()211222 160 m4.0 s .30 m/s50 m/sxxttvv−−===++(c) Since the train is at rest (v0= 0) when the clock starts, we find the value of t1from Eq. 2-11: 1011230 m/s6.0 s .5.0 m/satt=+¡==(d) The coordinate origin is taken to be the location at which the train was initially at rest (sox0= 0). Thus, we are asked to find the value of x1. Although any of several equations could be used, we choose Eq. 2-17: ()111130 m/s 6.0 s90 m .22xvvt==(e) The graphs are shown below, with SI units assumed.
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