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Unformatted text preview: A c c e le r a t io n V e c t o r
! Average acceleration = change of velocity during a given time interval !v Physics for Scientists & Engineers 1
Fall Semester 2006
Lecture 6 a!t = !t
!v dv # !t dt ! Instantaneous acceleration vector
a (t ) = lim a!t = lim
!t " 0 !t " 0 ! Acceleration vector is the second derivative of the position vector
a (t ) = d d ! d " d2 v = # x$ = 2 x dt dt % dt & dt
2 August 31, 2006 Physics for Scientists&Engineers 1 1 August 31, 2006 Physics for Scientists&Engineers 1 E x a m p le : V e lo c it y & A c c e le r a t io n
! Graph of x(t ) = 17.2 m ! (10.1 m)(t / s)+(1.1 m)(t /s)2 and
v(t ) = !10.1 m/s+(2.2 m/s)(t / s) E x a m p le : 1 0 0 m S p r in t
! Carl Lewis’ World Record, 1991 World Championship vi ! f = #x x f " xi = #t t f " ti ai ! f = #v v f " vi = #t t f " t i Fit: v = 11.58 m/s ! Acceleration:
August 31, 2006 a(t ) = 2.2 m/s 2
Physics for Scientists&Engineers 1 3 August 31, 2006 Physics for Scientists&Engineers 1 4 In v e r s e R e la t io n s h ip s
! Start with the definition v=dx/dt and integrate both sides: Linear motion with Constant Acceleration
! What is that about?
• A lo t o f c a s e s w e h a v e a lre a d y lo o k e d a t ! v= dx dx ! # vdt = # dt = # dx = x " x0 dt dt
t 0 ! Use the integral formula and set a = constant: x(t ) = x0 + ! v(t ')dt '
! Can find the position, if we know the velocity as a function of time and the position at t=0 ! Similar integration to obtain the velocity from the acceleration t v(t ) = v0 + " a (t ')dt ' = v0 + a " dt ' = v0 + at ! v(t ) = v0 + at
0 0 t t T h e v e lo c it y is lin e a r in t im e ! Integrate velocity to find position vector: x(t ) = x0 + " v(t ')dt ' = x0 + " (v0 + at ')dt '
0 0 t t v(t ) = v0 + ! a (t ')dt '
0
Physics for Scientists&Engineers 1 2 1 = x0 + v0 " dt '+ a " t ' dt ' ! x(t ) = x0 + v0t + 2 at
0 0 t t T h e p o s it io n is q u a d r a t ic in t im e
5 August 31, 2006 Physics for Scientists&Engineers 1 6 August 31, 2006 A v e r a g e V e lo c it y
! If velocity depends linearly on time, what is the average velocity over the time interval from t0 to t? ! Answer:
v0!t 1t 1t = " v(t ')dt ' = " (v0 + at ')dt ' 0 t t0 vt at = 0 " dt '+ " t ' dt ' = v0 + 1 at 2 t0 t0 = 1 v0 + 1 (v0 + at ) 2 2 So, between 0 and t: = 1 (v0 + v(t )) 2
V(t) P o s it io n a n d A v e r a g e V e lo c it y
! Start from v0!t ! We get: = 1 (v0 + v(t )) and v(t ) = v0 + at 2 ! Now multiply both sides with t and add x0: v0!t = 1 (v0 + v0 + at ) = v0 + 1 at 2 2 x0 + v0!t t = x0 + v0t + 1 at 2 2
! Right hand side is exactly the expression for x(t) we obtained from integration: v0!t = 1 [v0 + v(t )] 2
Physics for Scientists&Engineers 1 7 x(t ) = x0 + v0t + 1 at 2 ! x(t ) = x0 + v0!t t 2
August 31, 2006 Physics for Scientists&Engineers 1 8 August 31, 2006 Expression for v2
! Solve v(t ) = v0 + at for the time and get:
t= v(t ) ! v0 a Expression for v2 (cont.)
2 2 v(t )v0 ! v0 1 v(t ) 2 + v0 ! 2v(t )v0 x(t ) = x0 + +2 a a 2 2 a[ x(t ) ! x0 ] = v(t )v0 ! v0 + 1 [v(t ) 2 + v0 ! 2v(t )v0 ] 2 2 a[ x(t ) ! x0 ] = 1 v(t ) 2 ! 1 v0 2 2 ! Substitute this result into the position expression: x(t ) = x0 + v0t + 1 at 2 2 " v(t ) ! v0 # 1 " v(t ) ! v0 # = x0 + v0 $ % + 2 a$ % a a & ' & ' 2 2 2 v(t )v0 ! v0 1 v(t ) + v0 ! 2v(t )v0 = x0 + +2 a a ! Subtract x0 from both sides and multiply with a:
August 31, 2006 Physics for Scientists&Engineers 1 9 2 ! Result:
2 v(t ) 2 ! v0 = 2a[ x(t ) ! x0 ] August 31, 2006 Physics for Scientists&Engineers 1 10 S u m m a r y : F iv e K in e m a t ic a l E q u a t io n s
! Onedimensional motion with constant acceleration: E x a m p le : A ir p la n e T a k e o ff ( 1 )
! Experiment: use strain gauge and measure acceleration during airplane takeoff ! Result: Constant acceleration, in good approximation v(t ) = v0 + at x(t ) = x0 + v0t + 1 at 2 2 v0!t = 1 [v0 + v(t )] 2 x(t ) = x0 + v0!t t
2 v(t ) 2 " v0 = 2a[ x(t ) " x0 ] a = 4.3 m/s2 ! Can solve practically all onedimensional problems August 31, 2006 Physics for Scientists&Engineers 1 11 August 31, 2006 Physics for Scientists&Engineers 1 12 E x a m p le : A ir p la n e T a k e o ff ( 2 )
Question 1: ! Assuming a constant acceleration of a = 4.3 m/s2, starting from rest, what is the takeoff speed of the airplane reached after 18 seconds? Answer 1: ! The airplane accelerates from a standing start: initial velocity is 0. E x a m p le : A ir p la n e T a k e o ff ( 3 )
Question 2: ! How far down the runway has this airplane moved by the time it takes off? v(t ) = v0 + at Answer 2:
x0 = 0 v0 = 0 a = 4.3m.s t1 = 18s
!2 a = 4.3ms v0 = 0 t1 = 18s !2 v(t ) = v0 + at
v (t ? !"1 )" #
2 x(t ) = x0 + v0t + 1 at 2 2
x (t ) ? !"1 " #
2 x(t ) = x0 + v0t + 1 at 2 2 v0!t = 1 [v0 + v(t )] 2 x(t ) = x0 + v0!t t
2 v(t ) " v0 = 2a[ x(t ) " x0 ] v0!t = 1 [v0 + v(t )] 2 x(t ) = x0 + v0!t t
2 v(t ) " v0 = 2a[ x(t ) " x0 ] x(t1 ) = x0 + v0t1 + 1 at12 = 1 ( 4.3 m/s 2 )(18 s)2 2 2 = 697 m (2,290 ft)
August 31, 2006 13 Physics for Scientists&Engineers 1 v(t1 ) = v0 + at1 = 0 + ( 4.3 m/s 2 )(18 s) = 77.4 m/s (175 mph)
August 31, 2006 Physics for Scientists&Engineers 1 M ain r u n w ay at L an sin g a i r p o r t i s 7 5 0 0 f t lo n g
14 R a c e o f T w o B a lls
! Two balls roll down incline to gain speed; after that: ! Question 1 (multiple choice):
• Ball 1 rolls on horizontal track • Ball 2 rolls on “roller coaster” track • Ball 1 will arrive first • Ball 2 will arrive first • Both balls will arrive at the same time • Ball 1 will arrive with higher speed • Ball 2 will arrive with higher speed • Both balls will arrive with the same speed R a c e o f T w o B a lls , q u a n t it a t iv e
! Let’s look at accelerations and velocities a(x) v(x) x ! Question 2 (multiple choice): a(x) v(x) x August 31, 2006 Physics for Scientists&Engineers 1 15 August 31, 2006 Physics for Scientists&Engineers 1 16 ...
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 Spring '08
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 Acceleration

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