DESCRIBING MOTION: KINEMATICS IN ONE DIMENSION

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1 DESCRIBING MOTION: KINEMATICS IN ONE DIMENSION Responses to Questions. A cr speedometer mesures only speed. It does not give ny informtion bout the direction, so it does not mesure velocity.. If the velocity of n object is constnt, then the speed nd the direction of trvel must lso be constnt. If tht is the cse, then the verge velocity is the sme s the instntneous velocity, becuse nothing bout its velocity is chnging. The rtio of displcement to elpsed time will not be chnging, no mtter the ctul displcement or time intervl used for the mesurement. 3. There is no generl reltionship between the mgnitude of speed nd the mgnitude of ccelertion. For exmple, one object my hve lrge but constnt speed. The ccelertion of tht object is then zero. Another object my hve smll speed but be gining speed nd therefore hve positive ccelertion. So in this cse the object with the greter speed hs the lesser ccelertion. Consider two objects tht re dropped from rest t different times. If we ignore ir resistnce, then the object dropped first will lwys hve greter speed thn the object dropped second, but both will hve the sme ccelertion of 9.80 m/s. 4. The ccelertions of the motorcycle nd the bicycle re the sme, ssuming tht both objects trvel in stright line. Accelertion is the chnge in velocity divided by the chnge in time. The mgnitude of the chnge in velocity in ech cse is the sme, 0 km/h, so over the sme time intervl the ccelertions will be equl. 5. Yes. For exmple, cr tht is trveling northwrd nd slowing down hs northwrd velocity nd southwrd ccelertion. 6. The velocity of n object cn be negtive when its ccelertion is positive. If we define the positive direction to be to the right, then n object trveling to the left tht is hving reduction in speed will hve negtive velocity with positive ccelertion. If gin we define the positive direction to be to the right, then n object trveling to the right tht is hving reduction in speed will hve positive velocity nd negtive ccelertion. Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. -

2 - Chpter 7. If north is defined s the positive direction, then n object trveling to the south nd incresing in speed hs both negtive velocity nd negtive ccelertion. Or if up is defined s the positive direction, then n object flling due to grvity hs both negtive velocity nd negtive ccelertion. 8. Yes. Remember tht ccelertion is chnge in velocity per unit time, or rte of chnge in velocity. So velocity cn be incresing while the rte of increse goes down. For exmple, suppose cr is trveling t 40 km/h nd one second lter is going 50 km/h. One second fter tht, the cr s speed is 55 km/h. The cr s speed ws incresing the entire time, but its ccelertion in the second time intervl ws lower thn in the first time intervl. Thus its ccelertion ws decresing even s the speed ws incresing. Another exmple would be n object flling WITH ir resistnce. Let the downwrd direction be positive. As the object flls, it gins speed, nd the ir resistnce increses. As the ir resistnce increses, the ccelertion of the flling object decreses, nd it gins speed less quickly the longer it flls. 9. If the two crs emerge side by side, then the one moving fster is pssing the other one. Thus cr A is pssing cr B. With the ccelertion dt given for the problem, the ensuing motion would be tht cr A would pull wy from cr B for time, but eventully cr B would ctch up to nd pss cr A. 0. If there were no ir resistnce, the bll s only ccelertion during flight would be the ccelertion due to grvity, so the bll would lnd in the ctcher s mitt with the sme speed it hd when it left the bt, 0 km/h. Since the ccelertion is the sme through the entire flight, the time for the bll s speed to chnge from 0 km/h to 0 on the wy up is the sme s the time for its speed to chnge from 0 to 0 km/h on the wy down. In both cses the bll hs the sme mgnitude of displcement.. () If ir resistnce is negligible, the ccelertion of freely flling object stys the sme s the object flls towrd the ground. Tht ccelertion is 9.80 m/s. Note tht the object s speed increses, but since tht speed increses t constnt rte, the ccelertion is constnt. In the presence of ir resistnce, the ccelertion decreses. Air resistnce increses s speed increses. If the object flls fr enough, the ccelertion will go to zero nd the velocity will become constnt. Tht velocity is often clled the terminl velocity.. Averge speed is the displcement divided by the time. Since the distnces from A to B nd from B to C re equl, you spend more time trveling t 70 km/h thn t 90 km/h, so your verge speed should be less thn 80 km/h. If the distnce from A to B (or B to C) is x km, then the totl distnce trveled is x. The totl time required to trvel this distnce is x/70 plus x/90. Then d x (90)(70) = = = = km/h 79 km/h. t x/70 + x/ Yes. For exmple, rock thrown stright up in the ir hs constnt, nonzero ccelertion due to grvity for its entire flight. However, t the highest point it momentrily hs zero velocity. A cr, t the moment it strts moving from rest, hs zero velocity nd nonzero ccelertion. 4. Yes. Any time the velocity is constnt, the ccelertion is zero. For exmple, cr trveling t constnt 90 km/h in stright line hs nonzero velocity nd zero ccelertion. 5. A rock flling from cliff hs constnt ccelertion IF we neglect ir resistnce. An elevtor moving from the second floor to the fifth floor mking stops long the wy does NOT hve constnt ccelertion. Its ccelertion will chnge in mgnitude nd direction s the elevtor strts nd stops. The dish resting on tble hs constnt (zero) ccelertion. Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

3 Describing Motion: Kinemtics in One Dimension The slope of the position versus time curve is the object s velocity. The object strts t the origin with constnt velocity (nd therefore zero ccelertion), which it mintins for bout 0 s. For the next 0 s, the positive curvture of the grph indictes the object hs positive ccelertion; its speed is incresing. From 30 s to 45 s, the grph hs negtive curvture; the object uniformly slows to stop, chnges direction, nd then moves bckwrds with incresing speed. During this time intervl, the ccelertion is negtive, since the object is slowing down while trveling in the positive direction nd then speeding up while trveling in the negtive direction. For the finl 5 s shown, the object continues moving in the negtive direction but slows down, which gives it positive ccelertion. During the 50 s shown, the object trvels from the origin to point 0 m wy, nd then bck 0 m to end up 0 m from the strting position. 7. Initilly, the object moves in the positive direction with constnt ccelertion, until bout t = 45 s, when it hs velocity of bout 37 m/s in the positive direction. The ccelertion then decreses, reching n instntneous ccelertion of 0 t bout t = 50 s, when the object hs its mximum speed of bout 38 m/s. The object then begins to slow down but continues to move in the positive direction. The object stops moving t t = 90 s nd stys t rest until bout t = 08 s. Then the object begins to move in the positive direction gin, t first with lrger ccelertion, nd then with lesser ccelertion. At the end of the recorded motion, the object is still moving to the right nd gining speed. Responses to MisConceptul Questions. (, b, c, d, e, f, g) All of these ctions should be prt of solving physics problems.. (d) It is common misconception tht positive ccelertion lwys increses the speed, s in nd (c). However, when the velocity nd ccelertion re in opposite directions, the speed will decrese. 3. (d) Since the velocity nd ccelertion re in opposite directions, the object will slow to stop. However, since the ccelertion remins constnt, it will stop only momentrily before moving towrd the left. 4. (c) Students commonly confuse the concepts of velocity nd ccelertion in free-fll motion. At the highest point in the trjectory, the velocity is chnging from positive (upwrd) to negtive (downwrd) nd therefore psses through zero. This chnging velocity is due to constnt downwrd ccelertion. 5. () Since the distnce between the rocks increses with time, common misconception is tht the velocities re incresing t different rtes. However, both rocks fll under the influence of grvity, so their velocities increse t the sme rte. 6. (c) Since the distnces re the sme, common error is to ssume tht the verge speed will be hlfwy between the two speeds, or 40 km/h. However, it tkes the cr much longer to trvel the 4 km t 30 km/h thn t 50 km/h. Since more time is spent t 30 km/h, the verge speed will be closer to 30 km/h thn to 50 km/h. 7. (c) A common misconception is tht the ccelertion of n object in free fll depends upon the motion of the object. If there is no ir resistnce, the ccelertions for the two blls hve the sme mgnitude nd direction throughout both of their flights. 8. (b, c) Ech of the given equtions is bsed on Eqs. d. Answer () hs the ccelertion replced properly with g, but the initil velocity is downwrd nd s such should be negtive. Answer (d) is Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

4 -4 Chpter incorrect becuse the initil velocity hs been inserted for the verge velocity. Answers nd (c) hve the correct signs for ech vrible nd the known vlues re inserted properly. 9. () Incresing speed mens tht the slope must be getting steeper over time. In grphs nd (e), the slope remins constnt, so these re crs moving t constnt speed. In grph (c), s time increses x decreses. However, the rte t which it decreses is lso decresing. This is cr slowing down. In grph (d), the cr is moving wy from the origin, but gin it is slowing down. The only grph in which the slope is incresing with time is grph (). Solutions to Problems. The distnce of trvel (displcement) cn be found by rerrnging Eq. for the verge velocity. Also note tht the units of the velocity nd the time re not the sme, so the speed units will be converted. h =Δx/ Δt Δ x = Δ t = (95 km/h) (. 0 s) = km = 53 m 3600 s. The verge speed is given by Eq., using d to represent distnce trveled. = d/ Δ t = 35 km/.75 h = 85.5 km/h 3. The verge velocity is given by Eq.. Δx 8.5 cm 4.8 cm 3.7 cm = = = = 0.57 cm/s Δt 4.5 s (.0 s) 6.5 s The verge speed cnnot be clculted. To clculte the verge speed, we would need to know the ctul distnce trveled, nd it is not given. We only hve the displcement. 4. The verge velocity is given by Eq.. Δx 4.cm 84cm..6cm = = = = 4. cm/s Δ t 6. s 3.0 s 3. s The negtive sign indictes the direction. 5. The time of trvel cn be found by rerrnging the verge velocity eqution. =Δx/ Δt Δ t =Δ x/ = (3.5 km)/(5 km/h) = 0.4 h = 8.4 min 6. () The speed of sound is intimted in the problem s mile per 5 seconds. The speed is clculted s follows: distnce mi 60 m speed= = = 3m/s 300m/s time 5 s mi The speed of 3 m/s would imply the sound trveling distnce of 966 meters (which is pproximtely km) in 3 seconds. So the rule could be pproximted s km every 3 seconds. 7. The time for the first prt of the trip is clculted from the initil speed nd the first distnce, using d to represent distnce. d d 80 km = Δ t = = =.895 h = 3.7 min Δt 95 km/h Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

5 Describing Motion: Kinemtics in One Dimension -5 The time for the second prt of the trip is now clculted. Δ t =Δttotl Δ t = 4.5 h.895 h =.605 h = 56.3 min The distnce for the second prt of the trip is clculted from the verge speed for tht prt of the trip nd the time for tht prt of the trip. d = d = Δ t = (65 km/h)(. 605 h) = 69.3 km 70 km Δt () The totl distnce is then dtotl = d + d = 80 km km = km 350 km. The verge speed is NOT the verge of the two speeds. Use the definition of verge speed, Eq.. dtotl km = = = 77.6 km/h 78 km/h Δ t 45h. totl 8. The distnce trveled is 38 m + (38 m) = 57 m, nd the displcement is 38 m (38 m) = 9 m. The totl time is 9.0 s +.8 s = 0.8 s. () distnce 57 m Averge speed = 5.3 m/s time elpsed = 0.8 s = displcement 9 m Averge velocity = vg =.8 m/s time elpsed = 0.8 s = 9. The distnce trveled is 300 m (8 lps 400 m/lp). Tht distnce probbly hs either 3 or 4 significnt figures, since the trck distnce is probbly known to t lest the nerest meter for competition purposes. The displcement is 0, becuse the ending point is the sme s the strting point. d 300 m min () Averge speed = = = 3.68 m/s Δt 4.5 min 60 s Averge velocity = =Δx/ Δ t = 0m/s 0. The verge speed is the distnce divided by the time. 9 d 0 km yr d 5 5 = = =.4 0 km/h 0 km/h t yr 365.5d 4h. Both objects will hve the sme time of trvel. If the truck trvels distnce d truck, then the distnce the cr trvels will be dcr = dtruck + 0 m. Using the eqution for verge speed, = d/ Δ t, solve for time, nd equte the two times. dtruck dcr dtruck dtruck + 0 m Δ t = = = 75 km/h 95 km/h truck cr Solving for d truck gives d (75 km/h) truck = (0 m) m. (95km/h 75km/h) = The time of trvel is dtruck m 60 min Δ t = = = 0.63 min = 37.8 s 38 s truck 75, 000 m/h h Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

6 -6 Chpter dcr m + 0 m 60 min Also note tht Δ t = = = 0.63min = 37.8 s. cr 95, 000 m/h h ALTERNATE SOLUTION: The speed of the cr reltive to the truck is 95 km/h 75 km/h = 0 km/h. In the reference frme of the truck, the cr must trvel 0 m to ctch it. 0. km 3600 s Δ t = = 37.8 s 0 km/h h. The distnce trveled is 500 km (50 km outgoing, 50 km return, keep significnt figures). The displcement ( Δx) is 0 becuse the ending point is the sme s the strting point. To find the verge speed, we need the distnce trveled (500 km) nd the totl time elpsed. Δx Δx 50 km During the outgoing portion, =, so Δ t = = =.63 h. During the return portion, Δ t 95 km/h Δx Δx 50 km =, so Δ t = = = h. Thus the totl time, including lunch, is Δ t 55 km/h Δ ttotl = Δ t +Δ tlunch +Δ t = 8.77 h. Δxtotl 500 km = = = 6 km/h Δt 8.77 h totl To find the verge velocity, use the displcement nd the elpsed time. =Δx/ Δ t = 0 3. Since the locomotives hve the sme speed, they ech trvel hlf the distnce, 4.5 km. Find the time of trvel from the verge speed. d d 4.5 km 60 min = Δ t = = = h =.645 min.6 min 99 s Δt 55 km/h h 4. () The re between the concentric circles is equl to the length times the width of the spirl pth. π ( R R ) π[ (0.058 m) (0.05m) ] 3 π 6 πr R = w = = m 5400 m w.6 0 m 3 s min m = min 75 min. m 60 s 5. The verge speed of sound is given by sound =Δx/ Δ t, so the time for the sound to trvel from the Δx 6.5 m end of the lne bck to the bowler is sound Δ t = = = s. Thus the time for the sound 340 m/s bll to trvel from the bowler to the end of the lne is given by Δ tbll = Δttotl Δ tsound =.80 s s =.755 s. The speed of the bll is s follows: Δx 6.5 m bll = = = m/s 6.00 m/s Δt.755 s bll Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

7 Describing Motion: Kinemtics in One Dimension For the cr to pss the trin, the cr must trvel the length of the trin AND the distnce the trin trvels. The distnce the cr trvels cn thus be written s either dcr = cr t = (95 km/h) t or dcr = trin + trint =.30 km + (75 km/h) t. To solve for the time, equte these two expressions for the distnce the cr trvels..30 km (95 km/h) t =.30 km + (75 km/h) t t = = h = 3.9 min 0 km/h Note tht this is the sme s clculting from the reference frme of the trin, in which the cr is moving t 0 km/h nd must trvel the length of the trin. The distnce the cr trvels during this time is d = (95 km/h)(0.065 h) = 6.75 km 6. km. If the trin is trveling in the opposite direction from the cr, then the cr must trvel the length of the trin MINUS the distnce the trin trvels. Thus the distnce the cr trvels cn be written s either dcr = (95 km/h) t or dcr =.30 km (75 km/h) t. To solve for the time, equte these two expressions for the distnce the cr trvels..30 km 3 (95 km/h) t =.30 km (75 km/h) t t = = h 8 s 70 km/h The distnce the cr trvels during this time is 3 d = (95 km/h)( h) = 0.73 km. 7. The verge ccelertion is found from Eq m h (95 km/h) Δ 95 km/h 0 km/h km 3600s = = = = Δt 4.3 s 4.3 s 6. m/s 8. () Δ 9.00 m/s 0.00 m/s The verge ccelertion of the sprinter is = = = Δt.38 s We chnge the units for the ccelertion. km 3600 s 4 = (6. 5 m/s ) = km/h 000 m h 6.5 m/s. 9. The initil velocity of the cr is the verge velocity of the cr before it ccelertes. Δx 0 m = = = 4 m/s = 0 Δt 5.0 s The finl velocity is = 0, nd the time to stop is 4.0 s. Use Eq. to find the ccelertion m/s = 0 + t = = = 6.0 m/s t 4.0 s Thus the mgnitude of the ccelertion is 6.0 m/s, or g (6.0 m/s ) 0.6 g s. = 9.80 m/s 0. We ssume tht the speedometer cn red to the nerest km/h, so the vlue of 0 km/h hs three significnt digits. The time cn be found from the verge ccelertion, =Δ/ Δ t. Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

8 -8 Chpter m/s (55 km/h) Δ 0 km/h 65 km/h 3.6 km/h Δ t = = = = s 8.5 s.8 m/s.8 m/s. () Δx 385 m 5 m = = =. m/s Δt 0.0s 3.0s Δ 45.0 m/s.0 m/s = = =.00 m/s Δt 0.0s 3.0s. The ccelertion cn be found from Eq. c. 0 0 (8m/s) = 0 + x ( x0) = = = 4.5m/s ( x x0 ) (88 m) 0 m/s 4 m/s 3. By definition, the ccelertion is = = =.67 m/s m/s. t 6.0 s The distnce of trvel cn be found from Eq. b. 0 0 x x = t+ t = (4 m/s)(6.0 s) + (.67 m/s )(6.0 s) = 05 m 0 m It cn lso be found from Eq. 7 nd Eq m/s + m/s x x0 = Δ t = Δ t = (6.0 s) = 05 m 0 m 4. Assume tht the plne strts from rest. The distnce is found by solving Eq. c for x x 0. 0 (35 m/s) 0 = 0 + x x0 x x0 = = = ( ) 04. m.0 0 m (3.0 m/s ) 5. For the bsebll, 0 = 0, x x0 = 3.5 m, nd the finl speed of the bsebll (during the throwing motion) is = 43 m/s. The ccelertion is found from Eq. c. 0 (43 m/s) 0 = 0 + x ( x0) = = = 64 m/s 60 m/s ( x x0 ) (3.5 m) 6. The sprinter strts from rest. The verge ccelertion is found from Eq. c. 0 (.5 m/s) 0 = 0 + x ( x0) = = = m/s 3.67 m/s ( x x0 ) (8.0 m) Her elpsed time is found by solving Eq. for time. 0.5 m/s 0 = 0 + t t = = = 3.3 s m/s 7. The words slows down uniformly imply tht the cr hs constnt ccelertion. The distnce of trvel is found from combining Eqs. 7 nd m/s + 0 m/s x x0 = t = (8.00 s) = m Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

9 Describing Motion: Kinemtics in One Dimension The finl velocity of the cr is zero. The initil velocity is found from Eq. c with = 0 nd solving for 0. Note tht the ccelertion is negtive. 0 x x0 0 x x0 = + ( ) = ( ) = 0 ( 4.00 m/s )(65 m) = 3 m/s 9. The finl velocity of the driver is zero. The ccelertion is found from Eq. c with = 0 nd solving for. 000 m h 0 (95 km/h) km 3600 s 0 = = = 435. m/s 440 m/s ( x x ) (0.80 m) 0 Converting to g s : 435. m/s = = (9.80 m/s )/ g 44 g s. 30. () The finl velocity of the cr is 0. The distnce is found from Eq. c with n ccelertion of = 0.50 m/s nd n initil velocity of 85 km/h. 000 m h 0 (75 km/h) 0 km 3600s x x0 = = = 434m 430m ( 0.50 m/s ) The time to stop is found from Eq m h 0 (75 km/h) km 3600 s 0 t = = = 4.67 s 4 s ( 0.50m/s ) (c) Tke x0 = x( t = 0) = 0. Use Eq. b, with The first second is from = 0.50 m/s nd n initil velocity of 75 km/h. t = 0s to t = s, nd the fifth second is from t = 4s to t = 5s. m/s x(0) = 0; x() = 0 + (75 km/h) (s) + ( 0.50 m/s )(s) = 0.58m 3.6 km/h x() x(0) = 0.58 m m m/s x(4) = 0 + (75 km/h) (4 s) + ( 0.50 m/s )(4 s) = m 3.6 km/h m/s x(5) = 0 + (75 km/h) (5 s) + ( 0.50 m/s )(5 s) = 97.9 m 3.6 km/h x(5) x(4) = 97.9 m m = 8.59 m 9 m 3. The origin is the loction of the cr t the beginning of the rection time. The initil speed of the cr is 000 m h (95 km/h) = 6.39 m/s. The loction where the brkes re pplied is found from km 3600s the eqution for motion t constnt velocity. x0 = 0 t R = (6.39 m/s)(0.40 s) = 0.56 m This is now the strting loction for the ppliction of the brkes. In ech cse, the finl speed is 0. Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

10 -0 Chpter () Solve Eq. c for the finl loction. = 0 + x ( x 0 ) 0 0 (6.39 m/s) x0 x = + = 0.56 m + = 6.63 m 30 m ( 3.0 m/s ) Solve Eq. c for the finl loction with the second ccelertion. 0 0 (6.39 m/s) 0 x = x + = 0.56 m + = 69 m ( 6.0 m/s ) 3. Clculte the distnce tht the cr trvels during the rection time nd the decelertion. Δ x = 0 Δ t = (8.0 m/s)(0.350 s) = 6.3 m 0 0 (8.0m/s) 0 x x = + Δ Δ = = = 44.4m ( 3.65 m/s ) Δ x = 6.3 m m = 50.7 m Since she is only 0.0 m from the intersection, she will NOT be ble to stop in time. She will be 30.7 m pst the intersection. 33. Use the informtion for the first 80 m to find the ccelertion nd the informtion for the full motion to find the finl velocity. For the first segment, the trin hs 0 = 0 m/s, = 8 m/s, nd displcement of x x0 = 80 m. Find the ccelertion from Eq. c. 0 (8 m/s) 0 = 0 + x ( x0) = = = 0.90 m/s ( x x0) (80 m) Find the speed of the trin fter it hs trveled the totl distnce (totl displcement of x x0 = 55 m) using Eq. c. 0 x x0 0 x x0 = + ( ) = + ( ) = (0.90 m/s )(55 m) = m/s 34. Clculte the ccelertion from the velocity time dt using Eq., nd then use Eq. b to clculte the displcement t t =.0 s nd t = 6.0 s. The initil velocity is 0 = 65 m/s. 0 6 m/s 85 m/s = = = 7.7 m/s x = x0 + 0t+ t t 0.0 s x(6.0 s) x(.0 s) = [( x + (6.0 s) + (6.0 s) ) ( x + (.0 s) + (.0 s) )] 0 = (6.0 s.0 s) + [(6.0 s) (.0 s) ] = (85 m/s)(4.0 s) + (7.7 m/s )(3 s ) = 463. m 460 m 35. During the finl prt of the rce, the runner must hve displcement of 00 m in time of 80 s (3.0 min). Assume tht the strting speed for the finl prt is the sme s the verge speed thus fr. Δx 8800 m = = = 5.43 m/s = 0 Δ t (7 60) s Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

11 Describing Motion: Kinemtics in One Dimension - The runner will ccomplish this by ccelerting from speed 0 to speed for t seconds, covering distnce d, nd then running t constnt speed of for (80 t ) seconds, covering distnce d. We hve these reltionships from Eq. nd Eq. b. = + t d = t + t d = (80 t) = ( + t)(80 t) = d+ d = 0t + t t t = 0 + t t 00 m ( )(80 ) 00 m t t 00 m = (80 s)(5.43 m/s) + (80 s)(0.0 m/s ) (0.0 m/s ) 36 ± 36 4(0.0)(.4) 0.0t 36t+.4 = 0 t = = s, 6.8 s (0.0) Since we must hve t < 80 s, the solution is t = 6.3 s. 36. () The trin s constnt speed is trin = 5.0 m/s, nd the loction of the empty box cr s function of time is given by xtrin = trin t = (5.0 m/s) t. The fugitive hs 0 = 0 m/s nd =.4 m/s until his finl speed is 6.0 m/s. The elpsed time during the ccelertion is m/s tcc = = = 4.86 s. Let the origin be the loction of the fugitive when he strts to.4 m/s run. The first possibility to consider is, Cn the fugitive ctch the empty box cr before he reches his mximum speed? During the fugitive s ccelertion, his loction s function of time is given by Eq. b, x = x + t+ t = (.4 m/s ) t. For him to ctch the trin, we must hve fugitive 0 0 trin fugitive x = x (5.0 m/s) t = (.4 m/s ) t. The solutions re t = 0s, 7.s. Thus the fugitive cnnot ctch the cr during his 4.86 s of ccelertion. Now the eqution of motion of the fugitive chnges. After the 4.86 s of ccelertion, he runs with constnt speed of 6.0 m/s. Thus his loction is now given (for times t > 5s) by the following: fugitive x = (.4 m/s )(4.86 s) + (6.0 m/s)( t 4.86 s) = (6.0 m/s) t.86 m So now, for the fugitive to ctch the trin, we gin set the loctions equl. xtrin = xfugitive (5.0 m/s) t = (6.0 m/s) t.86 m t =.86 s 3 s The distnce trveled to rech the box cr is given by the following: x fugitive ( t = 5.0 s) = (6.0 m/s)(.86 s).86 m = 64 m 37. For the runners to cross the finish line side-by-side, they must both rech the finish line in the sme mount of time from their current positions. Tke Mry s current loction s the origin. Use Eq. b. For Slly: = t+ ( 0.40) t t 5t+ 85 = 0 5 ± 5 4(85) t = = s, 0.94 s The first time is the time she first crosses the finish line, so tht is the time to be used for the problem. Now find Mry s ccelertion so tht she crosses the finish line in tht sme mount of time. For Mry: 4t 4(4.059) = 0 + 4t+ t = = = 0.70 m/s t (4.059) Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

12 - Chpter 38. Define the origin to be the loction where the speeder psses the police cr. Strt timer t the instnt tht the speeder psses the police cr nd find nother time tht both crs hve the sme displcement from the origin. For the speeder, trveling with constnt speed, the displcement is given by the following: m/s Δ xs = st = (35 km/h) ( t) = (37.5 t) m 3.6 km/h For the police cr, the displcement is given by two components. The first prt is the distnce trveled t the initilly constnt speed during the second of rection time. m/s Δ xp = p(.00 s) = (95 km/h) (.00 s) = 6.39 m 36. km/h The second prt of the police cr displcement is tht during the ccelerted motion, which lsts for ( t.00) s. So this second prt of the police cr displcement, using Eq. b, is given s follows: p p p Δ x = ( t.00) + ( t.00) = [(6.39 m/s)( t.00) + (. 60 m/s )( t.00) ] m So the totl police cr displcement is the following: Δ xp = Δ xp+δ xp = ( ( t.00) +.30( t.00) ) m Now set the two displcements equl nd solve for the time ( t.00) +.30( t.00) = 37.5 t t 0.55t+.00 = ± (0.55) 4.00 t = = s, 0.5 s The nswer tht is pproximtely 0 s corresponds to the fct tht both vehicles hd the sme displcement of zero when the time ws 0. The reson it is not exctly zero is rounding of previous vlues. The nswer of 0.5 s is the time for the police cr to overtke the speeder. s check on the nswer, the speeder trvels Δ xs = (37.5 m/s)(0.5 s) = 394 m, nd the police cr trvels Δ x p = [ (9.5) +.30(9.5) ] m = 394 m. 39. Choose downwrd to be the positive direction, nd tke y 0 = 0 t the top of the cliff. The initil velocity is 0 = 0, nd the ccelertion is with x replced by y. 0 0 = 9.80 m/s. The displcement is found from Eq. b, y = y + t+ t y 0 = 0 + (9.80 m/s )(3.55 s) y = 6.8 m 40. Choose downwrd to be the positive direction, nd tke y 0 = 0 to be t the top of the Empire Stte Building. The initil velocity is 0 = 0, nd the ccelertion is = 9.80 m/s. () The elpsed time cn be found from Eq. b, with x replced by y. y (380 m) y y0 = 0t+ t t = = = s 8.8 s 9.80 m/s The finl velocity cn be found from Eq.. = 0 + t = 0 + (9.80 m/s )(8.806 s) = 86 m/s Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

13 Describing Motion: Kinemtics in One Dimension Choose upwrd to be the positive direction, nd tke y 0 = 0 to be the height from which the bll ws thrown. The ccelertion is = 9.80 m/s. The displcement upon ctching the bll is 0, ssuming it ws cught t the sme height from which it ws thrown. The strting speed cn be found from Eq. b, with x replced by y. y = y + t+ t = y y0 t 0 t = = = ( 9.80 m/s )(3.4 s) = 6.66 m/s 7 m/s t The height cn be clculted from Eq. c, with finl velocity of = 0 t the top of the pth. 0 0 (6.66 m/s) 0 y y0 y y0 = + ( ) = + = 0+ = 4 m ( 9.80 m/s ) 4. Choose upwrd to be the positive direction, nd tke y 0 = 0 to be t the height where the bll ws hit. For the upwrd pth, 0 = 5 m/s, = 0 t the top of the pth, nd = 9.80 m/s. () The displcement cn be found from Eq. c, with x replced by y. 0 0 (5 m/s) 0 y y0 y y0 = + ( ) = + = 0+ = 3m ( 9.80 m/s ) The time of flight cn be found from Eq. b, with x replced by y, using displcement of 0 for the displcement of the bll returning to the height from which it ws hit. (c) y = y + t+ t = 0 t( + t) = 0 0 (5 m/s) t = 0, t = = = 5. s 9.80 m/s The result of t = 0 s is the time for the originl displcement of zero (when the bll ws hit), nd the result of t = 5. s is the time to return to the originl displcement. Thus the nswer is t = 5. seconds. This is n estimte primrily becuse the effects of the ir hve been ignored. There is nontrivil mount of ir effect on bsebll s it moves through the ir tht s why pitches like the curve bll work, for exmple. So ignoring the effects of ir mkes this n estimte. Another effect is tht the problem sys lmost stright up, but the problem ws solved s if the initil velocity ws perfectly upwrd. Finlly, we ssume tht the bll ws cught t the sme height s which it ws hit. Tht ws not stted in the problem either, so tht is n estimte. 43. Choose downwrd to be the positive direction, nd tke y 0 = 0 to be t the mximum height of the kngroo. Consider just the downwrd motion of the kngroo. Then the displcement is y =.45 m, the ccelertion is = 9.80 m/s, nd the initil velocity is 0 = 0. Use Eq. b to clculte the time for the kngroo to fll bck to the ground. The totl time is then twice the flling time. y y = y0 + 0t + t = 0 y = t t fll = t y (.45m) = = =.09 s (9.80 m/s ) totl Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

14 -4 Chpter 44. Choose upwrd to be the positive direction, nd tke y 0 = 0 to be t the floor level, where the jump strts. For the upwrd pth, y =. m, = 0 t the top of the pth, nd () The initil speed cn be found from Eq. c, with x replced by y. = + y ( y) y y0 y = 9.80 m/s. = ( ) = = ( 9.80 m/s )(. m) = m/s 4.8 m/s The time of flight cn be found from Eq. b, with x replced by y, using displcement of 0 for the displcement of the jumper returning to the originl height y = y + t+ t = 0 t( + t) = 0 0 (4.897 m/s) t = 0, t = = = 0.99 s 9.80 m/s The result of t = 0 s is the time for the originl displcement of zero (when the jumper strted to jump), nd the result of t = 0.99 s is the time to return to the originl displcement. Thus the nswer is t = 0.99 seconds. 45. Choose downwrd to be the positive direction, nd tke y 0 = 0 to be t the height where the object ws relesed. The initil velocity is 0 = 0, nd the ccelertion is = 9.80 m/s. () The speed of the object will be given by Eq. with 0 = 0, so = t = (9.80 m/s ) t. This is the eqution of stright line pssing through the origin with slope of 9.80 m/s. The distnce fllen will be given by Eq. b with 0 = 0, so 0 0 y = y + t+ t = (4.90 m/s ) t. This is the eqution of prbol, with its vertex t the origin, opening upwrd. 46. Choose upwrd to be the positive direction, nd y 0 = 0 to be the height from which the stone is thrown. We hve 0 = 4.0 m/s, = 9.80 m/s, nd y y0 = 3.0 m. () The velocity cn be found from Eq. c, with x replced by y. 0 y y 0 = + ( ) = 0 =± 0 + y =± (4.0 m/s) + ( 9.80 m/s )(3.0 m) =± 7.9 m/s Thus the speed is = 7.9 m/s. Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

15 Describing Motion: Kinemtics in One Dimension -5 The time to rech tht height cn be found from Eq. b. (4.0 m/s) ( 3.0 m) y = y0 + 0t+ t t + t+ = m/s 9.80 m/s ± (4.898) 4(.653) t 4.898t+.653 = 0 t = = t = 4.8 s, 0.60 s (c) There re two times t which the object reches tht height once on the wy up ( t = 0.60s) nd once on the wy down ( t = 4.8s). 47. Choose downwrd to be the positive direction, nd tke y 0 = 0 to be the height from which the object is relesed. The initil velocity is 0 = 0, nd the ccelertion is = g. Then we cn clculte the position s function of time from Eq. b, with x replced by y, s yt () = gt. At the end of ech second, the position would be s follows: y(0) = 0; y() = g; y() = g() = 4 y(); y(3) = g(3) = 9 y() The distnce trveled during ech second cn be found by subtrcting two djcent position vlues from the bove list. d() = y() y(0) = y(); d() = y() y() = 3 y(); d(3) = y(3) y() = 5 y() We could do this in generl. Let n be positive integer, strting with 0. = gn ( + n+ n) = g(n+ ) yn ( ) = gn yn ( + ) = gn ( + ) dn ( + ) = yn ( + ) yn ( ) = gn ( + ) gn = g(( n+ ) n ) The vlue of (n + ) is lwys odd, in the sequence, 3, 5, 7,. 48. () Choose upwrd to be the positive direction, nd y 0 = 0 t the ground. The rocket hs 0 = 0, = 3. m/s, nd y = 775 m when it runs out of fuel. Find the velocity of the rocket when it runs out of fuel from Eq. c, with x replced by y. = + y ( y) 775 m m =± 0 + y y0 =± + = ( ) 0 (3. m/s )(775 m) m/s m/s The positive root is chosen since the rocket is moving upwrd when it runs out of fuel. Note tht the vlue hs significnt figures. The time to rech the 775 m loction cn be found from Eq m m/s m = 0 + t775 m t775 m = = =.0 s s 3. m/s (c) For this prt of the problem, the rocket will hve n initil velocity 0 = m/s, n ccelertion of = 9.80 m/s, nd finl velocity of = 0 t its mximum ltitude. The ltitude reched from the out-of-fuel point cn be found from Eq. c. = + y ( 775m) y 775 m m (70.43 m/s) mx = 775 m + = 775 m + = 775 m + 53 m = 08 m 030 m ( 9.80 m/s ) Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

16 -6 Chpter (d) The time for the costing portion of the flight cn be found from Eq m/s = 775 m + tcost tcost = = = 7.9 s 9.80 m/s Thus the totl time to rech the mximum ltitude is t =.0 s s = 9.0 s 9 s. (e) For this prt of the problem, the rocket hs 0 = 0 m/s, (f) = 9.80 m/s, nd displcement of 08 m (it flls from height of 08 m to the ground). Find the velocity upon reching the Erth from Eq. c. = + y ( y) y y0 =± + ( ) =± 0 + ( 9.80 m/s )( 08 m) = 4.95 m/s 4 m/s The negtive root ws chosen becuse the rocket is moving downwrd, which is the negtive direction. The time for the rocket to fll bck to the Erth is found from Eq m/s 0 = 0 + t tfll = = = 4.48 s 9.80 m/s Thus the totl time for the entire flight is t = 9.0 s s = s 44 s. 49. Choose downwrd to be the positive direction, nd tke y 0 = 0 to be the height where the object ws relesed. The initil velocity is 0 = 5.40 m/s, the ccelertion is = 9.80 m/s, nd the displcement of the pckge will be y = 05 m. The time to rech the ground cn be found from Eq. b, with x replced by y. 0 y ( 5.40 m/s) (05 m) y = y0 + 0t+ t t + t = 0 t + t = m/s 9.80 m/s.0 ± (.0) 4(.43) t.0t.43 = 0 t = = 5. s, 4. s The correct time is the positive nswer, t = 5. s. 50. () Choose y = 0 to be the ground level nd positive to be upwrd. Then y 0 = 5 m, = g, nd t = 0.83 s describe the motion of the blloon. Use Eq. b. 0 0 y y0 t 0 5 m ( 9.80 m/s )(0.83 s) y = y + t+ t 0 = = = 4.0 m/s 4 m/s t (0.83 s) So the speed is 4 m/s. Consider the chnge in velocity from being relesed to being t Roger s room, using Eq. c. 0 ( 4.0 m/s) 0 y y = + Δ Δ = = = 0.0 m ( 9.8 m/s ) Thus the blloons re coming from two floors bove Roger, or the fifth floor. Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

17 Describing Motion: Kinemtics in One Dimension Choose upwrd to be the positive direction nd y 0 = 0 to be the loction of the nozzle. The initil velocity is 0, the ccelertion is = 9.80 m/s, the finl loction is y =.8 m, nd the time of flight is t =.5 s. Using Eq. b nd substituting y for x gives the following: 0 0 y t.8 m ( 9.80 m/s )(.5 s) y = y + t+ t 0 = = =.53 m/s m/s t.5 s 5. Choose upwrd to be the positive direction nd y 0 = 0 to be the level from which the bll ws thrown. The initil velocity is 0, the instntneous velocity is = 4 m/s, the ccelertion is = 9.80 m/s, nd the loction of the window is y = 8 m. () Using Eq. c nd substituting y for x, we hve = + y ( y) y y0 =± ( ) =± (4 m/s) ( 9.80 m/s )(8 m) = 3.43 m/s 3 m/s Choose the positive vlue becuse the initil direction is upwrd. At the top of its pth, the velocity will be 0, so we cn use the initil velocity s found bove, long with Eq. c. 0 0 (3.43 m/s) 0 y y0 y y0 = + ( ) = + = 0+ = 8m ( 9.80 m/s ) (c) We wnt the time elpsed from throwing (velocity 0 = 3.43 m/s) to reching the window (velocity = 4 m/s). Using Eq., we hve the following: 0 4 m/s 3.43 m/s = 0 + t t = = = 0.96 s 0.96 s 9.80 m/s (d) We wnt the time elpsed from the window ( 0 = 4 m/s) to reching the street ( = 3.43 m/s). Using Eq., we hve: m/s 4 m/s = 0 + t t = = = 3.89 s 3.8 s 9.80 m/s The totl time from throwing to reching the street gin is 096s s. = 48s Choose downwrd to be the positive direction nd y 0 = 0 to be the height from which the stone is dropped. Cll the loction of the top of the window y w, nd the time for the stone to fll from relese to the top of the window is t w. Since the stone is dropped from rest, using Eq. b with y substituting for x, we hve y w = y0 + 0t+ t = gt w. The loction of the bottom of the window is y w +. m, nd the time for the stone to fll from relese to the bottom of the window is t s. Since the stone is dropped from rest, using Eq. b, we hve the following: w w +. m = = ( w + 0.3s) y y t t g t Substitute the first expression for y w into the second one nd solve for the time. Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

18 -8 Chpter gtw gtw gt w gt w tw. = g( t w(0.3) + (0.3) ). = tw(0.3) g+ g(0.3). g(0.3) tw = = s +. m = ( s) +. = ( + (0.3) + (0.3) ) (0.3) g Use this time in the first eqution to find the desired distnce. y w gt w = = (9.80 m/s )(0.569 s) =.587 m.6 m 54. For the flling rock, choose downwrd to be the positive direction nd y 0 = 0 to be the height from which the stone is dropped. The initil velocity is 0 = 0 m/s, the ccelertion is = g, the finl position is y = H, nd the time of fll is t. Using Eq. b with y substituting for x, we hve 0 0 H = y + t+ t = gt. For the sound wve, use the constnt speed eqution tht Δx H H s = =, which cn be rerrnged to give t = T, where T = 3.4 s is the totl time Δt T t s elpsed from dropping the rock to hering the sound. Insert this expression for t into the eqution for H from the stone, nd solve for H. H g gt 0 s s s 5 4 H = g T H + H + gt = H.098H = 0 H = 5.7 m,.59 0 m H If the lrger nswer is used in t = T, negtive time of fll results, so the physiclly correct nswer is H = 5 m. s 55. Slightly different nswers my be obtined since the dt come from reding the grph. () The gretest velocity is found t the highest point on the grph, which is t t 48 s. The indiction of constnt velocity on velocity vs. time grph is slope of 0, which occurs from t = 90 s to t 08 s. (c) The indiction of constnt ccelertion on velocity vs. time grph is constnt slope, which occurs from t = 0 s to t 4 s, gin from t 65 s to t 83 s, nd gin from t = 90 s to t 08 s. (d) The mgnitude of the ccelertion is gretest when the mgnitude of the slope is gretest, which occurs from t 65 s to t 83 s. 56. Slightly different nswers my be obtined since the dt come from reding the grph. We ssume tht the short, nerly horizontl portions of the grph re the times tht shifting is occurring, nd those times re not counted s being in certin ger. Δ 4 m/s 4 m/s () The verge ccelertion in nd ger is given by = = =.5 m/s. Δt 8 s 4 s The verge ccelertion in 4th ger is given by 4 4 Δ4 44 m/s 37 m/s = = = 0.64 m/s. Δt 7 s 6 s Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

19 Describing Motion: Kinemtics in One Dimension Slightly different nswers my be obtined since the dt come from reding the grph. () The instntneous velocity is given by the slope of the tngent line to the curve. At t = 0.0 s, 3 m 0 the slope is pproximtely (0) = 0.3 m/s. 0.0 s 0 At t = 30.0 s, the slope of the tngent line to the curve, nd thus the instntneous velocity, is 0 m 8 m pproximtely (30) =. m/s. 35 s 5 s (c) (d) (e) x(5) x(0).5 m 0 The verge velocity is given by = = = 0.30 m/s. 5.0s 0s 5.0s x(30) x(5) 6 m 9 m The verge velocity is given by = = =.4 m/s s 5.0 s 5.0 s x(50) x(40) 0 m 9.5 m The verge velocity is given by = = = 0.95 m/s s 40.0 s 0.0 s 58. Slightly different nswers my be obtined since the dt come from reding the grph. () The indiction of constnt velocity on position versus time grph is constnt slope, which occurs from t = 0s to t 8s. (c) The gretest velocity will occur when the slope is the highest positive vlue, which occurs t bout t = 7 s. The indiction of 0 velocity on position versus time grph is slope of 0, which occurs t bout t = 38 s. (d) The object moves in both directions. When the slope is positive, from t = 0 s to t = 38 s, the object is moving in the positive direction. When the slope is negtive, from t = 38 s to t = 50 s, the object is moving in the negtive direction. 59. The vs. t grph is found by tking the slope of the x vs. t grph. Both grphs re shown here. 60. Choose the upwrd direction to be positive nd y 0 = 0 to be the level from which the object ws thrown. The initil velocity is 0 nd the velocity t the top of the pth is = 0. The height t the top of the pth cn be found from Eq. c with x replced by y. Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

20 -0 Chpter 0 = 0 + y ( y0) y y0 = From this we see tht the displcement is inversely proportionl to the ccelertion, so if the ccelertion is reduced by fctor of 6 by going to the Moon, nd the initil velocity is unchnged, then the displcement increses by fctor of We re treting the vlue of 30 g s s if it hd significnt figures. The initil velocity of the cr is m/s 0 = (95 km/h) = 6.39 m/s. Choose x 0 = 0 to be loction t which the decelertion 3.6 km/h begins. We hve = 0 nd = 30g = 94 m/s. Find the displcement from Eq. c. 0 0 (6.39 m/s) 0 x x0 x x0 = + ( ) = + = 0 + =.8 m. m ( 94 m/s ) 6. () For the free-flling prt of the motion, choose downwrd to be the positive direction nd y 0 = 0 to be the height from which the person jumped. The initil velocity is 0 = 0, ccelertion is = 9.80 m/s, nd the loction of the net is y = 8.0 m. Find the speed upon reching the net from Eq. c with x replced by y. = + y ( y) 0 0 =± 0 + y ( 0) =± (9.80 m/s )(8.0 m) = 8.78 m/s The positive root is selected since the person is moving downwrd. For the net-stretching prt of the motion, choose downwrd to be the positive direction, nd y 0 = 8. 0 m to be the height t which the person first contcts the net. The initil velocity is 0 = 8.78 m/s, the finl velocity is = 0, nd the loction t the stretched position is y = 9.0 m. Find the ccelertion from Eq. c with x replced by y. 0 0 (8.78 m/s) = 0 + y ( y0) = = = 76 m/s ( y y0 ) (.0 m) This is bout 8 g s. For the ccelertion to be smller, in the bove eqution we see tht the displcement would hve to be lrger. This mens tht the net should be loosened. 63. Choose downwrd to be the positive direction nd y 0 = 0 to be t the strt of the pelicn s dive. The pelicn hs n initil velocity of 0 = 0, n ccelertion of = g, nd finl loction of y = 4.0 m. Find the totl time of the pelicn s dive from Eq. b, with x replced by y. y (4.0 m) y = y0 + 0t+ t y = t t dive = = =.69 s 9.80 m/s The fish cn tke evsive ction if he sees the pelicn t time of.69 s 0.0 s =.49 s into the dive. Find the loction of the pelicn t tht time from Eq. b. 0 0 y = y + t+ t = (9.80 m/s )(.49 s) = 0.9 m Thus the fish must spot the pelicn t minimum height from the surfce of the wter of 4.0 m 0.9 m = 3. m. Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

21 Describing Motion: Kinemtics in One Dimension The initil velocity is 0 = 5 km/h, the finl velocity is = 65 km/h, nd the displcement is x x 0 = 4.0 km = 4000 m. Find the verge ccelertion from Eq. c. 0 x x 0 = + ( ) m/s [(65 km/h) (5 km/h) ] 3.6 km/h = = = m/s ( ) (4000 m) 0 x x0 m/s 65. The speed limit is 40 km/h =. m/s. 3.6 km/h () For your motion, you would need to trvel ( )m = 60m to get the front of the cr to the third stoplight. The time to trvel the 60 m is found using the distnce nd the speed limit. Δx 60 m Δ x = Δt Δ t = = = 4.40 s. m/s No, you cnnot mke it to the third light without stopping, since it tkes you longer thn 3.0 seconds to rech the third light. The second cr needs to trvel 65 m before the third light turns red. This cr ccelertes from 0 = 0 to mximum of =. m/s with =.00 m/s. Use Eq. to determine the durtion of tht ccelertion. 0. m/s 0 m/s = 0 + t tcc = = = s.00 m/s The distnce trveled during tht time is found from Eq. b. 0 cc 0 cc cc ( x x ) = t + t = 0 + (.00 m/s )(5.556 s) = m Since s hve elpsed, there re = s remining to cler the intersection. The cr trvels nother s t speed of. m/s, covering distnce of Δ xconstnt speed = vg t = (. m/s)(7.444 s) = 8.70 m. Thus the totl distnce is m m = 3.57 m. No, the cr cnnot mke it through ll three lights without stopping. The cr hs to trvel nother 5.43 m to cler the third intersection nd is trveling t speed of. m/s. Thus the front of the cr would cler the intersection time Δx 5.43 m t = = = 4.6 s fter the light turns red.. m/s d 66. The verge speed for ech segment of the trip is given by =, Δ t d 00 km For the first segment, Δ t = = =.97 h. 70 km/h d 800 km For the second segment, Δ t = = =.88 h. 990 km/h so d Δ t = for ech segment. Thus the totl time is Δ ttot = Δ t +Δ t =.97 h h = h 5.7 h. Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

22 - Chpter The verge speed of the plne for the entire trip is dtot 00 km km = = = 85.9 km/h 850 km/h. Δ ttot h Note tht Eq. d does NOT pply in this sitution. 67. () Choose downwrd to be the positive direction nd y 0 = 0 to be the level from which the cr ws dropped. The initil velocity is 0 = 0, the finl loction is y = H, nd the ccelertion is = g. Find the finl velocity from Eq. c, replcing x with y = + y ( y) =± + y ( y) = ± gh The speed is the mgnitude of the velocity, = gh. Solving the bove eqution for the height, we hve tht = 35 km/h, the corresponding height is s follows: H =. Thus for collision of g m/s (35 km/h) 3.6 km/h H = = = 4.83 m 4.8 m g (9.80 m/s ) (c) For collision of = 95 km/h, the corresponding height is the following: m/s (95 km/h) 3.6 km/h H = = = m 36 m g (9.80 m/s ) 68. Choose downwrd to be the positive direction nd y 0 = 0 to be t the roof from which the stones re dropped. The first stone hs n initil velocity of 0 = 0 nd n ccelertion of = g. Eqs. nd b (with x replced by y) give the velocity nd loction, respectively, of the first stone s function of time. 0 t gt y y0 0t t y gt = + = = + + = The second stone hs the sme initil conditions, but its elpsed time is t = t.30 s, so it hs velocity nd loction equtions s follows: gt y gt = (.30 s) = (.30 s) The second stone reches speed of =.0 m/s t time given by.0 m/s t = = t.30 s t =.30 s + =.30 s + =.54 s g g 9.80 m/s The loction of the first stone t tht time is gt y = = (9.80 m/s )(.54 s) = 3. m The loction of the second stone t tht time is y g t = (.30 s) = (9.80 m/s )(.54 s.30 s) = 7.34 m Thus the distnce between the two stones is y y = 3. m 7.34 m = 3.88 m 3.9 m. Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

23 Describing Motion: Kinemtics in One Dimension For the motion in the ir, choose downwrd to be the positive direction nd y 0 = 0 to be t the height of the diving bord. Then diver hs 0 = 0 (ssuming the diver does not jump upwrd or downwrd), = g = 9.80 m/s, nd y = 40. m when reching the surfce of the wter. Find the diver s speed t the wter s surfce from Eq. c, with x replced by y. 0 y y0 x 0 y y0 = + ( ) = ± + ( ) = 0 + (9.80 m/s )(4.0 m) = 8.85 m/s For the motion in the wter, gin choose down to be positive, but redefine y 0 = 0 to be t the surfce of the wter. For this motion, 0 = 8.85 m/s, = 0, nd y y0 =.0 m. Find the ccelertion from Eq. c, with x replced by y. 0 0 (8.85 m/s) = 0 + y ( y0) = = = 9.6 m/s 0 m/s ( y y0 ) x (.0 m) The negtive sign indictes tht the ccelertion is directed upwrd. 70. First consider the uphill lie, in which the bll is being putted down the hill. Choose x 0 = 0 to be the bll s originl loction nd the direction of the bll s trvel s the positive direction. The finl velocity of the bll is = 0, the ccelertion of the bll is =.8 m/s, nd the displcement of the bll will be x x0 = 6.0 m for the first cse nd x x0 = 8.0 m for the second cse. Find the initil velocity of the bll from Eq. c. = = 0 + x ( x0) 0 = x ( x0) = = 0 (.8 m/s )(6.0 m) 4.65 m/s 0 (.8 m/s )(8.0 m) 5.37 m/s The rnge of cceptble velocities for the uphill lie is 4.65 m/s to 5.37 m/s, spred of 0.7 m/s. Now consider the downhill lie, in which the bll is being putted up the hill. Use very similr setup for the problem, with the bsic difference being tht the ccelertion of the bll is now =.6 m/s. Find the initil velocity of the bll from Eq. c. = = 0 + x ( x0) 0 = x ( x0) = = 0 (.6 m/s )(6.0 m) 5.59 m/s 0 (.6 m/s )(8.0 m) 6.45 m/s The rnge of cceptble velocities for the downhill lie is 5.59 m/s to 6.45 m/s, spred of 0.86 m/s. Becuse the rnge of cceptble velocities is smller for putting down the hill, more control in putting is necessry, so putting the bll downhill (the uphill lie ) is more difficult. 7. Choose upwrd to be the positive direction nd y 0 = 0 to be t the throwing loction of the stone. The initil velocity is 0 = 5.5 m/s, the ccelertion is = 9.80 m/s, nd the finl loction is y = 75 m. () Using Eq. b nd substituting y for x, we hve the following: y = y + t+ t (4.9 m/s ) t (5.5 m/s) t 75 m = ± (5.5) 4(4.9)( 75) t = = 5.80 s,.638 s (4. 9) The positive nswer is the physicl nswer: t = 5.80 s. Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

24 -4 Chpter (c) Use Eq. to find the velocity just before hitting. = 0 + t = 5.5 m/s + ( 9.80 m/s )(5.80 s) = 4.4 m/s = 4.4 m/s The totl distnce trveled will be the distnce up plus the distnce down. The distnce down will be 75 m more thn the distnce up. To find the distnce up, use the fct tht the speed t the top of the pth will be 0. Then using Eq. c we hve the following: 0 0 (5.5m/s) 0 y y0 y y0 = + ( ) = + = 0+ =.6m ( 9.80 m/s ) Thus the distnce up is.6 m, the distnce down is 87.6 m, nd the totl distnce trveled is 99.5 m. 7. This problem cn be nlyzed s series of three one-dimensionl motions: the ccelertion phse, the constnt-speed phse, nd the decelertion phse. The mximum speed of the trin is s follows: m/s (95 km/h) = 6.39 m/s 3.6 km/h In the ccelertion phse, the initil velocity is 0 = 0, the ccelertion is velocity is =. m/s, nd the finl = 6.39 m/s. Find the elpsed time for the ccelertion phse from Eq m/s 0 = 0 + t tcc = = = 3.99 s. m/s Find the displcement during the ccelertion phse from Eq. b. 0 cc 0 ( x x ) = t+ t = 0 + (. m/s )(3.99 s) = 36.5 m In the decelertion phse, the initil velocity is 0 = 6.39 m/s, the ccelertion is =.0 m/s, nd the finl velocity is = 0. Find the elpsed time for the decelertion phse from Eq m/s = 0 + t tdec = = = 3.0 s.0 m/s Find the distnce trveled during the decelertion phse from Eq. b. 0 dec 0 ( x x ) = t+ t = (6.39 m/s)(3. 0 s) + (.0 m/s )(3.0 s) = 74. m The totl elpsed time nd distnce trveled for the ccelertion/decelertion phses re: tcc + tdec = 3.99 s s = 37.9 s ( x x ) + ( x x ) = 36.5 m m = 49 m 0 cc 0 dec () If the sttions re spced 3.0 km = 3000 m prt, then there is totl of 5,000 m m = intersttion segments. A trin mking the entire trip would thus hve totl of 5 intersttion segments nd 4 stops of s ech t the intermedite sttions. Since 49 m is trveled during ccelertion nd decelertion, 3000 m 49 m = 509 m of ech segment is trveled t n verge speed of = 6.39 m/s. The time for tht 509 m is given by Δ x = Δt Δx 509 m Δ t = = = s. Thus totl intersttion segment will tke 6.39 m/s constnt speed Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

25 Describing Motion: Kinemtics in One Dimension s s = 3.6 s. With 5 intersttion segments of 3.6 s ech, nd 4 stops of s ech, the totl time is given by t 30km. = 5(3.6 s) + 4( s) = 749 s =.5 min. If the sttions re spced 5.0 km = 5000 m prt, then there is totl of 5,000 m m = intersttion segments. A trin mking the entire trip would thus hve totl of 3 intersttion segments nd stops of s ech t the intermedite sttions. Since 49 m is trveled during ccelertion nd decelertion, 5000 m 49 m = 4509 m of ech segment is trveled t n verge speed of = 6.39 m/s. The time for tht 4509 m is given by d = t d 4509 m t = = = s. Thus totl intersttion segment will tke 6.39 m/s 37.9 s s = s. With 3 intersttion segments of s ech, nd stops of s ech, the totl time is given by t 50km. = 3(08.05 s) + ( s) = 668 s =. min. m/s 73. The cr s initil speed is 0 = (35 km/h) = 9.7 m/s. 3.6 km/h Cse I: trying to stop. The constrint is, with the brking decelertion of the cr ( = 5.8m/s ), cn the cr stop in 8-m displcement? The.0 seconds hs no reltion to this prt of the problem. Using Eq. c, the distnce trveled during brking is s follows: 0 0 (9. 7 m/s) 0 ( x x ) = = = 8.4 m ( 5.8 m/s ) She cn stop the cr in time. Cse II: crossing the intersection. The constrint is, with the ccelertion of the cr 65 km/h 45 km/h m/s = = m/s, cn she get through the intersection 6. 0 s 3.6 km/h (trvel 43 m) in the.0 seconds before the light turns red? Using Eq. b, the distnce trveled during the.0 s is 0 0 ( x x ) = t+ t = (9.7 m/s)(.0 s) + ( m/s )(.0 s) =.3 m She should stop. 74. The criticl condition is tht the totl distnce covered by the pssing cr nd the pproching cr must be less thn 500 m so tht they do not collide. The pssing cr hs totl displcement composed of severl individul prts. These re (i) the 0 m of cler room t the rer of the truck, (ii) the 0-m length of the truck, (iii) the 0 m of cler room t the front of the truck, nd (iv) the distnce the truck trvels. Since the truck trvels t speed of = 8 m/s, the truck will hve displcement of Δ xtruck = (8 m/s) t. Thus the totl displcement of the cr during pssing is Δ x = 40 m + (8 m/s) t. pssing cr To express the motion of the cr, we choose the origin to be t the loction of the pssing cr when the decision to pss is mde. For the pssing cr, we hve n initil velocity of 0 = 8 m/s nd n ccelertion of =.0 m/s. Find Δ xpssing from Eq. b. cr pssing c 0 0 cr Δ x = x x = t+ t = (8 m/s) t+ (0. 60 m/s ) t Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

26 -6 Chpter Set the two expressions for Δ x equl to ech other in order to find the time required to pss. pss pssing cr tpss tpss t pss t pss 80 s.55 s 40 m + (8 m/s) = (8 m/s) + (0.60 m/s ) 40 m = (0.60 m/s ) t = = 060. Clculte the displcements of the two crs during this time. Δ x = 40 m + (8 m/s)(.55 s) = 47.9 m pssing cr Δ x = t = (5 m/s)(.55 s) = m pproching cr pproching cr Thus the two crs together hve covered totl distnce of 47.9 m m = m, which is more thn llowed. The cr should not pss. 75. Choose downwrd to be the positive direction nd y 0 = 0 to be t the height of the bridge. Agent Bond hs n initil velocity of 0 = 0, n ccelertion of = g, nd will hve displcement of y = 5 m 3.5 m =.5 m. Find the time of fll from Eq. b with x replced by y. y (.5m) y = y0 + 0t+ t t = = =.53 s 9.80 m/s If the truck is pproching with = 5 m/s, then he needs to jump when the truck is distnce wy given by d = t = (5 m/s)(.53 s) = 38.3 m. Convert this distnce into poles. d = (38. 3 m) ( pole/5 m) =.53 poles So he should jump when the truck is bout 5. poles wy from the bridge. 76. The speed of the conveyor belt is found from Eq. for verge velocity. Δx. m Δ x = Δ t = = = m/min 0.43 m/min Δ t.8 min The rte of burger production, ssuming the spcing given is center to center, cn be found s follows: burger m burgers = m min min 77. Choose downwrd to be the positive direction nd the origin to be t the top of the building. The brometer hs y 0 = 0, 0 = 0, nd = g = 9.8 m/s. Use Eq. b to find the height of the building, with x replced by y. y = y + t+ t = (9.8m/s ) t y 0 0 t=.0 y t=.3 = (9.8 m/s )(.0 s) = 9.6 m = (9.8 m/s )(. 3 s) = 5.9 m The difference in the estimtes is 6.3 m. If we ssume the height of the building is the verge of the 6.3 m two mesurements, then the % difference in the two vlues is % 30%..75 m = Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

27 Describing Motion: Kinemtics in One Dimension -7 The intent of the method ws probbly to use the chnge in ir pressure between the ground level nd the top of the building to find the height of the building. The very smll difference in time mesurements, which could be due to humn rection time, mkes 6.3-m difference in the height. This could be s much s floors in error. 78. () The two bicycles will hve the sme velocity t ny time when the instntneous slopes of their x vs. t grphs re the sme. Tht occurs ner the time t A s mrked on the grph. x B Bicycle A hs the lrger ccelertion, becuse its grph is concve upwrd, indicting positive ccelertion. Bicycle B hs no ccelertion becuse its grph hs constnt slope. t (c) The bicycles re pssing ech other t the times t when the two grphs cross, becuse they both hve the sme position t tht time. The grph with the steepest slope is the fster bicycle, so it is the one tht is pssing t tht instnt. So t the first crossing, bicycle B is pssing bicycle A. At the second crossing, bicycle A is pssing bicycle B. (d) Bicycle B hs the highest instntneous velocity t ll times until the time t, where both grphs hve the sme slope. For ll times fter t, bicycle A hs the highest instntneous velocity. The lrgest instntneous velocity is for bicycle A t the ltest time shown on the grph. (e) The bicycles pper to hve the sme verge velocity. If the strting point of the grph for prticulr bicycle is connected to the ending point with stright line, the slope of tht line is the verge velocity. Both pper to hve the sme slope for tht verge line. 79. To find the verge speed for the entire rce, we must tke the totl distnce divided by the totl time. If one lp is distnce of L, then the totl distnce will be 0 L. The time elpsed t given constnt 9L speed is given by t = d/, so the time for the first 9 lps would be t =, nd the time for 96.0 km/h the lst lp would be t = L/, where is the verge speed for the lst lp. Write n expression for the verge speed for the entire rce, nd then solve for. dtotl 0L = = = 00.0 km/h = = 45.0 km/h t 9L L t km/h 00.0 km/h 96.0 km/h 80. Assume tht y 0 = 0 for ech child is the level t which the child loses contct with the trmpoline surfce. Choose upwrd to be the positive direction. () The second child hs 0 = 4.0 m/s, = g = 9.80 m/s, nd = 0m/s t the mximum height position. Find the child s mximum height from Eq. c, with x replced by y. y = 0 + y ( y 0 ) 0 0 (4.0 m/s) y0 = + = 0 + = m 0.8 m ( 9.80 m/s ) Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

28 -8 Chpter Since the first child cn bounce up to one-nd--hlf times higher thn the second child, the first child cn bounce up to height of.5(0.863 m) =.4 m = y y0. Eq. c is gin used to find the initil speed of the first child. = + y ( y) y y0 =± ( ) = 0 ( 9.80 m/s )(.4 m) = m/s 4.9 m/s The positive root ws chosen since the child ws initilly moving upwrd. (c) To find the time tht the first child ws in the ir, use Eq. b with totl displcement of 0, since the child returns to the originl position. 0 0 y = y + t + t 0 = (4.898 m/s) t + ( 9.80 m/s ) t t = 0 s, s The time of 0 s corresponds to the time the child strted the jump, so the correct nswer is.0 s. 8. Choose downwrd to be the positive direction nd the origin to be t the loction of the plne. The prchutist hs 0 = 0, = g = 9.80 m/s, nd will hve y y0 = 300 m 450 m = 750 m when she pulls the ripcord. Eq. b, with x replced by y, is used to find the time when she pulls the ripcord y = y + t+ t t = ( y y )/ = (750 m)/(9.80 m/s ) = 3.69 s 3.7 s The speed is found from Eq km/h = 0 + t = 0 + (9. 80 m/s )(3.69 s) = 3.6 m/s 30 m/s 840 km/h m/s This is well over 500 miles per hour! 8. As shown in Exmple 5, the speed with which the bll ws thrown upwrd is the sme s its speed on returning to the ground. From the symmetry of the two motions (both motions hve speed = 0 t top, hve sme distnce trveled, nd hve sme ccelertion), the time for the bll to rise is the sme s the time for the bll to fll,.4 s. Choose upwrd to be the positive direction nd the origin to be t the level where the bll ws thrown. For the bll, = 0 t the top of the motion, nd = g. Find the initil velocity from Eq.. = 0 + t 0 = t = 0 ( 9.80 m/s )(.4 s) = 3.7 m/s 4 m/s 83. () Multiply the reding rte times the bit density to find the bit reding rte.. m bit 6 N = = bits/s s m The number of excess bits is N N N N 0 = bits/s.4 0 bits/s =.9 0 bits/s 6 N N0.9 0 bits/s = = 0.67 = 67% N bits/s Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.

29 Describing Motion: Kinemtics in One Dimension -9 Solutions to Serch nd Lern Problems. The two conditions re tht the motion needs to be ner the surfce of the Erth nd tht there is no ir resistnce. An exmple where the second condition is not even resonble pproximtion is tht of prchuting. The ir resistnce cused by the prchute results in the ccelertion not being constnt, with vlues much different thn 9.8 m/s.. The sounds will not occur t equl time intervls becuse the longer ny prticulr bolt flls, the higher its speed. With equl distnces between bolts, ech successive bolt, hving fllen longer time when its predecessor reches the plte, will hve higher verge velocity nd thus trvel the interbolt distnce in shorter periods of time. Thus the sounds will occur with smller nd smller intervls between sounds. To her the sounds t equl intervls, the bolts would hve to be tied t distnces corresponding to equl time intervls. The first bolt (cll it bolt #0) is touching the plte. Since ech bolt hs n initil speed of 0, the distnce of fll nd time of fll for ech bolt re relted to ech other by d = Thus for bolt #, d gt =. For bolt #, we wnt t = t, so d = gt = g( t ) = 4( gt ) = 4 d. Likewise, t 3 = 3 t, which leds to d3 = 9 d; t 4 = 4 t, which leds to d4 = 6 d, nd so on. If the distnce from the bolt initilly on the pn to the next bolt is d, then the distnce from tht bolt to the next one is 3 d, the distnce to the next bolt is 5 d, nd so on. The ccompnying tble shows these reltionships in simpler formt. 3. Tke the origin to be the loction where the speeder m/s psses the police cr. The speeder s constnt speed is speeder = (40 km/h) = m/s, 3.6 km/h nd the loction of the speeder s function of time is given by xspeeder = speedertspeeder = (38.89 m/s) tspeeder. The police cr hs n initil velocity of 0 = 0 m/s nd constnt ccelertion of police. The loction of the police cr s function of time is given by Eq. b. () police = 0 + = police police x t t t The position vs. time grphs would qulittively look like the grph shown here. The time to overtke the speeder occurs when the speeder hs gone distnce of 850 m. The time is found using the speeder s eqution from bove. 850 m 850 m = (38.89 m/s) tspeeder tspeeder = =.86 s s m/s x Speeder i gt i Police cr t t. Copyright 04 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist.