, 1.0 s apart. How long ater the rst damond begns to all wll they wll be 10.0 m apart? Usng the ree all equatons: y = y0 + 0 t ½ g t See the dagram below or explanaton o the problem Note: y10 = y0 = 0.0, t10 = t0 = 0.0, t = t1 1.0, and 10 = 0 = 0.0 Frst Damond Second Damond y10 = 0.0, t10 = 0.0 y0 = 0.0, t0 = 0.0 y = y, t = t m y1 = y 10.0, t1 y1= ½ g t1, y = ½ g t, y1 = y 10, and t = t1 1 y1 = ½ g t1 = y 10 = ½ g t 10 = ½ g (t1 1) 10.0 ½ g t1 = ½ g (t1 1) 10 ½ g t1 = ½ g t1 + g t1 ½ g 10 g t1 ½ g = 10 t1 = (10+ ½ g)/ g = (10+4.9)/9.8 =14.9 /9.8 = 1.5 s t1 = 1.5 s. 8

21 Pro. Dr. I. Nasser Term171 October 16, 017 PHYS101 - Chapter (Instructor: Dr. Al Al-Shukr) Extra Problems (Chapter ):. Two cars A and B trael on a straght lne. The dsplacement o car A s gen by xa(t) =.60 t t, where t s n seconds and xa n m. The dsplacement o car B s gen by xb(t) =.80 t t 3. At what tme the two cars wll hae the same acceleraton? a..67 s b. 6.7 s c. 7.6 s d s e s 3. A ball s thrown rom ground straght upward wth a elocty o 6 m/s. How long does t take the ball to strke the ground? a. 5.3 s b..7 s c. 1.6 s d. 0.8 s e. 7.5 s 4. A ball s thrown rom ground straght upward. I t stays n lght or 6.0 s, nd the ntal elocty and the maxmum heght t reaches. Also nd the aerage elocty or the entre lght. a. 0 = 9.4 m/s, h = 44.1 m, ae = 0 5. Two automobles, 150 klometers apart, are traelng toward each other. One automoble s mong at 60 km/h and the other s mong at 40 km/h. In how many hours wll they meet? a. 1.5 b..0 c. 1.0 d..5 e The graph shown n the Fgure represents the straght-lne moton o a car. Fnd ts acceleraton at t = 6 s. a. 3.0 m/s b m/s c m/s d. 5.0 m/s 7. A car traels along a straght lne at a constant elocty o 18 m/s or.0 s and then accelerate at 6.0 m/s or a perod o 3.0 s. The aerage elocty o the car durng the whole 5.0 s s: 9

22 Pro. Dr. I. Nasser Term171 October 16, 017 a. 13 m/s b. 18 m/s c. 10 m/s d. 16 m/s 8. The elocty as a uncton o tme or a partcle mong along the x-axs s shown n Fgure 1. The moton clearly has two derent parts: the rst part s rom t = 0 to t =.0 s, and the second part s rom t =.0 s to t = 6.0 s. Whch one o the ollowng statements s correct? a. At t = 4.0 s the acceleraton s -5.0 m/s b. At t = 4.0 s the acceleraton s zero c. From t =0 to t=6.0 s, the dsplacement s zero d. From t =0 to t=6.0 s, the dsplacement s -0 m e. At t = 1.0 s the acceleraton s 10 m/s 9. A partcle moes along the x axs. Its poston s gen by the equaton x = t t 3 wth x n meters and t n seconds. The aerage acceleraton rom t = 0 to t =.0 s s: a. 6.0 m/s b. 3.0 m/s c..0 m/s d. 4.0 m/s 10. An arrow s shot straght up wth an ntal speed o 98 m/s. I rcton s neglected, how hgh the arrow can reach? a. 490 m b. 980 m c. 50 m d. 98 m 11. A stone s thrown ertcally downward rom the top o a 40 m tall buldng wth an ntal speed o 1.0 m/s. Ater.0 s the stone wll hae traeled a dstance o a. m b. 38 m c. 40 m d. 5 m e. 15 m 1. A stone and a ball are thrown ertcally upward wth derent ntal speeds: 0 m/s or the stone and 10 m/s or the ball. I the maxmum heght reached by the ball s H then the maxmum heght reached by the stone s: a. 4 H b. H c. H d. H/ e. H/4 10

23 Pro. Dr. I. Nasser Term171 October 16, A partcle starts rom the orgn at t = 0 and moes along the poste x-axs. A graph o the elocty o the partcle as a uncton o tme s show n Fgure 1. The aerage elocty o the partcle between t = 0.0 s and 5.0 s s: a. 1.4 m/s b. 1.0 m/s c. -.0 m/s d. 0 m/s 14. At a trac lght, a truck traelng at 10 m/s passes a car as t starts rom rest. The truck traels at a constant elocty and the car accelerates at 4.0 m/s. How much tme does the car take to catch up wth the truck? a. 5.0 s b..0 s c. 15 s d. 0 s e. 5 s 15. The coordnate o a partcle n meters s gen by x(t) =.0 t.0 t, where the tme t s n seconds. The partcle s momentarly at rest at tme t equal to: a s b s c..0 s d. 1.3 s 16. An object starts rom rest at the orgn and moes along the x-axs wth a constant acceleraton o 4 m/s. Its aerage elocty as t goes rom x = m to x = 18 m s: a. 8 m/s b. m/s c. 6 m/s d. 5 m/s 17. Two cars are a dstance d apart and traelng toward each other. One car s mong at 60. km/h and the other s mong at 40. km/h. I they meet ater 90 mnutes o traelng, nd the dstance d. a. 150 km b. 10 km c. 100 km d. 50 km 18. A helcopter at heght h (m) rom the surace o the sea s descendng at a constant speed o (m/s). The tme t takes to reach the surace o the sea can be ound rom: a. h = t b. h = ½ g t c. h = ½ g t d. h = t ½ g t e. h = t ½ g t 11

24 Pro. Dr. I. Nasser Term171 October 16, A partcle starts rom rest at t = 0 s. Its acceleraton as a uncton o tme s shown n Fg. 1. What s ts speed at the end o the 6.0 s? a. 4.0 m/s b. 0 m/s c. 1 m/s d..0 m/s e. -1 m/s 0. The poston o a partcle x(t) as a uncton o tme (t) s descrbed by the equaton: x(t) = t t 3, where x s n m and t s n s. What s the maxmum poste poston o the partcle on the x-axs? a. 4.0 m b. 3.0 m c..0 m d. 1.0 m 1. A stone s thrown ertcally downward rom a buldng wth an ntal speed o.0 m/s. It reaches the ground ater 5.0 s. What s the heght o the buldng? a. 130 m b. 60 m c. 180 m d. 10 m 1

25 Pro. Dr. I. Nasser Term171 October 16, 017 Chapter, selected problems Q1. The poston ersus tme or a certan partcle mong along the x-axs s shown n Fgure. The aerage elocty n the tme nteral 4.0 s to 7.0 s s: A) 1.7 m/s Q. A ball s thrown drectly downward rom a heght o 30.0 m. It takes 1.79 s to reach the ground. Fnd the magntude o the ntal elocty.a) 7.99 m/s Q3. A stone s thrown outward rom pont A at the top o a 58.8 m hgh cl wth an upward elocty component o 19.6 m/s (see Fgure). Assume that t lands on the ground, at pont B, below the cl, and that the ground below the cl s lat. How long was the stone n the ar? [Neglect the ar resstance]. A) 6.00 s Q4. Fgure 3 llustrates the moton o a partcle startng rom rest and mong along an x-axs wth a constant acceleraton. The gure s ertcal scalng s set by xs = 1 m. The partcle s acceleraton s A).0 m/s Q6. A stone s thrown ertcally upwards wth an ntal speed o 4.0 m/s rom a wndow whch s 8.0 m aboe the ground. Wth what speed wll the stone ht the ground? (Neglect ar resstance) A) 13 m/s Q6. A man s runnng n a straght lne (along the x-axs). The graph n Fgure shows the man s elocty as a uncton o tme. Durng the rst 1.0 s, the total dstance traeled s A) 46.0 m Q7. A rock s thrown ertcally upward rom ground leel at tme t =0.0 s. At t =1.5 s t passes the top o a tall tower, and then 1.0 s later t reaches ts maxmum heght. What s the heght o the tower? A) 6 m Q8: A ball moes n a straght lne along the x-axs and Fgure shows ts elocty as a uncton o tme t. What s the ball aerage elocty and aerage speed, respectely, oer a perod o 3.00 s. A) m/s,.33 m/s Q9: A car traels n a straght lne. Frst, t starts rom rest at pont A and accelerates at a rate o 5.00 m/s untl t reaches a speed o 100 m/s at pont B. The car then slows down at a constant rate o 8.00 m/s untl t stops at pont C. Fnd the tme the car takes or ths trp (rom pont A to pont C). A) 3.5 s Q10: Fgure 1 shows the elocty Vx (m/s) o a partcle mong along the x- axs. I x =.0 m at t = 1.0 s, what s the poston, measured n meters, o the partcle at t = 6.0 s? A)