ch02-p090 - 90. We take +x in the direction of motion. We...

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() 21 12 2 2 160 m 4.0 s . 30 m/s 50 m/s xx tt vv −= = = ++ (c) Since the train is at rest ( v 0 = 0) when the clock starts, we find the value of t 1 from Eq. 2-11: 10 1 1 2 30 m/s 6.0 s . 5.0 m/s a t t =+ ¡ == (d) The coordinate origin is taken to be the location at which the train was initially at rest (so x 0 = 0). Thus, we are asked to find the value of x 1 . Although any of several equations could be used, we choose Eq. 2-17: ( ) 1 1 11 30 m/s 6.0 s 90 m . 22 xv v t = = (e) The graphs are shown below, with SI units assumed.
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