KINEMATICS IN ONE DIMENSION PREVIEW Kinemaics is he sudy of how hings move how far (disance and displacemen), how fas (speed and velociy), and how fas ha how fas changes (acceleraion). We say ha an objec moving in a sraigh line is moving in one dimension, and an objec which is moving in a curved pah (like a projecile) is moving in wo dimensions. We relae all hese quaniies wih a se of equaions called he kinemaic equaions. QUICK REFERENCE Imporan Terms acceleraion he rae of change in velociy acceleraion due o graviy he acceleraion of a freely falling objec in he absence of air resisance, which near he earh s surface is approximaely 1 m/s. acceleraion-ime graph plo of he acceleraion of an objec as a funcion of ime average acceleraion he acceleraion of an objec measured over a ime inerval average velociy he velociy of an objec measured over a ime inerval; he displacemen of an objec divided by he change in ime during he moion consan (or uniform) acceleraion acceleraion which does no change during a ime inerval consan (or uniform) velociy velociy which does no change during a ime inerval displacemen change in posiion in a paricular direcion (vecor) disance he lengh moved beween wo poins (scalar) free fall moion under he influence of graviy iniial velociy he velociy a which an objec sars a he beginning of a ime inerval 19
insananeous he value of a quaniy a a paricular insan of ime, such as insananeous posiion, velociy, or acceleraion kinemaics he sudy of how moion occurs, including disance, displacemen, speed, velociy, acceleraion, and ime. posiion-ime graph he graph of he moion of an objec ha shows how is posiion varies wih ime speed he raio of disance o ime velociy raio of he displacemen of an objec o a ime inerval velociy-ime graph plo of he velociy of an objec as a funcion of ime, he slope of which is acceleraion, and he area under which is displacemen Equaions and Symbols v vo a = v = v + a 1 Δx = ( vo + v) 1 Δx = vo + a v o = v o + aδx where Δx = displacemen (final posiion iniial posiion) v = velociy or speed a any ime v o = iniial velociy or speed = ime a = acceleraion
DISCUSSION OF SELECTED SECTIONS Displacemen Disance d can be defined as oal lengh moved. If you run around a circular rack, you have covered a disance equal o he circumference of he rack. Disance is a scalar, which means i has no direcion associaed wih i. Displacemen Δx, however, is a vecor. Displacemen is defined as he sraigh-line disance beween wo poins, and is a vecor which poins from an objec s iniial posiion x o oward is final posiion x f. In our previous example, if you run around a circular rack and end up a he same place you sared, your displacemen is zero, since here is no disance beween your saring poin and your ending poin. Displacemen is ofen wrien in is scalar form as simply Δx or x. Speed and Velociy Average speed is defined as he amoun of disance a moving objec covers divided by he amoun of ime i akes o cover ha disance: average speed = v = disance elapsed ime d = where v sands for speed, d is for disance, and is ime. Average velociy is defined a lile differenly han average speed. While average speed is he oal change in disance divided by he oal change in ime, average velociy is he displacemen divided by he change in ime. Since velociy is a vecor, we mus define i in erms of anoher vecor, displacemen. Ofenimes average speed and average velociy are inerchangeable for he purposes of he AP Physics B exam. Speed is he magniude of velociy, ha is, speed is a scalar and velociy is a vecor. For example, if you are driving wes a 5 miles per hour, we say ha your speed is 5 mph, and your velociy is 5 mph wes. We will use he leer v for boh speed and velociy in our calculaions, and will ake he direcion of velociy ino accoun when necessary. Acceleraion Acceleraion ells us how fas velociy is changing. For example, if you sar from res on he goal line of a fooball field, and begin walking up o a speed of 1 m/s for he firs second, hen up o m/s, for he second second, hen up o 3 m/s for he hird second, you are speeding up wih an average acceleraion of 1 m/s for each second you are walking. We wrie Δv 1m / s m a = = = 1m / s / s = 1 Δ 1s s In oher words, you are changing your speed by 1 m/s for each second you walk. If you sar wih a high velociy and slow down, you are sill acceleraing, bu your acceleraion would be considered negaive, compared o he posiive acceleraion discussed above. 1
Usually, he change in speed Δv is calculaed by he final speed v f minus he iniial speed v o. The iniial and final speeds are called insananeous speeds, since hey each occur a a paricular insan in ime and are no average speeds. Applicaions of he Equaions of Kinemaics for Consan Acceleraion Kinemaics is he sudy of he relaionships beween disance and displacemen, speed and velociy, acceleraion, and ime. The kinemaic equaions are he equaions of moion which relae hese quaniies o each oher. These equaions assume ha he acceleraion of an objec is uniform, ha is, consan for he ime inerval we are ineresed in. The kinemaic equaions lised below would no work for calculaing velociies and displacemens for an objec which is acceleraing erraically. Forunaely, he AP Physics B exam generally deals wih uniform acceleraion, so he kinemaic equaions lised above will be very helpful in solving problems on he es. Freely Falling Bodies An objec is in free fall if i is falling freely under he influence of graviy. Any objec, regardless of is mass, falls near he surface of he Earh wih an acceleraion of 9.8 m/s, which we will denoe wih he leer g. We will round he free fall acceleraion g o 1 m/s for he purpose of he AP Physics B exam. This free fall acceleraion assumes ha here is no air resisance o impede he moion of he falling objec, and his is a safe assumpion on he AP Physics B es unless you are old differenly for a paricular quesion on he exam. Since he free fall acceleraion is consan, we may use he kinemaic equaions o solve problems involving free fall. We simply need o replace he acceleraion a wih he specific free fall acceleraion g in each equaion. Remember, anyime a velociy and acceleraion are in opposie direcions (like when a ball is rising afer being hrown upward), you mus give one of hem a negaive sign.
Example 1 A girl is holding a ball as she seps ono a all elevaor on he ground floor of a building. The girl holds he ball a a heigh of 1 meer above he elevaor floor. The elevaor begins acceleraing upward from res a 3 m/s. Afer he elevaor acceleraes for 5 seconds, find (a) he speed of he elevaor (b) he heigh of he floor of he elevaor above he ground. A he end of 5 s, he girl les go of he ball from a heigh of 1 meer above he floor of he elevaor. If he elevaor coninues o accelerae upward a 3 m/s, describe he moion of he ball (c) relaive o he girl s hand, (d) relaive o he ground. (e) Deermine he ime afer he ball is released ha i will make conac wih he floor. (f) Wha is he heigh above he ground of he ball and floor when hey firs make conac? Soluion: (a) v = vo + a = + ( 3m / s )( 5s) = 15 m / s upward (b) y v 1 1 = o + a = + ( 3m / s )( 5s) = 37. 5m (c) When he girl releases he ball, boh she and he ball are moving wih a speed of 15 m/s upward. However, he girl coninues o accelerae upward a 3 m/s, bu he ball ceases o accelerae upward, and he ball s acceleraion is direced downward a g = 1 m/s, ha is, i is in free fall wih an iniial upward velociy of 15 m/s. Therefore he ball will appear o he girl o fall downward wih an acceleraion of 3 m/s (- 1 m/s ) = 13 m/s downward, and will quickly fall below her hand. (d) Someone waching he ball from he ground would simply see he ball rising upward wih an iniial velociy of 15 m/s, and would wach i rise o a maximum heigh, a which poin i would be insananeously a res (provided i doesn srike he floor of he elevaor before i reaches is maximum heigh). (e) When he ball is released, i is raveling upward wih a speed of 15 m/s, has a downward acceleraion of 13 m/s relaive o he floor, and is a a heigh y = 1 m above he floor. The ime i akes o fall o he floor is 3
1 y = a 1 1m = =.4 s ( 13m / s ) (f) In his ime of.4 s, he elevaor floor has moved up a disance of 1 1 Δ y = ae = ( 3m / s )(.4s) =. 4 m Thus, he ball and elevaor floor collide a a heigh above he ground of 37.5 m +.4 m = 37.74 m. Graphical Analysis of Velociy and Acceleraion Le s ake some ime o review how we inerpre he moion of an objec when we are given he informaion abou i in graphical form. On he AP Physics B exam, you will need o be able o inerpre hree ypes of graphs: posiion vs.ime, velociy vs. ime, and acceleraion vs. ime. Posiion vs. ime Consider he posiion vs. ime graph below: x (m) Δx x (m) P Δx Δ Δ (s) (s) Δ x The slope of he graph on he lef is, and is herefore velociy. The curved graph on Δ he righ indicaes ha he slope is changing. The slope of he curved graph is sill velociy, even hough he velociy is changing, indicaing he objec is acceleraing. The insananeous velociy a any poin on he graph (such as poin P) can be found by drawing a angen line a he poin and finding he slope of he angen line. 4
Velociy vs. ime Consider he velociy vs. ime graph below: v (m/s) Δv v (m/s) Δ v As shown in he figure on he lef, he slope of a velociy vs. ime graph is, and is Δ herefore acceleraion. As shown on he figure on he righ, he area under a velociy vs. m ime graph would have unis of ( s) = m, and is herefore displacemen. s Acceleraion vs. ime Δ (s) Area Since he AP Physics B exam generally deals wih consan acceleraion, any graph of acceleraion vs. ime on he exam would likely be a sraigh horizonal line: (s) a (m/s ) +5 m/s a (m/s ) (s) -5 m/s (s) This graph on he lef ells us ha he acceleraion of his objec is posiive. If he objec were acceleraing negaively, he horizonal line would be below he ime axis, as shown in he graph on he righ. 5
Example Consider he posiion vs. ime graph below represening he moion of a car. Assume ha all acceleraions of he car are consan. G H I J x(m) C D E F B K (s) A On he axes below, skech he velociy vs. ime and acceleraion vs. ime graphs for his car. v(m/s) (s) a(m/s ) (s) 6
Soluion: The car sars ou a a disance behind our reference poin of zero, indicaed on he graph as a negaive displacemen. The velociy (slope) of he car is iniially posiive and consan from poins A o C, wih he car crossing he reference poin a B. Beween poins C and D, he car goes from a high posiive velociy (slope) o a low velociy, evenually coming o res (v = ) a poin D. A poin E he car acceleraes posiively from res up o a posiive consan velociy from poins F o G. Then he velociy (slope) decreases from poins G o H, indicaing he car is slowing down. I is beween hese wo poins ha he car s velociy is posiive, bu is acceleraion is negaive, since he car s velociy and acceleraion are in opposie direcions. The car once again comes o res a poin H, and hen begins gaining a negaive velociy (moving backward) from res a poin I, increasing is speed negaively o a consan negaive velociy beween poins J and K. A K, he car has reurned o is original saring posiion. The velociy vs. ime graph for his car would look like his: v(m/s) B C F G A D E H I (s) J K The acceleraion vs. ime graph for his car would look like his: a(m/s ) E F A B C D G H I J K (s) 7
REVIEW QUESTIONS For each of he muliple choice quesions below, choose he bes answer. Unless oherwise noed, use g = 1 m/s and neglec air resisance. 1. Which of he following saemens is rue? (A) Displacemen is a scalar and disance is a vecor. (B) Displacemen is a vecor and disance is a scalar. (C) Boh displacemen and disance are vecors. (D) Neiher displacemen nor disance are vecors. (E) Displacemen and disance are always equal.. Which of he following is he bes saemen for a velociy? (A) 6 miles per hour (B) 3 meers per second (C) 3 km a 45 norh of eas (D) 4 km/hr (E) 5 km/hr souhwes 3. A jogger runs 4 km in.4 hr, hen 8 km in.8 hr. Wha is he average speed of he jogger? (A) 1 km/hr (B) 3 km/hr (C) 1 km/hr (D).1 km/hr (E) 1 km/hr 5. A bus saring from a speed of +4 m/s slows o 6 m/s in a ime of 3 s. The average acceleraion of he bus is (A) m/s (B) 4 m/s (C) 6 m/s (D) m/s (E) 6 m/s 6. A rain acceleraes from res wih an acceleraion of 4 m/s for a ime of s. Wha is he rain s speed a he end of s? (A).5 m/s (B) 4 m/s (C).5 m/s (D).8 m/s (E) 8 m/s 7. A fooball player sars from res 1 meers from he goal line and acceleraes away from he goal line a 5 m/s. How far from he goal line is he player afer 4 s? (A) 6 m (B) 3 m (C) 4 m (D) 5 m (E) 6 m 4. A moorcycle sars from res and acceleraes o a speed of m/s in a ime of 8 s. Wha is he moorcycle s average acceleraion? (A) 16 m/s (B) 8 m/s (C) 8 m/s (D).5 m/s (E).4 m/s 8
8. A ball is dropped from res. Wha is he acceleraion of he ball immediaely afer i is dropped? (A) zero (B) 5 m/s (C) 1 m/s (D) m/s (E) 3 m/s Quesions 9 11: A ball is hrown sraigh upward wih a speed of +1 m/s. 1. Which wo of he following pairs of graphs are equivalen? (A) x v 9. Wha is he ball s acceleraion jus afer i is hrown? (A) zero (B) 1 m/s upward (C) 1 m/s downward (D) 1 m/s upward (E) 1 m/s downward (B) x v 1. How much ime does i ake for he ball o rise o is maximum heigh? (A) 4 s (B) 1 s (C) 1 s (D) s (E) 1. s (C) (D) x x v v 11. Wha is he approximae maximum heigh he ball reaches? (A) 4 m (B) 17 m (C) 1 m (D) 7 m (E) 5 m (E) x v 9
Quesions 13 14: Consider he velociy vs ime graph below: 13. A which ime(s) is he objec a res? (A) zero (B) 1 s (C) 3 s o 4 s (D) 4 s only (E) 8 s 14. During which inerval is he speed of he objec decreasing? (A) o 1 s (B) 1 s o 3 s (C) 3 s o 4 s (D) 4 s o 8 s (E) he speed of he objec is never decreasing in his graph 3
Free Response Quesion Direcions: Show all work in working he following quesion. The quesion is worh 15 poins, and he suggesed ime for answering he quesion is abou 15 minues. The pars wihin a quesion may no have equal weigh. 1. (15 poins) A car on a long horizonal rack can move wih negligible fricion o he lef or o he righ. During he ime inervals when he car is acceleraing, he acceleraion is consan. The acceleraion during oher ime inervals is also consan, bu may have a differen value. Daa is aken on he moion of he car, and recorded in he able below. Displacemen Velociy ime x(m) v(m/s) (s) - 4-1 - - 3 1 6 1 7 9 1 31
(a) Plo hese daa poins on he v vs graph below, and draw he bes-fi sraigh lines beween each daa poin, ha is, connec each daa poin o he one before i. The acceleraion is consan or zero during each inerval lised in he daa able. (b) Lis all of he imes beween = and = 1 s a which he car is a res. (c) i. During which ime inerval is he magniude of he acceleraion of he car he greaes? ii. Wha is he value of his maximum acceleraion? (d) Find he displacemen of he car from x = a a ime of 1 s. (e) On he following graph, skech he acceleraion vs. ime graph for he moion of his car from = o = 1 s. 3
ANSWERS AND EXPLANATIONS TO CHAPTER REVIEW QUESTIONS Muliple Choice 1. B Displacemen is he sraigh-line lengh from an origin o a final posiion and includes direcion, whereas disance is simply lengh moved.. E Velociy is a vecor and herefore direcion should be included. 3. A Average speed is oal disance divided by oal ime. The oal disance covered by he jogger is 1 km and he oal ime is 1. hours, so he average speed is 1 km/hr. 4. D Δv m / s m a = = =.5 Δ 8s s 5. E v a = f v o 6m / s 4 m / s = 3s m = 6 s 6. E v f = vi + a = + / ( 4m/ s )( s) = 8 m s 33
7. D m x x v 1 1 f = o + o + a = (1 m) + + 5 5 s ( 4s) = m 8. C The acceleraion due o graviy is 1 m/s a all poins during he ball s fall. 9. C Afer he ball is hrown, he only acceleraion i has is he acceleraion due o graviy, 1 m/s. 1. E A he ball s maximum heigh, v f =. Thus, v = v g = f o m s = 1 / 1 m / s = 1. s 11. D 1 1 m y = g = 1 7 s ( 1. s) = 7. m m 1. B Boh of hese graphs represen moion ha begins a a high posiive velociy, and slows down o zero velociy. 13. B The line crosses he axis (v = ) a a ime of 1 second. 14. A The objec begins wih a high negaive (backward) velociy a =, hen is speed decreases o zero by a ime of 1 s. 34
Free Response Quesion Soluion (a) 4 poins (b) poins The car is a res when he velociy is zero, ha is, when he graph crosses he ime axis. Thus, v = a 5 s, 9 s, and 1 s, as well as all poins beween 9 and 1 s. (c) i. 1 poin The acceleraion can be found by finding he slope of he v vs graph in a paricular inerval. The slope (acceleraion) is maximum (seepes) in he ime inerval from o 1 s. ii. poins Acceleraion = slope of v vs graph = m / s 1s s ( 4m / s) = m / s (d) 3 poins The displacemen of he car from x = can be found by deermining he area under he graph. Noe ha he area is negaive from o 5 s, and posiive from 5 s o 9 s. Don forge he iniial displacemen of m a =. Area from o 5 s = 1 squares = - 1 m. Area from 5 o 1 s =.5 squares = +.5 m Toal displacemen from x = is m 1 m +.5 m = - 5.5 m. 35
(e) 3 poins 36