Unformatted text preview: = p ( − 40) 2 + 2(1 . 0)(200 − 950) = √ 100 = 10 m / s using Eq 2-16 again. SpeciFcally, 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 Fnally 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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- Acceleration, Velocity, -x direction, unprimed train points