(a) From v(t) = 0 we find it is (momentarily) at rest at t= 0. (b) We obtain x(0) = 4.0 m (c) and (d) Requiring x(t) = 0 in the expression x(t) = 4.0 – 6.0t2leads to t= ±0.82 s for the times when the particle can be found passing through the origin. (e) We show both the asked-for graph (on the left) as well as the “shifted” graph which is relevant to part (f). In both cases, the time axis is given by –3 ≤t≤3 (SI units understood).(f) We arrived at the graph on the right (shown above) by adding 20
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.