UNIVERSITY PHYSICS I CHAPTER 4 Kinematics ii Motion in two and Three Dimensions 84.1 The position velocity, and acceleration vectors in two dimensions 1. The position vector and displacement Let the plane of the two-dimensional motion be the x-y plane of a Cartesian coordinate system. r(t=x(t)i+y(t) 本s r(t+4-r(t) =|x(t+∠r)-x(t)i +[y(t+4)-y(t)lj
1 1. The position vector and displacement Let the plane of the two –dimensional motion be the x-y plane of a Cartesian coordinate system. y t t y t j x t t x t i r t t r t r r r r t x t i y t j f i ˆ [ ( ) ( )] ˆ [ ( ) ( )] ( ) ( ) ˆ ( ) ˆ ( ) ( ) + + − = + − = + − = − = + ∆ ∆ ∆ ∆ r r r r r r x y O ∆s §4.1 The position, velocity, and acceleration vectors in two dimensions
4.1 The position, velocity, and acceleration vectors in two dimensions 2. The path(trajectory) of a particle x=x(t) y=y(t) Eliminating y=y(x) Example 1: r(t)=A cos ati+ B sin atj x=x(t=Acos @t V= y(t)=b sin at 十 A2B 84.1 The position velocity, and acceleration vectors in two dimensions 3. Speed and velocity vector a Average speed and instantaneous speed As ave ∠t as d v= m →0 4t dt b Average velocity and instantaneous velocity m布r(t+A)-F(t) Notes≠ ∠t Ive
2 2. The path (trajectory) of a particle ( ) ( ) y y t x x t = = Eliminating t y = y(x) Example 1: 1 ( ) sin ( ) cos ˆ sin ˆ ( ) cos 2 2 2 2 + = = = = = = + B y A x y y t B t x x t A t r t A ti B tj ω ω ω ω r §4.1 The position, velocity, and acceleration vectors in two dimensions §4.1 The position, velocity, and acceleration vectors in two dimensions 3. Speed and velocity vector a. Average speed and instantaneous speed t s t s v t s v t d d lim 0s ave = = = → ∆ ∆ ∆ ∆ ∆ b. Average velocity and instantaneous velocity t r t t r t t r v ∆ ∆ ∆ ∆ ( ) ( ) ave r r r r + − = = x y O ∆s Note: ave ave v v r ≠
84.1 The position, velocity and acceleration vectors in two dimensions v(t=lim r(+△)-r(t)dr( M→0s c. The components of the velocity in Cartesian coordinate system dr(t) dx(t):, dj A v(t) =v,(t)i+v,(t) dt d dt Magnitude:y=/(6)=[v2(0)+22(0v ds+ dt Direction: tangent to the path followed by the particle; or 0= tan-y The angle between v and x-axis 84.1 The position velocity, and acceleration vectors in two dimensions v(t) rajectory t+△n v(t+At v(t+At) 4. Acceleration vector Average acceleration Av(0 v(t+4r)-v( t
3 t r t t r t t r t v t t d ( ) ( ) d ( ) ( ) lim 0s r r r r = ∆ + ∆ − = ∆ → c. The components of the velocity in Cartesian coordinate system j v t i v t j t y t i t x t t r t v t x y ˆ ( ) ˆ ( ) ˆ d d ( ) d d ( ) d d ( ) ( ) = = + = + r r r §4.1 The position, velocity, and acceleration vectors in two dimensions Magnitude: t s v v t v t v t x y d d ( ) [ ( ) ( )] 2 2 1 2 = = + = r Direction: tangent to the path followed by the particle; or x y v v 1 tan− θ = The angle between v and x − axis r §4.1 The position, velocity, and acceleration vectors in two dimensions 4. Acceleration vector a. Average acceleration v(t) r v(t + ∆t) r v(t) r ∆ t v t t v t t v t a ∆ ∆ ∆ ∆ ( ) ( ) ( ) ave r r r r + − = = x y O v(t) r v(t + ∆t) r
84.1 The position, velocity and acceleration vectors in two dimensions b. Instantaneous acceleration (t)=li Av(t dv(t a→0st dt 4v() dv,()a dv, (t 十 dt v(t+4) =a(t)i+a,(t)j Magnitude: a=a(D)=[a2()+a2(0/2 , Direction: 6=tan The angle between a and x-axis 84.1 The position, velocity, and acceleration vectors in two dimensions C. The direction of acceleration: 4v=v-v vI> B B < g
4 §4.1 The position, velocity, and acceleration vectors in two dimensions a t i a t j j t v t i t v t t v t t v t a t x y x y t ˆ ( ) ˆ ( ) ˆ d d ( ) ˆ d d ( ) d ( ) d ( ) ( ) lim 0s = + = + = = → r r r ∆ ∆ ∆ b. Instantaneous acceleration Magnitude: 2 2 1 2 a a(t) [a (t) a (t)] x y = = + r Direction: x y a a1 tan− θ = The angle between a and x − axis r v(t) r v(t + ∆t) r v(t) r ∆ c. The direction of acceleration： B A v v v r v r ∆ = − B A v v r r > A v r v r ∆ a α r g r v r A v r B v r B A v v r r < B v r v r ∆ a r g r α v r A v r B A v v r r = B v r v r ∆ a r α v r a r A v r B v r §4.1 The position, velocity, and acceleration vectors in two dimensions
84.1 The position, velocity and acceleration vectors in two dimensions Example 2: If we know the position vector of a particle F=2ni+(2-t2 Fined the trajectory of the particle; the position vector at fs and ts the velocity and the acceleration of the particle at instant t=2s. Solution: (itrajetory r= 2t y=2 Eliminate t, we can get y=2-te 4-parabola (2) position vector: t=0s,x=0p=2 t=2s,x=4 r′=4i-2j 84.1 The position, velocity, and acceleration vectors in two dimensions J 2 Q r’=4i-2j The magnitude: r=r=2(m) F=√42+(-2)2=447m) The direction: The angle between r and x-axis 8=arcfe2 90 The angle between rand x-axis o'=arctg-=-26 32
5 Solution: (1)trajetory ⎩ ⎨ ⎧ = − = 2 2 2 y t x t (2) position vector： t = 0s，x = 0 y= 2 t = 2s，x = 4 y = -2 r i j r j r r r r r 4 2 2 ′ = − = Example 2: If we know the position vector of a particle r ti t j r r r 2 (2 ) 2 = + − (SI) Fined the trajectory of the particle; the position vector at t=s and t=2s; the velocity and the acceleration of the particle at instant t=2s. 4 2 2 x Eliminate t, we can get y = − —parabola §4.1 The position, velocity, and acceleration vectors in two dimensions The magnitude： 4 ( 2) 4.47(m) 2(m) 2 2 ′ = ′ = + − = = = r r r r r r o y x Q r r r′ r 2 P -2 4 θ′ θ 4 2 2 x y = − r i j r j r r r r r 4 2 2 ′ = − = r r The direction： 26 32 4 2 arctg 90 0 2 arctg = − ′ − ′ = = = o o θ The angle between and x-axis θ The angle between and r ′ x-axis r §4.1 The position, velocity, and acceleration vectors in two dimensions
84.1 The position, velocity and acceleration vectors in two dimensions ()The velocity: F=2t i+2-t dr v =2i-2t j Magnitude: V =2 v,=-2t V"x+v=2√ +t t=2n2=2√5=4.47ms The angle between velocity and x-axis 4 a= tan tan 2 84.1 The position, velocity, and acceleration vectors in two dimensions v=,=2i-2tj dνd2r T he magnitu 2 Direction of the acceleration
6 (3)The velocity: 2 4 tan tan 1 1 − = = − − x y v v α The angle between velocity and x-axis: ( ) -1 2 2 2 2 2 2 2 5 4.47 m s 2 1 2 2 2 2 d d 2 2 = = = ⋅ = + = + = = − = = − = + − t v v v v t v v t i t j t r v r t i t j x y x y r r r r r r r Magnitude: §4.1 The position, velocity, and acceleration vectors in two dimensions ( ) j t r t v a i t j t r v r t i t j ˆ 2 d d d d ˆ 2 ˆ 2 d d ˆ 2 ˆ 2 2 2 2 = = = − = = − = + − r r r r r r (3)The acceleration: Direction of the acceleration: The magnitude: a = a = 2 r j ˆ − §4.1 The position, velocity, and acceleration vectors in two dimensions
84.1 The position, velocity and acceleration vectors in two dimensions Exercise: The position of a small bumper car in an amusement park ride is described as a function of time by the coordinates x=0.2t2+5.0t+0.5 (SD) y=-1.0r+10.0t+20(S1) Find (a) the position vector at t1.0 s and t3.0 s (b) the displacement vector between these tim e (c)average velocity over the period from 1.0 s to 3.0s, and the velocity at t=3.0 s. (d the magnitude and direction of the acceleration at tl0s and t3.0 s 84.1 The position, velocity, and acceleration vectors in two dimensions 5. Perpendicular motions are independent of each other We can analyze the motion along each coordinate axis separately Experimental result
7 Exercise: The position of a small bumper car in an amusement park ride is described as a function of time by the coordinates 1.0 10.0 2.0 (SI) 0.2 5.0 0.5 (SI) 2 2 = − + + = + + y t t x t t Find (a) the position vector at t=1.0 s and t=3.0 s. (b) the displacement vector between these time. (c) average velocity over the period from 1.0 s to 3.0 s, and the velocity at t=3.0 s. (d) the magnitude and direction of the acceleration at t=1.0 s and t=3.0 s. §4.1 The position, velocity, and acceleration vectors in two dimensions 5. Perpendicular motions are independent of each other §4.1 The position, velocity, and acceleration vectors in two dimensions We can analyze the motion along each coordinate axis separately. Experimental result
84.3 motion in three dimension 84.2 Two dimensional motion with a constant acceleration (self-study) 84.3 motion in three dimension r(t=x(t)i+y(t)j+z(t k △r(r)=△x(t)i+△y()j+△()k △ △F △ν ave △t ave △t ave △ d dr dy d dt dt 84. 4 relative velocity addition and accelerations 1. Reference frame A coordinate system with a set of synchronized clocks, all ticking at the same rate, is called reference system Coordinate +synchronized clocks Different choices for the reference frame lead to different description of motion, but the underlying physics is nonetheless the same 8
8 §4.2 Two dimensional motion with a constant acceleration (self-study) §4.3 motion in three dimension §4.3 motion in three dimension 2 2 ave ave ave d d d d , d d , d d , , ˆ ( ) ˆ ( ) ˆ ( ) ( ) ˆ ( ) ˆ ( ) ˆ ( ) ( ) t r t v a t r v t s v t v a t r v t s v r t x t i y t j z t k r t x t i y t j z t k r r r r r r r r r r r = = = = ∆ ∆ = ∆ ∆ = ∆ ∆ = ∆ = ∆ + ∆ + ∆ = + + §4.4 relative velocity addition and accelerations 1. Reference frame A coordinate system with a set of synchronized clocks, all ticking at the same rate, is called reference system. Different choices for the reference frame lead to different description of motion, but the underlying physics is nonetheless the same. Coordinate +synchronized clocks
84. 4 relative velocity addition and accelerations 2. relative velocity addition and accelerations r=R+r dr dr dr dtdt dt Vpo+voo R Po-ap0ta if voo=constant, then a 84. 4 relative velocity addition and accelerations 3. Some topics of discussion OIs there really inertial reference frame? @Are space and time absolute or not? @The measurement of the time interva @The measurement of the space distance
9 §4.4 relative velocity addition and accelerations 2. relative velocity addition and accelerations r = R+ r′ r r PO PO O O PO PO O O a a a v v v ′ ′ ′ ′ = + = + r r r r r r t r t R t r d d d d d d ′ = + r r r PO PO O O a a v ′ ′ = = r r r then if constant, O r r r′ r R r P O′ §4.4 relative velocity addition and accelerations 3. Some topics of discussion 1Is there really inertial reference frame? 2Are space and time absolute or not? 3The measurement of the time interval; 4The measurement of the space distance
84. 4 relative velocity addition and accelerations Exercise 1 a train travels due south at 28 m/s(relative to the ground in a rain that is blown to the south by the wind. The path of each raindrop makes an angle of 64 with the vertical, as measured by an observer stationary on the earth. An observer on the train however. see perfectly vertical tracks of rain on the windowpane. Determine the speed of the rain drops relative to the ground 84. 4 relative velocity addition and accelerations Solution: =v+卩 g From the problem, we have vsin64°=p,=28m/s 28 33.3m/s sin64°0.84
10 Exercise 1: A train travels due south at 28 m/s(relative to the ground) in a rain that is blown to the south by the wind. The path of each raindrop makes an angle of 64°with the vertical, as measured by an observer stationary on the earth. An observer on the train, however, see perfectly vertical tracks of rain on the windowpane. Determine the speed of the rain drops relative to the ground. §4.4 relative velocity addition and accelerations Solution: rt v r tg v r rg v r tg v r o 64 o 64 rg v r rt v r From the problem, we have sin64 = = 28m/s rg tg v v o 33.3m/s 0.84 28 sin64 = = = o tg rg v v §4.4 relative velocity addition and accelerations rg rt tg v v v r r r = +