# Lecture 5: Ch 2

Lecture 5: Ch 2

Sep. 24th, 2008Lecture 5

Today’s outlines• Motion at Constant Acceleration• Freely Falling Objects• Variable Acceleration; Integral Calculus

P. 28Example (again, using board):x= 8− 6 t+tvadx= =− 6+2tdtdv= =dt2xt ( = 0) = 8 mvt ( = 0) =− 6 ( m/ s) < 028642xCCU Physics 2 - 16x = tdx=dtnntn −1Plott1 2 3 4 5 6

P. 28§ 2-5 Motion at Constant Accelerationx=c + ct+ ct ⇒ x = cCCU Physics 2 - 16n21 2 3 0 1dx n −= nt= dx122dt30 2v c c t vdt = + ⇒ = c2dv d xa= = = 2 c2 3⇒ c3=dt dt1x = x0+ v0t+at2v= v + at02va20= ==xv0− xt+ v2xvt0=+tt12at

2-5 Motion at Constant AccelerationWe can also combine these equations so as toeliminate t:We now have all the equations we need to solveconstant-acceleration problems.

2-7 Freely Falling ObjectsNear the surface of the Earth, all objectsexperience approximately the same accelerationdue to gravity.This is one of the mostcommon examples ofmotion with constantacceleration.

2-7 Freely Falling ObjectsIn the absence of airresistance, all objectsfall with the sameacceleration, althoughthis may be tricky totell by testing in anenvironment wherethere is air resistance.

2-7 Freely Falling ObjectsThe acceleration due togravity at the Earth’ssurface is approximately9.80 m/s 2 . At a givenlocation on the Earth andin the absence of airresistance, all objects fallwith the same constantacceleration.

2-8 Variable Acceleration; Integral Calculusthe area under the graph ofthe acceleration vs. time isthe change in velocitya(t)( ) at it t= 2=∫t=t1atdt () ≡ lim at ( ) Δt,Δ t=t=t∫2=Δ→ t 0or N →∞i N∑i=1[ ( , )]Area a t t( = )vt t2( )1 1i1 2t− tN2 1( ) ( )a()t dt = dv = v t −v t∫t= tv t=t2 1

Position as the integral of velocitythe area under the graph ofthe velocity vs. time is thedisplacementvt ()≡dxdtt=t∫2( = )xt t2( )1 1( ) ( )v()t dt = dx = x t − x t∫t= tx t=t2 1

Some Basic integration Formulasd n n 11. x nx−=dxd2. ( sin x)= cos xdxd3. ( cos x)= −sinxdxd4. ( ex)ex=dxd 15. ( ln x ) =dx x∫∫∫∫∫abbabaab1nbedx xaCCU Physics 2 - 7bn−1nx dx = xa(cos xdx ) = sin x(sin xdx )=ebxa1dx = ln xxbaba=−cosxba

2-9 Graphical Analysis andNumerical IntegrationSimilarly, the velocity may be written as the areaunder the a-t curve.However, if the velocity or acceleration is notintegrable, or is known only graphically,numerical integration may be used instead.

How aboutguess∫abln( xdx= ) ?try xln(x)− xcheck[ ln( ) − ] [ ln( )]d x x x d x x dx= −dx dx dx⎛dx ⎞ d ln( x)= ⎜ ln( x) x 1dx⎟ + −⎝ ⎠ dx= ln( x) + 1− 1=ln( x)work!!

Exampleddx⎛sinx⎞ ⎜ =x⎟⎝ ⎠?CCU Physics 2 - 7d ⎛sin x⎞ d(1/ x) 1 dsin(x)sin( x)dx⎜x⎟ = +⎝ ⎠ dx x dx1 cos( x)1= sin( x) ( − ) + = xcos x−sinx2 2x x x[ ]

Exampledx2dx3x =cos?2 2d x dx 1= +dx33cos x dx cos xx23−3 2d(1/ cos x)dy dyy≡ cos( x)⇒=dx dy dxdy d cos( x)= =−sin( x)dx dx⇒=2d x 2x=dx3 3x x2xcos3cos3+ xx 22cos5−2+xcos ( x)sin( x)2dx3(1/ cos )dx53 −2( − y )[ −sin( x)]2

P. 40Example (Group problem): The acceleration of an object is givenby a = 3 + 6t. The position of the object is 2 m at t = 1 sec , andthe velocity of the object is 3 m/s at t = 2 sec . Write expressions forthe position and velocity of the object as functions of time.∫v( t)dv=∫v (2) 2ta( t)dtt⎛3t′ 6t′vt () − 3= ∫ (3 + 6t′) dt′= ⎜ +2⎝ 1 2∫x( t)tdx = v( t′ ) dt′∫x (1) 12⎞⎟⎠t2CCU Physics 2 - 28vt t t2() = 3 + 3−153 2t2⎛ 3t′ 3t′⎞x() t − 2 = ∫ (3t′ + 3t′ − 15) dt′ = 15t′1⎜ + − ⎟⎝ 3 2 ⎠3 3 2 29x() t = t + t − 15t+2 2t1

Example: non-constant accelerationvelocity dependentConsider an object released at time t = 0 with aninitial velocity v (t = 0) = 0 and acceleratingaccording toat () = c − cvt ()0 11. Find the velocity as function of time.2. What is the velocity as t >> 1? (c 1 >0)Hint: the example in Tuesday’s lecture is useful1 − t⎡ ⎤x= v0⎢t+ e ββ ⎥⎣ ⎦dxv = = vdt(1 0−e −βt)dva== v d0βe −βtt

dvdva = = c − c v ⇒ = dt0 1dt c0 − c1vv( t)tdv′1∫ = dt′, dx= ln ( x)c( 0) 0− cv t1v′∫ ∫x=0v1→− ln ( c0 − c1v′) = tc1 v( t= 0) = 0dyy≡c −c v′⇒ = −cdv′1 ⎛ c0 − c1v()t ⎞− dv′dy / c⎜ln[] ⎟ = t1= −c1 ⎝ c0⎠ c − c v′y0 10 1 1ordv′ =−dyc1⇒or11 (v(tc vt− ) = ec0c= −c0) 11−ct1(−ct)1e

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