p02_034

# Fundamentals of Physics,Vol 1 (Chapters 1 - 20)

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34. 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 v 0 o = 72 km/h = 20 m/s) 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 v 0 =0) at x 0 = ( v 0 ) 2 ( v 0 o ) 2 2 a 0 = 0 20 2 2 = 200 m . The speed of the other train, when it reaches that location, is v = p v 2 o +2 a x
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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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