Kinematics Motion in 1 Dimension and Graphs

Kinemaics Moion in 1 Dimension and Graphs Lana Sheridan De Anza College Sep 27, 2017

Las ime moion in 1-dimension some kinemaic quaniies graphs

Overview velociy and speed acceleraion more graphs

Kinemaics Par I: Moion in 1 Dimension Velociy How posiion changes wih ime. (insananeous) velociy average velociy v = dx d v avg = x speed and direcion insananeous speed v or v speedomeer speed average speed d disance divided by ime Can velociy be negaive?

Quesion Quick Quiz 2.1 1 Under which of he following condiions is he magniude of he average velociy of a paricle moving in one dimension smaller han he average speed over some ime inerval? A A paricle moves in he +x direcion wihou reversing. B A paricle moves in he x direcion wihou reversing. C A paricle moves in he +x direcion and hen reverses he direcion of is moion. D There are no condiions for which his is rue. 1 Serway & Jewe, page 24.

Comparing Posiion and Velociy vs. Time Graphs aper 2 Moion in One Dimension x (m) 60 40 20 0 20 40 60 0 10 20 30 40 (s) 50 60 40 The blue line beween posiions and approaches he green angen line as poin is moved closer o poin. a b 2 y graph of represens in he n he veriin he quanhorizonal Figure 2.3 (a) Graph represening he moion of he car in Figure 2.1. (b) An enlargemen of he upper-lef-hand corner of he graph. v = lim x( + ) x() = lim + 0 0 represens he velociy of he car a poin. Wha we have done is deermine he insananeous velociy a ha momen. In oher words, he insananeous velociy v x equals he limiing value of he raio Dx/D as D approaches zero: 1 Dx x = dx d

n One Dimension Velociy vs. Time Graphs v x The area of he shaded recangle is equal o he displacemen in he ime inerval n. v xn,avg i n f x = lim v der he curve in he velociy ime graph. n = v d 0 Therefore, in he limi n S `, or D n i n displacemen is where x represens he change in posiion (displacemen) in he ime inerval i o f. Dx 5 lim Dn S 0 a v xn,avg D n (2 is called a dummy variable. n f

n Velociy One Dimensionvs. Time Graphs v x The area of he shaded recangle is equal o he displacemen in he ime inerval n. v xn,avg i f Or we can wrie n der he curve in he velociy ime x() graph. = vtherefore, d in he limi n S `, or D n displacemen is i if he objec sars a posiion x = 0 when = i. Dx 5 lim Dn S 0 a v xn,avg D n (2 is called a dummy variable. n

Some Examples Traveling wih consan velociy: a car doing exacly he speed limi on a sraigh road Voyager I (nearly)

Some Examples Traveling wih consan velociy: a car doing exacly he speed limi on a sraigh road Voyager I (nearly) Traveling wih consan speed: a car doing exacly he speed limi on a road wih curves a plane raveling in a perfecly circular orbi disance, d = 2πr and v = 2πr T, for orbial period, T

Concepual Quesion y rankings, remember ha zero is greaer han a negaive value. If wo values are equal, show ha hey are equal in your ranking.) c Concepual Quesions 1. denoes answer available in Suden So 1. If he average velociy of an objec is zero in some ime inerval, wha can you say abou he displacemen of he objec for ha inerval? 2. Try he following experimen away from raffic where you can do i safely. Wih he car you are driving moving slowly on a sraigh, level road, shif he ransmission ino neural and le he car coas. A he momen he car comes o a complee sop, sep hard on he brake and noice wha you feel. Now repea he same experimen on a fairly genle, uphill slope. Explain he difference in wha a person riding in he car feels in he wo cases. (Brian Popp suggesed he idea for his 1 Serway & Jewe, page 50. 6. Y g w a w e 7. ( b i z 8. ( b

v 2 v v 2 v leraion Velociy vs. Time Graphs ding o he expresseconds. o 5 2.0 s. The acceleraion a is equal o he slope of he green angen line a 2 s, which is 20 m/s 2. v x (m/s) 40 30 20 10 0 (s) 2.9 (Example 2.6) ociy ime graph for a moving along he x axis ng o he expression 2 5 2. 10 20 30 0 1 2 3 4 v x 5 40 2 5 2 5 40 2 5(0) 2 5 140 m/s v x 5 40 2 5 2 5 40 2 5(2.0) 2 5 120 m/s

leraion Velociy vs. Time Graphs ding o he expresseconds. o 5 2.0 s. The acceleraion a is equal o he slope of he green angen line a 2 s, which is 20 m/s 2. v x (m/s) 40 30 20 10 0 (s) 2.9 (Example 2.6) ociy ime graph for a moving along he x axis ng o he expression 2 5 2. 10 20 30 0 1 2 3 4 v x 5 40 2 5 2 5 40 2 5(0) 2 5 140 m/s The slope a any poin of he velociy-ime curve is he v x 5 40 2 5 2 acceleraion 5 a40 ha 2 5(2.0) ime. 2 5 120 m/s v 2 v v 2 v

Acceleraion acceleraion average acceleraion a = dv d = d2 x d 2 a avg = v Acceleraion is also a vecor quaniy.

Acceleraion acceleraion average acceleraion a = dv d = d2 x d 2 a avg = v Acceleraion is also a vecor quaniy. If he acceleraion vecor is poined in he same direcion as he velociy vecor (ie. boh are posiive or boh negaive), he paricle s speed is increasing. If he acceleraion vecor is poined in he opposie direcion as he velociy vecor (ie. one is posiive he oher is negaive), he paricle s speed is decreasing. (I is deceleraing.)

Acceleraion and Velociy-Time Graphs Acceleraion is he slope of a velociy-ime curve. Unis: meers per second per second, m/s 2

Acceleraion and Velociy-Time Graphs Acceleraion is he slope of a velociy-ime curve. Unis: meers per second per second, m/s 2 In general, acceleraion can be a funcion of ime a().

Example Suppose a paricle has a velociy described by: v = (3 + 4) i (m/s) Wha is he acceleraion of his paricle? Wha is he displacemen of his paricle over he inerval = 0 o = 3 s?

Example Suppose a paricle has a velociy described by: v = (3 + 4) i (m/s) Wha is he acceleraion of his paricle? Wha is he displacemen of his paricle over he inerval = 0 o = 3 s? a = dv d = 4 i m s 2

Example Suppose a paricle has a velociy described by: v = (3 + 4) i (m/s) Wha is he acceleraion of his paricle? Wha is he displacemen of his paricle over he inerval = 0 o = 3 s? a = dv d = 4 i m s 2 x = 27 i m

Acceleraion vs. Time Graphs angen nd 5 mly, so e 2.8b. is con and alue of e slope a ha graph se unie slope o zero. mean- a b x v x a x he anf accelre 2.8c. 0 and c

Quesion Wha does he area under an acceleraion-ime graph represen?

Consan Acceleraion Graphs x i x Slope v xf x Slope v xi Slope v xf x i aslope v xi x i vslope x v a xi Slope a x v a x v xi Slope a x v x v xi Slope a x b v xi ab x a x b a x va xi x v xi v xf Slope 0 Slope 0 Slope 0 a x v xi a x a x a x v xf v xf 2.6 Analysis Mod If he acceleraion of a pari Under o analyze. 2.6 Analysis Consan A very common Mod Ac ha in which he accelera aunder x,avg over any Consan ime inerval A If he acceleraion of a paricl o analyze. A very common a a any insan wihin he in ha If he in acceleraion which he acceleraio ou he moion. This of a siua paric ao x,avg analyze. over any A ime very common inerval isa model: ha any in insan he paricle which wihin he acceleraio he under iner ou generae a x,avg he over moion. several any ime This equaions inerval siuaioi model: If we a any insan he replace paricle a wihin x,avg under by a he ine co x i generae, we find ou he moion. several ha equaions This siuaio h model: If we he replace paricle a x,avg under by a x inc, generae we find several ha equaions h or If we replace a x,avg by a x in, we find ha v or This powerful expression v e xf or if we know he objec s This velociy ime powerful expression graph for hi en v The if we graph know is he a sraigh objec s line xf velociy ime This slope powerful is consisen graph expression wih for his a x en 5c The ive, if we graph which know is indicaes a he sraigh objec s a line, pos in velociy ime slope is of consisen he line graph in wih for Figure a his x 5 ive, The san, which graph he indicaes graph is a sraigh of a acceler posii line,

bes describes he moion. Maching Velociy o Acceleraion Graphs hs v x v x v x s e d), a b c a x a x a x d e f

Summary velociy and acceleraion graphs Colleced Homework! Posed oday. Due in class Friday, Oc 6. Quiz Sar of class Friday, Sep 29. (Uncolleced) Homework Serway & Jewe, Se yeserday: Ch 1, onward from page 14. Problems: 9, 45, 57, 67, 71 Se yeserday: Ch 2, onward from page 49. Obj. Q: 1; CQ: Concep. Q: 1; Probs: 1, 3, 7, 11 New: Ch 2, onward from page 49. Concepual Q: 4, 5