Unformatted text preview: B (0s) + v B (0s)t + ½ a B t 2 (substituting our results we get:) x B (t=8s) – x B (t=0s) = (-10/3) m/s * (8s) + ½ (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 + ½ 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) + ½ (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...
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- Spring '08
- SEPTA Regional Rail, Jaguar Racing, 2k, 3 m, 0 m/s, 3 m/s