HW1-phy2048-Spring-2014-solution

00 s 492 ms 2 b although several of the equations in

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Unformatted text preview: e value it had in part (c). Then, |v| is larger than that value for t > 3 s. (e) Yes, since v smoothly changes from negative values (consider the t = 1 result) to positive (note that as t → + ∞, we have v → + ∞). One can check that v = 0 when t = 2 s. (f) No. In fact, from v = –12 + 6t, we know that v > 0 for t > 2 s. 2P22. In this solution, we make use of the notation x(t) for the value of x at a particular t. The notations v(t) and a(t) have similar meanings. x = ct2 – bt3 where c = 3 and b = 2. (a) Its displacement at t=4 is Δx = x2 – x1, where x1 = 0 and x2 = –80 m. Thus, Δx = −80 m . The velocity is given by v = 2ct – 3bt2 = (6.0 m/s2)t – (6.0 m/s3)t2. 4 PHY2...
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