ch07-p006 - 6 We apply the equation x(t = x0 v0t 1 at 2...

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6. We apply the equation 2 1 00 2 () x tx v ta t =+ + , found in Table 2-1. Since at t = 0 s, x 0 = 0 and 0 12 m/s v = , the equation becomes (in unit of meters) 2 1 2 () 12 x tt a t . With 10 m x = when 1.0 s t = , the acceleration is found to be 2 4.0 m/s a =− . The fact that 0 a < implies that the bead is decelerating. Thus, the position is described by 2 () 12 2 .0 x ttt . Differentiating
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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