John is rollerblading down a long, straight path. At time zero, there is a mailbox about 1 min front of him. In the 5 s time perio that follows, John's velocity is given by the velocity versus time graph in the gure. Taking the mailbox to mark the zero location, with positions beyond the mailbox as positive, plot his position versus time in the given position versus time graph. 4 3 3 2 2 q l E l 5 Time (s) \\E Q 0 Ill|,' g O...... 2' 1 2 3 4 5 g ' D4 l -1 2 0.0 1.0 2.0 3.0 4.0 5.0 Assuming that all the numbers given are exact, what is John's position at a time of 4.49 s? Enter your answer to at least three significant digits. position at 4.49 s: mvelocity 4 distance 4 1 3 2 . + (2,2) (2, 5) 2 345 1345 time time Between t ( 2 , 5 ) , V = 2 t = 2 V = 2 , distance = 1 distance, d = t=5 V= 2 , distance= GIIN At + = 2 velocity decreases and become zero . t= 1 , V decrease and be come -2 , so distance decrease and d= - 2 After + ( 0 , 1) , d = =2 it - 0, do-d t= 4.495, the distance is d= 210 = 0.4454 m 4.49 Position at 4.49 s: 0. 445 m CS Scanned with CamScanner y