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-=-26 325 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