4x At m- 2m 05 s 3m 0.6m/s 55 (VA) -5m -5s S = - = slope (VB) stope = - 5s -Sm = slope A - 3 m -6m X(m) 5 4 3 2 1 -5 0 -2 -3 (912) X=2m 1 2 & C 3 4 5 xi B (5,5) X-Sm 6 + X=0 tc = 11 s Xc = -5 m vc = ?-1.6 m/s 7 8 9 10 11 t(s) X=-5m (11,-5) XF tA=0 XA = 2 m V = ?0.6m/s OK A AB+velocity The above position vs. time graph shows a motion of a walking man in x direction. Within your group: Describe/discuss the motion qualitatively. Then answer the following questions. Below is a pictorial representation for the motion (NOT a motion diagram). (VP) Bu=-velocity slope Final-Initial m Final-initial s = 4x 4t 4. What is the velocity at t = 6 s? (Show calculations) (1 point) Constant tp X= 5 s constant = slope +6m-5m 11115-55 XB = 5 m VB =?.6 m/s 3. What is the velocity at t = 2.5 s? (Show calculations or explain) (1 point) 1. Is the velocity constant from Os to 5s? Yes No (1 point) 2. How do you know that the velocity is constant from Os to 5s? Give your reason. (2 points) -10 = 1.6 6 (Vc)