Mechanics Acceleraion The Kinemaics Equaions Lana Sheridan De Anza College Sep 27, 2018
Las ime kinemaic quaniies graphs of kinemaic quaniies
Overview acceleraion he kinemaics equaions (consan acceleraion) applying he kinemaics equaions
Quesion: Average Velociy vs Average Speed 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.
Quesion: Average Velociy vs Average Speed 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.
Insananeous Velociy and Posiion-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
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.
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().
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,
Reurning o Velociy vs Time Graphs PROBLEMS 49 The he area moorcycle under a velociy-ime during each graph of he has following a special segmens inerpreaion: of he i is moion: he displacemen (a) A, (b) B, and of he (c) objec C. over he ime inerval considered. 15 Velociy, v (m/s) 10 5 A B C O 5 10 15 20 25 Time, (s) FIGURE 2 31 Problem 32 x = v avg 33. A person on horseback moves according o he velociy-
Reurning o Velociy vs Time Graphs PROBLEMS 49 The he area moorcycle under a velociy-ime during each graph of he has following a special segmens inerpreaion: of he i is moion: he displacemen (a) A, (b) B, and of he (c) objec C. over he ime inerval considered. 15 Velociy, v (m/s) 10 5 A B C O 5 10 15 20 25 Time, (s) FIGURE 2 31 Problem 32 x = (25 m + 100 m + 75 m)i = 200 m i 33. A person on horseback moves according o he velociy-
Area under Velociy vs. Time Graphs CHAPTER 2 KINEMATICS 65 v- and x- graphs for he same objec: Area under v- graph = x. Slope of x- curve = v.
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 n der he curve in he velociy ime x = lim graph. v n Therefore, = v d in he limi n S `, or D 0 n displacemen is n i where x represens he change in posiion (displacemen) in he Dx 5 lim Dn S 0 a v xn,avg D n (2 ime inerval i o f. n f
Velociy vs. Time Graphs n One Dimension 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 D S 0 a v xn,avg D n (2
Quesion Wha does he area under an acceleraion-ime graph represen?
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 acceleraion Homework - CHANGED! Read Ch 2. Ch 2, Quesions: 1, 2, 4, 5; Problems: 19, 21, 90 (will be se on Monday: Ch 2, Problems: 23, 25, 31, 35, 41, 69, 73)