ch02-p086 - 2 2 2 o 2 40 m/s 2 1.0 m/s 200 m 950 m 10 m/s v...

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86. We take the moment of applying brakes to be t = 0. The deceleration is constant so that Table 2-1 can be used. Our primed variables (such as 0 72 km/h = 20 m/s v ′ = ) refer to one train (moving in the + x direction and located at the origin when t = 0) and unprimed variables refer to the other (moving in the – x direction and located at x 0 = +950 m when t = 0). We note that the acceleration vector of the unprimed train points in the positive direction, even though the train is slowing down; its initial velocity is v o = –144 km/h = –40 m/s. Since the primed train has the lower initial speed, it should stop sooner than the other train would (were it not for the collision). Using Eq 2-16, it should stop (meaning 0 v ′ = ) at () () 22 2 0 2 0( 2 0 m / s ) 200 m . 22 m / s vv x a ′′ == =
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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 = + Δ = − + − = 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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