4.6 One Dimensional Kinematics and Integration
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1 4.6 One Dimensional Kinemaics and Inegraion When he acceleraion a( of an objec is a non-consan funcion of ime, we would like o deermine he ime dependence of he posiion funcion x( and he x -componen of he velociy v(. Because he acceleraion is non-consan we no longer can use Eqs. (4.4. and ( Insead we shall use inegraion echniques o deermine hese funcions Change of Velociy as he Indefinie Inegral of Acceleraion Consider a ime inerval < <. Recall ha by definiion he derivaive of he velociy v( is equal o he acceleraion a(, dv( = a(. (4.5. d
2 Inegraion is defined as he inverse operaion of differeniaion or he ani-derivaive. For our example, he funcion v( is called he indefinie inegral of a( wih respec o, and is unique up o an addiive consan C. We denoe his by wriing v( + C = a( d. (4.5. The symbol...d means he inegral, wih respec o, of, and is hough of as he d inverse of he symbol.... Equivalenly we can wrie he differenial dv( = a(d, d called he inegrand, and hen Eq. (4.5. can be wrien as v( + C = dv(, (4.5.3 which we inerpre by saying ha he inegral of he differenial of funcion is equal o he funcion plus a consan. Example 4.6 Non-consan acceleraion Suppose an objec a ime = 0 has iniial non-zero velociy v0 and acceleraion a( = b, where b is a consan. Then dv( = b d = d(b 3 / 3. The velociy is hen v( + C = d(b 3 / 3 = b 3 / 3. A = 0, we have ha v0 + C = 0. Therefore C = v0 and he velociy as a funcion of ime is hen v( = v0 + (b 3 / Area as he Indefinie Inegral of Acceleraion Consider he graph of a posiive-valued acceleraion funcion a( vs. for he inerval, shown in Figure 4.4a. Denoe he area under he graph of a( over he inerval by A. Figure 4.4a: Area under he graph of acceleraion over an inerval a( a( a( Area = A c
3 a( a( c a( Figure 4.4b: Inermediae value Theorem. The shaded regions above and below he curve have equal areas. a( c The Inermediae Value Theorem saes ha here is a leas one ime such ha he c area A is equal o A = a( c (. (4.5.4 In Figure 4.4b, he shaded regions above and below he curve have equal areas, and hence he area A under he curve is equal o he area of he recangle given by a( c (. a( a( a( A + A + Figure 4.5 Area funcion is addiive We shall now show ha he derivaive of he area funcion is equal o he acceleraion and hererfore we can wrie he area funcion as an indefinie inegral. From Figure 4.5, he area funcion saisfies he condiion ha +Δ +Δ A + A = A. (4.5.5 Le he small incremen of area be denoed by ΔA = A +Δ A = A +Δ. By he Inermediae Value Theorem 3
4 ΔA = a( c Δ, (4.5.6 where c + Δ. In he limi as Δ 0, da ΔA = lim = lim a( = a(, (4.5.7 d Δ 0 Δ c c wih he iniial condiion ha when =, he area A = 0 is zero. Because v( is also an inegral of a(, we have ha A = a( d = v( + C. (4.5.8 When =, he area A = 0 is zero, herefore v( + C = 0, and so C = v(. Therefore Eq. (4.5.8 becomes A = v( v( = a( d. (4.5.9 When we se =, Eq. (4.5.9 becomes A = v( v( = a( d. (4.5.0 The area under he graph of he posiive-valued acceleraion funcion for he inerval can be found by inegraing a( Change of Velociy as he Definie Inegral of Acceleraion Le a( be he acceleraion funcion over he inerval i f. Recall ha he velociy v( is an inegral of a( because dv( / d = a(. Divide he ime inerval [ i, f ] ino n equal ime subinervals Δ = ( f i / n. For each subinerval [ j, j+ ], where he index j =,,...,n, = i and n+ = f, le be a ime such ha j j+. Le c j c j j=n S n = a( c j Δ. (4.5. j= S n is he sum of he blue recangle shown in Figure 4.6a for he case n = 4. The Fundamenal Theorem of Calculus saes ha in he limi as n, he sum is equal o he change in he velociy during he inerval [ i, f ] 4
5 lim S n = lim n n j=n a( j= cj Δ = v( f v(i. a( a(c 4 a(c a( a(c 4 a(c a(c a(c a(c a(c ( c c 3 c 3 4 c 4 5 c Figure 4.6a Graph of a( vs. c 3 c 3 4 c 4 5 Figure 4.6b Graph of a( vs. The limi of he sum in Eq. (4.5. is a number, which we denoe by he symbol f j=n a( d lim a( n i j= cj Δ = v( f v(i, (4.5.3 and is called he definie inegral of a( from i o f. The imes i and f are called he limis of inegraion, i he lower limi and f he upper limi. The definie inegral is a linear map ha akes a funcion a( defined over he inerval [i, f ] and gives a number. The map is linear because f f f (a ( + a ( d = a ( d + a ( d i i, (4.5.4 i Suppose he imes c, j =,...,n, are seleced such ha each c saisfies he Inermediae j j Value Theorem, Δv j v( j+ v( j = dv(c j d Δ = a(c Δ, j (
6 where a( is he insananeous acceleraion a, (Figure 4.6b. Then he sum of he c j c j changes in he velociy for he inerval [ i, f ] is j=n Δv j = (v( v( + (v( 3 v( + + (v( n+ v( n = v( n+ v( j= (4.5.6 = v( f v( i. where v( f = v( n+ and v( i = v(. Subsiuing Eq. (4.5.5 ino Equaion (4.5.6 yields he exac resul ha he change in he x -componen of he velociy is give by his finie sum. j=n j=n v( f v( i = Δv j = a( c j Δ. (4.5.7 j= j= We do no specifically know he inermediae values a( c j and so Eq. (4.5.7 is no useful as a calculaing ool. The saemen of he Fundamenal Theorem of Calculus is ha he limi as n of he sum in Eq. (4.5. is independen of he choice of he se of c j. Therefore he exac resul in Eq. (4.5.7 is he limi of he sum. Thus we can evaluae he definie inegral if we know any indefinie inegral of he inegrand a(d = dv(. Addiionally, provided he acceleraion funcion has only non-negaive values, he limi is also equal o he area under he graph of a( vs. for he ime inerval, [ i, f ]: f f = i i A a( d. (4.5.8 In Figure 4.4, he red areas are an overesimae and he blue areas are an underesimae. As N, he sum of he red areas and he sum of he blue areas boh approach zero. If here are inervals in which a( has negaive values, hen he summaion is a sum of signed areas, posiive area above he -axis and negaive area below he -axis. We can deermine boh he change in velociy for he ime inerval [ i, f ] and he area under he graph of a( vs. for [ i, f ] by inegraion echniques insead of limiing argumens. We can urn he linear map ino a funcion of ime, insead of jus giving a number, by seing f =. In ha case, Eq. (4.5.3 becomes 6
7 = v( v( i = a( d. (4.5.9 = i Because he upper limi of he inegral, f =, is now reaed as a variable, we shall use he symbol as he inegraion variable insead of Displacemen as he Definie Inegral of Velociy We can repea he same argumen for he definie inegral of he x -componen of he velociy v( vs. ime. Because x( is an inegral of v( he definie inegral of v( for he ime inerval [ i, f ] is he displacemen = f x( f x( i = v( d. (4.5.0 = i If we se f =, hen he definie inegral gives us he posiion as a funcion of ime = x( = x( i + v( d. (4.5. = i Summarizing he resuls of hese las wo secions, for a given acceleraion a(, we can use inegraion echniques, o deermine he change in velociy and change in posiion for an inerval [ i, ], and given iniial condiions (x i,v i, we can deermine he posiion x( and he x -componen of he velociy v( as funcions of ime. Example 4.5 Non-consan Acceleraion Le s consider a case in which he acceleraion, a(, is no consan in ime, a( = b 0 + b + b. (4.5. The graph of he x -componen of he acceleraion vs. ime is shown in Figure 4.6 7
8 a( a( = b 0 + b + b b 0 slope = b + Figure 4.6 Non-consan acceleraion vs. ime graph. Denoe he iniial velociy a = 0 by v 0. Then, he change in he x -componen of he velociy as a funcion of ime can be found by inegraion: = = b 3 b v( v0 = a( d = d + +. (4.5.3 (b + b + b = b 0 0 =0 =0 3 The x -componen of he velociy as a funcion in ime is hen 3 b b v( = v0 + b ( Denoe he iniial posiion a = 0 by x 0. The displacemen as a funcion of ime is = x( x 0 = v( d. (4.5.5 =0 Use Equaion (4.5.7 for he x-componen of he velociy in Equaion (4.5.4 and hen inegrae o deermine he displacemen as a funcion of ime: x( x 0 = = =0 v( d = b b b 0 b b = v0 + b d = v =0 (4.5.6 Finally he posiion as a funcion of ime is hen b b b x( = x0 + v x, (
9 Example 4.6 Bicycle and Car A car is driving hrough a green ligh a = 0 locaed a x = 0 wih an iniial speed v c,0 = m s -. A ime = s, he car sars braking unil i comes o res a ime. The acceleraion of he car as a funcion of ime is given by he piecewise funcion where b = (6 m s -3. a c ( = 0; 0 < < = s, b( ; s < < (a Find he x -componen of he velociy and he posiion of he car as a funcion of ime. (b A bicycle rider is riding a a consan speed of v b,0 and a = 0 is 7 m behind he car. The bicyclis reaches he car when he car jus comes o res. Find he speed of he bicycle. Soluion: a In order o apply Eq. (4.5.9, we shall rea each sage separaely. For he ime inerval 0 < <, he acceleraion is zero so he x -componen of he velociy is consan. For he second ime inerval < <, he definie inegral becomes = c c = v ( v ( = b( d Because v c ( = v c0, he x -componen of he velociy is hen v c0 ; 0 < v ( = =. c v + b( d ; < c0 = Inegrae and subsiue he wo endpoins of he definie inegral, yields v ; 0 < c0 v ( =. c v + b( c0 ; < In order o use Eq. (4.5.5, we need o separae he definie inegral ino wo inegrals corresponding o he wo sages of moion, using he correc expression for he velociy for each inegral. The posiion funcion is hen 9
10 Upon inegraion we have = x c0 + v d ; 0 < c0 =0 x ( =. c = x ( + v + b( d; < c c0 = x c (0 + v c0 ; 0 < = x c ( =. x ( + v ( + b( 3 ; < c c0 6 = We chose our coordinae sysem such ha he iniial posiion of he car was a he origin, x = 0, herefore x = v. So afer subsiuing in he endpoins of he inegraion c0 c ( c0 inerval we have ha v ; 0 < c0 x c ( =. v + v + 3 c0 c0 ( b( ; < 6 (b We are looking for he insan ha he car has come o res. So we use our expression for he x -componen of he velociy he inerval <, where we se = and v = 0: c ( Solving for yields 0 = v ( = v + b(. c c0 = + v c0 b, where we have aken he posiive square roo. Subsiue he given values hen yields = s + ( m s ( 6 m s 3 = 3 s. The posiion of he car a is hen given by 0
11 x ( = v + v ( + b( 3 c c0 c0 6 x ( / b 3/ = v + v v / b + b( v c c0 c0 c0 c0 6 3/ (v c0 xc ( = v c0 + 3( b / where we used he condiion ha = v c0 / b. Subsiue he given values hen yields 4 (v c0 3/ 4 (( m s - 3/ x c ( = v c0 + = ( m s - (s + = 8 m. 3( b / 3((6 m s 3 / b Because he bicycle is raveling a a consan speed wih an iniial posiion x b0 = 7 m, he posiion of he bicycle is given by x b ( = 7 m + v b. The bicycle and car inersec a ime = 3 s, where x b ( = x c (. Therefore 7 m + v b (3 s = 8 m. So he speed of he bicycle is v b = 5 m s.
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4.5 Constant Acceleration
4.5 Consan Acceleraion v() v() = v 0 + a a() a a() = a v 0 Area = a (a) (b) Figure 4.8 Consan acceleraion: (a) velociy, (b) acceleraion When he x -componen of he velociy is a linear funcion (Figure 4.8(a)),
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