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