PHY183-Lecture6 - A c c e le r a t io n V e c t o r !...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ! One-dimensional 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 take-off ! 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 one-dimensional 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 ...
View Full Document

Ask a homework question - tutors are online