This preview shows page 1. Sign up to view the full content.
Unformatted text preview: B (0s) + v B (0s)t + a B t 2 (substituting our results we get:) x B (t=8s) x B (t=0s) = (-10/3) m/s * (8s) + (2/3 m/s 2 )(8s) 2 = (-16/3) m/s or 5.3 m down. This is the magnitude of the displacement. We need to check if this is the total distance traveled as well. To do this, since the pulley is decelerating, we need to find if/when the pulley turns around, that is, when v B (t) = 0 m/s. Differentiating x B (t) = x B (0s) + v B (0s)t + a B t 2 we get v B (t) = v B (0s) + a B t This = 0 when t = -v B (0s)/a B or t = 10/3 m/s / (2/3 m/s 2 ) = 5s Just as we feared, the pulley has gone down and is on its way up before 8s has passed. The displacement before turning around is x B (t=5s) = (-10/3) m/s * (5s) + (2/3 m/s 2 )(5s) 2 = -25/3 m From the calculations above, 3s later it has traveled 25/3 16/3 m or 9/3 m. Therefore, the total distance traveled is 25/3 + 9/3 m = 34/3 m = 11.3 m...
View Full Document
This note was uploaded on 04/14/2010 for the course TAM 212 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
- Spring '08