I. Give the solutions of the following problems 1. A particle leaves the origin at t=0 with an initial velocity vo=(3.6m/s)i. It experiences a constant acceleration a=(1. 2m/s )i-(1. 4m/s ).(a)At what time does the particle reach its maximum x coordinate? (b)What is the velocity of the particle at this time?()Where is the particle at this time? Solution:(a) According to the problem, we have a, =-1. 2m/s. and vo =3.6m/s.Then the position of the particle is x= -Xmaxdi-0, thus we get =3s, the particle reach its maximum x coordinate x=5.4. dy (b )Using a= dr ,we have dv= adt →节=0+(-1.21-14/dr=361+(-1.2n-141)=(36-1.2)-14 When t=3s, the velocity of the particle is v=4.2(m/s) (c) Using v=dr, we ha d →F=36-12)-140d=(36-062)2-072 When f3s the position vector of the particle is F=(361-0612)-0.72j=541-63j(m) 2. A young basketball player is attempting to make a shot. the ball leaves the hand the player at angle of 60 to the horizontal at an elevation of 2.0 m above the floor The skillful player makes the shot with the ball traveling precisely through the center 2.0m the hoop as indicated in Figure 3 loud cheers, calculate the speed at which Fig 3III. Give the Solutions of the Following Problems 1. A particle leaves the origin at t = 0 with an initial velocity v i 6m/s)ˆ (3. 0 = r . It experiences a constant acceleration a i j ˆ (1.4m/s ) ˆ ( 1.2m/s ) 2 2 = − − r . (a) At what time does the particle reach its maximum x coordinate? (b) What is the velocity of the particle at this time? (c) Where is the particle at this time? Solution: (a) According to the problem, we have 2 = −1.2m/s ax and 3.6m/s 0 =x v . Then the position of the particle is 3.6 0.6 0.6( 3) 5.4 2 1 2 2 2 x = v0xt + ax t = t − t = − t − + , when max x = x , 0 d d = t x , thus we get t=3s, the particle reach its maximum x coordinate x=5.4. (b) Using t v a d d v v = , we have v v i j t i ti tj t i tj v a t t v t v 4 ˆ 1. ˆ ) (3.6 1.2 ) 4 ˆ 1. 2 ˆ ( 1. 6ˆ )d 3. 4 ˆ 1. ˆ (-1.2 d d 0 0 0 0 ⇒ = + − = + − − = − − = ∫ ∫ ∫ v v v v v When t=3s，the velocity of the particle is (m/s) 2 ˆ v = 4. j v . (c) Using t r v d d v v = , we have r t i tj t t t i t j r v t t r t 7 ˆ 0. ˆ ]d (3.6 0.6 ) 4 ˆ 1. ˆ [(3.6 -1.2 ) d d 2 2 0 0 0 ⇒ = − = − − = ∫ ∫ ∫ v v v v When t=3s the position vector of the particle is (m) 3 ˆ 6. 4ˆ 5. 7 ˆ 0. ˆ (3.6 0.6 ) 2 2 r = t − t i − t j = i − j v 2. A young basketball player is attempting to make a shot. The ball leaves the hands the player at angle of 60° to the horizontal at an elevation of 2.0 m above the floor. The skillful player makes the shot with the ball traveling precisely through the center of the hoop as indicated in Figure 3. To loud cheers, calculate the speed at which 8.0m 3.0m 2.0m 60° Fig.3 y x