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Unformatted text preview: ( ) ( ) ( ) 2 2 2 o 2 40 m/s 2 1.0 m/s 200 m 950 m 10 m/s v v a x = + = + = using Eq 2-16 again. Specifically, its velocity at that moment would be 10 m/s since it is still traveling in the x direction when it crashes. If the computation of v had failed (meaning that a negative number would have been inside the square root) then we would have looked at the possibility that there was no collision and examined how far apart they finally were. A concern that can be brought up is whether the primed train collides before it comes to rest; this can be studied by computing the time it stops (Eq. 2-11 yields t = 20 s) and seeing where the unprimed train is at that moment (Eq. 2-18 yields x = 350 m, still a good distance away from contact)....
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- Spring '06