Version 001 Summer Reiew #03 tubmn (IBII20142015) 1 This print-out should he 35 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. Concept 20 P03 001 10.0 points An ocenic depth-sounding essel sureys the ocen bottom with ultrsonic wes tht trel 1530 m/s in sewter. How deep is the wter directly below the essel if the time dely of the echo to the ocen floor nd bck is 4 s? Correct nswer: 3060 m. Explntion: Let : = 1530 m/s nd t = 4 s. Thesoundtkes2storechtheocenfloor (nd 2 s to return), so d = t = (1530 m/s)(2 s) = 3060 m. Constnt Velocity 02 002 10.0 points An object is moing with constnt elocity. Which sitution is impossible in such circumstnce? 1. Four forces ct on the object. 2. One force cts on the object. 3. Three forces ct on the object. 4. Two forces ct on the object. 5. No forces ct on the object. Explntion: Two or more ectorscn sum to zero, but single nonzero ector cnnot sum to zero, nd Newton s Second Lw requires tht F = m = 0 in this cse. Displcement Cure 02 003 (prt 1 of 2) 10.0 points Consider moing object whose position x is plotted s function of the time t. The object moed in different wys during the time interls denoted I, II nd III on the figure. 6 4 2 x I II III 2 4 6 During these three interls, when ws the object s speed highest? Do not confuse the speed with the elocity. 1. Sme speed during ech of the three interls. 2. During interl I 3. During interl II 4. Sme speed during interls II nd III 5. During interl III Explntion: Theelocity istheslopeofthex(t)cure; the mgnitude = of this slope is the speed. The cure is steepest (in bsolute mgnitude) during the interl III nd tht is when the object hd the highest speed. 004 (prt 2 of 2) 10.0 points During which interl(s) did the object s elocity remin constnt? 1. During none of the three interls 2. During interl II only 3. During ech of the three interls t
Version 001 Summer Reiew #03 tubmn (IBII20142015) 2 4. During interl III only 5. During interl I only Explntion: ForechofthethreeinterlsI,IIorIII,the x(t) cure is liner, so its slope (the elocity ) is constnt. Between the interls, the elocity chnged in n brupt mnner, but it did remin constnt during ech interl. Aerge Speed on Trip 005 (prt 1 of 2) 10.0 points A person trels by cr from one city to nother. She dries for 20.9 min t 78 km/h, 8.05mint 114km/h, 39.9mint48.3km/h, nd spends 17.8 min long the wy eting lunch nd buying gs. Determine the distnce between the cities long this route. Correct nswer: 74.5845 km. Explntion: Let : t 1 = 20.9 min, 1 = 78 km/h, t 2 = 8.05 min, 2 = 114 km/h, t 3 = 39.9 min, nd 3 = 48.3 km/h. The totl time is Let : t other = 17.8 min. t = t 1 +t 2 +t 3 +t other = 20.9 min+8.05 min +39.9 min+17.8 min = 1.44417 h, so = x 74.5845 km = = 51.6454 km/h. t 1.44417 h Comprison of Aerge Velocity 007 10.0 points The position-ersus-time grph below describes the motion of three different bodies (lbeled 1, 2, 3). x x A x B A t A 1 2 3 Consider the erge elocities of the three bodies. Which of the following sttements is? t B B t x = x 1 +x 2 +x 3 = 1 t 1 + 2 t 2 + 3 t 3 = (78 km/h)(20.9 min) +(114 km/h)(8.05 min) +(48.3 km/h)(39.9 min) = 74.5845 km. 006 (prt 2 of 2) 10.0 points Determine the erge speed for the trip. Correct nswer: 51.6454 km/h. Explntion: 1. 1 > 2 > 3 2. 1 = 2 = 3 3. 1 < 2 < 3 4. 1 > 2 nd 3 > 2 Explntion: Aerge elocity is = displcement time = x B x A t B t A. All three bodies he exctly sme displcement in exctly sme time, so the erge elocities re exctly equl: 1 = 2 = 3.
Version 001 Summer Reiew #03 tubmn (IBII20142015) 3 Accelerting Object 008 10.0 points Ifthe ccelertionof nobject is zero t some instnt in time, wht cn be sid bout its elocity t tht time? 1. It is positie. 2. It is not chnging t tht time. 3. Unble to determine. 4. It is negtie. 5. It is zero. Explntion: The ccelertion = t = 0 = 0. 3. t 4. t 5. t 6. t Accelertion Time Grph 01 009 (prt 1 of 5) 10.0 points Consider toy cr which cn moe to the right (positie direction) or left on horizontl surfce long stright line. cr O + Wht is the ccelertion-time grph if the cr moes towrd the right (wy from the origin), speeding up t stedy rte? 1. t 2. None of these grphs is. 7. t 8. t Explntion: Since the cr speeds up t stedy rte, the ccelertion is constnt. 010 (prt 2 of 5) 10.0 points Wht is the ccelertion-time grph if the cr moes towrd the right, slowing down t stedy rte?
Version 001 Summer Reiew #03 tubmn (IBII20142015) 4 1. t 2. t 011 (prt 3 of 5) 10.0 points Wht is the ccelertion-time grph if the cr moes towrds the left (towrd the origin) t constnt elocity? 1. t 3. t 2. t 4. t 3. t 5. t 4. None of these grphs is. 5. t 6. t 6. t 7. t 7. t 8. None of these grphs is. Explntion: Since the cr slows down, the ccelertion is in the opposite direction.
Version 001 Summer Reiew #03 tubmn (IBII20142015) 5 8. t 7. t Explntion: Since the cr moes t constnt elocity, the ccelertion is zero. 012 (prt 4 of 5) 10.0 points Wht is the ccelertion-time grph if the cr moestowrdtheleft,speedinguptstedy rte? 8. t 1. t 2. t Explntion: The sme reson s Prt 1. 013 (prt 5 of 5) 10.0 points Wht is the ccelertion-time grph if the cr moes towrd the right t constnt elocity? 1. t 3. t 2. t 4. t 3. t 5. None of these grphs is. 6. t 4. t
Version 001 Summer Reiew #03 tubmn (IBII20142015) 6 5. t 6. t 7. None of these grphs is. 8. t Explntion: The sme reson s Prt 3. Accelertion of Vehicle 014 10.0 points Acrtrelinginstrightlinehselocity of 2.34 m/s t some instnt. After 5.46 s, its elocity is 12.2 m/s. Wht is its erge ccelertion in this time interl? Correct nswer: 1.80586 m/s 2. Explntion: Let : 1 = 2.34 m/s, t = 5.46 s, nd 2 = 12.2 m/s. The erge ccelertion is = t = f i t 12.2 m/s 2.34 m/s = 5.46 s = 1.80586 m/s 2. Accelertion s Time 01 015 (prt 1 of 4) 10.0 points Consider the plot below describing the ccelertion of prticle long stright line with n initil position of 25 m nd n initil elocity of 8 m/s. ccelertion (m/s 2 ) 4 3 2 1 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 9 time (s) Wht is the elocity t 1 s? Correct nswer: 5 m/s. Explntion: Let : 0 = 8 m/s, 1 = 3 m/s 2, nd t = 1 s. The chnge in elocity is the re t between the ccelertion cure nd the time xis. f = 0 + 1 t = 8 m/s+(3 m/s 2 )(1 s) = 5 m/s. 016 (prt 2 of 4) 10.0 points Wht is the position t 1 s?
Version 001 Summer Reiew #03 tubmn (IBII20142015) 7 Correct nswer: 18.5 m. Explntion: Let : x 0 = 25 m. so the position fter 7 s is x 2 = x 1 + 1 t 2 + 1 2 2( t 2 ) 2 = 31 m+(10 m/s)(1 s) x 1 = x 0 + 0 t+ 1 2 1 t 2 + 1 2 ( 7 m/s2 )(1 s) 2 = 37.5 m. = 25 m+( 8 m/s)(1 s) + 1 2 (3 m/s2 )(1 s) 2 = 18.5 m. 017 (prt 3 of 4) 10.0 points Wht is the elocity t 7 s? Correct nswer: 3 m/s. Explntion: Let : t 1 = 6 s, 2 = 7 m/s 2, nd t 2 = 1 s. The clcultion is done in two prts, ech with constnt ccelertion. The elocity fter 6 s is 1 = 0 + 1 t 1 = 8 m/s+(3 m/s 2 )(6 s) = 10 m/s, so the elocity fter 7 s is 2 = 1 + 2 t 2 = 10 m/s+( 7 m/s 2 )(1 s) = 3 m/s. 018 (prt 4 of 4) 10.0 points Wht is the position t 7 s? Correct nswer: 37.5 m. Explntion: The position fter 6 s is x 1 = x 0 + 0 t 1 + 1 2 1( t 1 ) 2 = (25 m)+( 8 m/s)(6 s) + 1 2 (3 m/s2 )(6 s) 2 = 31 m, Describing Motion 019 (prt 1 of 2) 10.0 points A cr initilly t rest on stright rod ccelertes ccording to the ccelertion s time plot. t 1 t 2 t 3 t 4 Wht is the best description of the motion of the cr? Tke forwrd to be the positie direction. 1. Thecr strtstrest, ccelertesto low speed, comes to stop, ccelertes bckwrds nd cruises in reerse. 2. The cr goes forwrd nd then goesbckwrd, ending where it strted. 3.Thecrstrtstrest,ccelertestohigh speed, cruises for short while, decelertes to lower speed, then cruises. 4. The cr goes bckwrd nd then goes forwrd. 5. The cr goes forwrd nd then goesbckwrd, ending behind where it strted. 6. The cr strts t rest, goes up to high speed, stops moing, trels bckwrd, nd stops. Explntion: t
Version 001 Summer Reiew #03 tubmn (IBII20142015) 8 Use the ccelertion to determine the elocity behior: 1) 0 = 0 (initilly t rest) 2) 0 < t < t 1 : = 0 (remins t rest) 3) t 1 < t < t 2 : > 0 (ccelertes forwrd) 4) t 2 < t < t 3 : = 0 (constnt speed) 5) t 3 < t < t 4 : < 0 (decelertes) 6) t 4 < t : = 0 (constnt speed) 020 (prt 2 of 2) 10.0 points Which of the following grphs describes the elocity s time of the cr? 1. t 3 t 4 t t 1 t 2 2. None of these grphs is. 3. t 2 t 3 t t 1 t 4 4. t 3 t 4 t t 1 t 2 5. t 2 t 3 t t 1 t 4 6. t 1 t t 2 t 3 t 4 7. t 3 t 4 t t 1 t 2 8. t 3 t t 1 t 2 t 4 9. t 1 t 2 t 3 t 4 Explntion: 1) 0 < t < t 1 : 0 = 0 2) t 1 < t < t 2 : elocity increses uniformly 3) t 2 < t < t 3 : constnt elocity 4) t 3 < t < t 4 : elocity decreses uniformly 5) t 4 < t : constnt elocity The elocity s time grph cn be found by plotting the re under the ccelertion s time cure. Hewitt CP9 03 E07 021 10.0 points Cn n object reerse its direction of trel while mintining constnt ccelertion? 1. No; if the ccelertion is constnt, the direction of the speed remins unchnged. 2. Yes; bll tossed upwrd reerses its direction of trel t its highest point. 3. No; the direction of the speed is lwys the sme s the direction of the ccelertion. t 4. Yes; bll thrown towrd wll bounces bck from the wll. 5. All re wrong. Explntion: Velocity nd ccelertion need not be in the
Version 001 Summer Reiew #03 tubmn (IBII20142015) 9 sme direction. When bll is tossed upwrd it experiences constnt ccelertion directed downwrd. Hewitt CP9 03 E13 022 10.0 points Which of the following isn exmpleof something tht undergoes ccelertion while moing t constnt speed? 1. None of these. An object tht undergoes n ccelertion hs to chnge its speed 2. A cr moing stright bckwrds on the rod 3. A footbll flying in the ir 4. A mn stnding in n eletor 5. A cr mking circle in prking lot Explntion: When n object moes in circulr pth t constnt speed its direction chnges, so its elocity chnges, mening it experiences ccelertion. Velocity s Time 08 023 (prt 1 of 5) 10.0 points The scle on the horizontl xis is 5 s nd on the erticl xis 5 m/s. The initil position is 14 m. elocity ( 5 m/s) 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 time ( 5 s) Wht is the initil elocity? Correct nswer: 0. Explntion: At time t = 0, the elocity is 0. 024 (prt 2 of 5) 10.0 points Wht is the finl elocity? Correct nswer: 25 m/s. Explntion: The finl elocity is red from the grph ndis5timestheerticlsclefctor(5m/s). f = 5(5 m/s) = 25 m/s. 025 (prt 3 of 5) 10.0 points Wht is the finl position? Correct nswer: 576.5 m. Explntion: The chnge in position is the re under the elocity s time grph, so p f = p o + 1 2 = p o + 1 2 = 576.5 m. (9t) (5) 9(5 s) 5(5 m/s) 026 (prt 4 of 5) 10.0 points Wht ccelertion is represented by the grph? Correct nswer: 0.555556 m/s 2. Explntion: The ccelertion is the slope of the elocity s time grph, so = t = 5 0 5(5 m/s) = 9t 0 9(5 s) = 0.555556 m/s 2. 027 (prt 5 of 5) 10.0 points In which direction is the motion? 1. Unble to determine.
Version 001 Summer Reiew #03 tubmn (IBII20142015) 10 2. forwrd 3. bckwrd Explntion: The elocity is positie, so the motion is directed forwrd. AP B 1993 MC 5 028 10.0 points An object is relesed from rest on plnet tht hs no tmosphere. The object flls freely for 3 m in the first second. Wht is the mgnitude of the ccelertion due to grity on the plnet? 1. 3.0 m/s 2 2. 12.0 m/s 2 3. 6.0 m/s 2 4. 10.0 m/s 2 5. 1.5 m/s 2 Explntion: s = 1 2 t2 Let : s = 3 m. = 2s 2(3 m) = t2 (1.0 s) 2 = 6 m/s2. 3. negtie negtie 4. positie positie 5. positie negtie Explntion: At the mximum eletion, the erticl elocity is zero. The ccelertion is due to grity, which lwys cts down. Accelertion of Flling Object 030 10.0 points If you drop n object, it ccelertes downwrd t 9.8 m/s 2 (in the bsence of ir resistnce). If, insted, you throw it downwrd, its downwrd ccelertion fter relese is 1. 9.8 m/s 2. 2. less thn 9.8 m/s 2. 3. more thn 9.8 m/s 2. Explntion: The ccelertion of ll freely flling objects is g = 9.8 m/s 2. Bll M 01 031 10.0 points A bll is thrown upwrd with n initil erticl speed of 0 to mximum height of h mx. AP M 1993 MC 19 029 10.0 points An object is shot erticlly upwrd into the ir with positie initil elocity. Wht ly describes the elocity nd ccelertion of the object t its mximum eletion? Velocity Accelertion 0 hmx 1. zero zero 2. zero negtie Wht is its mximum height h mx? The ccelertion of grity is g. Neglect ir resis-
Version 001 Summer Reiew #03 tubmn (IBII20142015) 11 tnce. 1. h mx = 2 0 4g 2. h mx = 32 0 4g 5 2 3. h mx = 0 2 2 g 4. h mx = 2 0 2 g 5. h mx = 2 0 g 6. h mx = 3 2 0 2g 7. h mx = 52 0 8g 3 2 8. h mx = 0 2 2 g 9. h mx = 2 0 2g Explntion: With up positie, = g nd for the upwrd motion t f = 0, so for constnt ccelertion, 2 f = 2 0 +2 ( y f y 0 ) 0 = 2 0 2g (h mx 0) h mx = 2 0 2g. Bll Drop on Olympus 032 (prt 1 of 2) 10.0 points The tllest olcno in the solr system is the 19 km tll Mrtin olcno, Olympus Mons. An stronut drops bll off the rim of the crter nd tht the free fll ccelertion of the bll remins constnt throughout the bll s 19 km fll t lue of 3.1 m/s 2. (We ssume thtthecrterissdeepstheolcnoistll, which is not usully the cse in nture.) Findthetimefortheblltorechthecrter floor. Correct nswer: 110.716 s. Explntion: Let : h = 19 km nd = 3.1 m/s 2. The distnce n object flls from rest under n ccelertion is h = 1 2 t2 2h t = = 2(19 km) (3.1 m/s 2 ) = 110.716 s. 033 (prt 2 of 2) 10.0 points Find the mgnitude of the elocity with which the bll hits the crter floor. Correct nswer: 343.22 m/s. Explntion: Since the object flls from rest, 2 = 2h = 2h = 2(3.1 m/s 2 )(19 km) = 343.22 m/s. Holt SF 02F 03 034 (prt 1 of 2) 10.0 points A tennis bll is thrown erticlly upwrd with n initil elocity of +7.0 m/s. Wht will the bll s elocity be when it returns to its strting point? The ccelertion of grity is 9.81 m/s 2. Correct nswer: 7 m/s. Explntion: Let : i = 7.0 m/s, = 9.81 m/s 2, nd y = 0 m. 2 f = 2 i +2 y = 2 i f = ± i = ±7 m/s.
Version 001 Summer Reiew #03 tubmn (IBII20142015) 12 Since the direction is down, the finl elocity is 7 m/s. 035 (prt 2 of 2) 10.0 points Howlongwilltheblltketorechitsstrting point? Correct nswer: 1.42712 s. Explntion: f = i + t f i = t t = f i = = 1.42712 s. 7 m/s (7 m/s) 9.81 m/s 2