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

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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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