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Unformatted text preview: Position, velocity and acceleration vectors ! Related through derivatives and integrals Physics for Scientists & Engineers 1
Fall Semester 2006 Lecture 7 x(t ) ! dx(t ) v(t ) = dt ! dv(t ) a (t ) = dt
1 August 31, 2006 x(t ) = x0 + " v(t ')dt '
0 t ! v(t ) = v0 + " a (t ')dt '
0 t ! a (t )
Physics for Scientists&Engineers 1 2 August 31, 2006 Physics for Scientists&Engineers 1 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: F r e e F a ll
! Particular motion with constant acceleration in 1 d y • a = g, with g = 9.81 m/s2 x
• Conventional notation: call vertical axis y axis v(t ) = v0 + at x(t ) = x0 + v0t + 1 at 2 v0!t = 1 [v0 + v(t )] 2 x(t ) = x0 + v0!t t
2 v(t ) 2 " v0 = 2a[ x(t ) " x0 ] 2 ! Kinematic equations for this case
U s e th o s e fo r c o n s ta n t a c c e le r a tio n : X > y a > g v(t ) = v0 " gt y (t ) = y0 + v0t " 1 gt 2 2 v0!t = [v(t ) + v0 ]
1 2 ! ! ! Fg = mg = !mgey
Newton’s Second law ! ! Fg = ma ! Can solve practically all onedimensional problems y (t ) = y0 + v0!t t
2 v(t ) 2 " v0 = "2 g[ y (t ) " y0 ] !mg = ma !g = a
4 August 31, 2006 Physics for Scientists&Engineers 1 3 August 31, 2006 Physics for Scientists&Engineers 1 R e m a r k s o n F r e e F a ll
! A ll o b j e c t s f a ll a t t h e s a m e r a t e , because a = g = constant
• Need to get rid of air resistance effects to see this • Experiment => calculation of g R e a c t io n T im e
! We can measure the human reaction time with a simple freefall experiment
• One person lets ruler go, the other tries to catch it as soon as possible • Measure displacement h = y0 ! y (t ) > 0 2 • From y (t ) = y0 + v0t ! 1 gt and 2 v0 = 0 we obtain • Reaction time: y (t ) = y0 ! 1 gt 2
2 v(t ) = v0 " gt y (t ) = y0 + v0t " 1 gt 2 2
g? !"# v0!t = 1 [v(t ) + v0 ] 2 y (t ) = y0 + v0!t t
2 v(t ) 2 " v0 = "2 g[ y (t ) " y0 ] ! I n s p a c e , t h e r e i s a lm o s t n o g r av it y
• Why all objects float around • Not net force: constant velocity, independent of the mass " h = gt
1 2 2 "t = 2h g August 31, 2006 Physics for Scientists&Engineers 1 5 August 31, 2006 Physics for Scientists&Engineers 1 6 E x a m p le : T o p F u e l R a c in g ( 1 ) E x a m p le : T o p F u e l R a c in g ( 2 )
! Accelerating from rest, a top fuel racer can reach 333.2 mph (=148.9 m/s, record established in 2003) at the end of a quarter mile (=402.3 m) run. For this example, we will assume constant acceleration. Question 1: ! What is the value of this acceleration? Answer 1: 2 ! It is convenient in this case to use v(t ) 2 = v0 + 2a[ x(t ) ! x0 ] and solve for the acceleration
a=
2 v(t ) 2 ! v0 (148.9 m/s) 2 = = 27.6 m/s 2 2[ x(t ) ! x0 ] 2(402.3 m)
Physics for Scientists&Engineers 1 8 August 31, 2006 Physics for Scientists&Engineers 1 7 August 31, 2006 E x a m p le : T o p F u e l R a c in g ( 4 )
Question 3: ! How long does it take to complete quarter mile race from a standing start? Answer 3: ! Because the final velocity is 148.9 m/s, the average velocity is v0!t = 1 (148.9 m/s + 0) = 74.45 m/s . Now we can use 2 B a ll t h r o w n v e r t ic a lly ( 1 )
A ball is thrown vertically upwards with an initial velocity of 27.0 m/s. Question 1: ! Neglecting air resistance, how long is the ball in the air? Answer 1: ! Even if the ball is first going up, this is a case of free fall!
a = ! g ; y0 = 0; v0 = 27.0m.s !1 x(t ) " x0 402.3 m = = 5.40 s v0!t 74.45 m/s ! Real answer, measured on the track: t ! 4.5 s. Assumption x(t ) = x0 + v0!t t # t =
of constant acceleration not quite satisfied.
August 31, 2006 Physics for Scientists&Engineers 1 9 v(t ) = v0 " gt y (t ) = y0 + v0t " 1 gt 2 2 v0!t = 1 [v0 + v(t )] 2 y (t ) = y0 + v0!t t
2 v(t ) 2 " v0 = "2 g[ y (t ) " y0 ]
10 t1 / y (t1 ) = 0 ?
" t1 =
August 31, 2006 2v0 2 * 27.0m.s !1 " t1 = = 5.51s g 9.81m.s !2
Physics for Scientists&Engineers 1 B a ll t h r o w n v e r t ic a lly ( 2 )
Question 2: ! What is the greatest height reached by the ball? Answer 2: ! What characterizes the point of maximum height? Question 3: ! In fact, the ball hits a bird on its way up when it has half of its initial velocity. At what altitude does that happens? Answer 3: ! Since we want to relate an information about the velocity to the altitude, let us use the last kinematical equation: B a ll t h r o w n v e r t ic a lly ( 3 ) v(t ) = 0 ! v " t2 = 0 g " y (t2 ) = " y (t2 ) = v 2g
2 0 v(t3 ) = v0 v2 2 ! 0 " v0 = "2 gy (t3 ) 2 4 ! y (t3 ) = 2 3v0 3(27.0m / s ) 2 = = 27.87 m 8 g 8 * 9.81m / s 2 (27.0m.s ) !1 2 2 * 9.81m.s !2 = 37.2m
11 ! We could have used v 27.0m.s !1 " t3 = 0 " t3 = = 1.38s " y (t3 ) 2g 2 * 9.81m.s !2
August 31, 2006 Physics for Scientists&Engineers 1 12 August 31, 2006 Physics for Scientists&Engineers 1 A q u a t h lo n ( 1 )
! The Aquathlon consists of two legs, a swim (distance b) followed by a run (distance a). A q u a t h lo n ( 2 )
Corresponding time t red = a 2 + b 2 / v1 = ( 3.354 km) / ( 3.5 km/h) = 0.958 h a = 3km; b = 1.5km ! The athletes swim with a speed v1 = 3.5 km/h and run with a speed 2 ) S w im st r aig h t t o sh o r e an d t h e n r u n tblue = b / v1 + a / v2 = (1.5 km ) / ( 3.5 km/h ) + ( 3 km ) / (14 km/h ) = 0.643 h
3 ) A r b it r ar y pat h b e t w e e n t h o se t w o e x t r e m e s: ch ar act e r iz e d b y !
2 " Swim a distance of a1 + b 2 and then run a distance of v2 = 14 km/h Which angle ! will result in the shortest finish time? ! 1) Red dashed line: shortest distance between start and finish a2 " T h e t o t al t im e is a 2 + b 2 = 1.5 2 + 32 km = 3.354 km
August 31, 2006 Physics for Scientists&Engineers 1 13 t (! ) = a12 + b 2 a2 (1 + tan 2 ! )b 2 a " b tan ! b a " b tan ! += + = + v1 v2 v1 v2 v1 cos ! v2
Physics for Scientists&Engineers 1 14 August 31, 2006 A q u a t h lo n ( 2 ) A q u a t h lo n ( 3 )
! T o f in d t h e m in im u m t im e r espect to ! t (! ) w e h av e t o t ak e t h e d e r iv at iv e w it h dt (! ) b tan ! b = " d! v1 cos! v2 cos 2 ! a n d f i n d t h e a n g le f o r w h i c h t h i s d e r i v a t i v e v a n i s h e s : v dt (! ) = 0 " v2 sin ! m = v1 " sin ! m = 1 d! v2
! T h e r e s u lt f o r t h e o p t i m a l p a t h d o e s n o t d e p e n d o n a a n d b t (! ) =
August 31, 2006 a12 + b 2 a2 (1 + tan 2 ! )b 2 a " b tan ! b a " b tan ! += + = + v1 v2 v1 v2 v1 cos ! v2
Physics for Scientists&Engineers 1 15 ! m = arcsin( 3.5 / 14 ) = 14.48°
B u t o n ly o n t h e r a t i o o f t h e s p e e d s i n t h e w a t e r a n d o n t h e s a n d
August 31, 2006 Physics for Scientists&Engineers 1 16 ...
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This note was uploaded on 03/01/2010 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.
 Spring '08
 Wolf
 Acceleration

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