6. We apply the equation21002()xtxvtat=++, found in Table 2-1. Since at t= 0 s, x0= 0 and012 m/sv=, the equation becomes (in unit of meters) 212() 12xttat.With 10 mx=when1.0 st=, the acceleration is found to be 24.0 m/sa=−. The fact that0a<implies that the bead is decelerating. Thus, the position is described by 2() 12 2.0xttt. Differentiating
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