# ch02-p036 - 36 We assume the train accelerates from rest v0...

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36. We assume the train accelerates from rest ( v 0 0 = and x 0 0 = ) at a 1 2 134 =+ .m / s until the midway point and then decelerates at a 2 2 =− . m / s until it comes to a stop v 2 0 = bg at the next station. The velocity at the midpoint is v 1 which occurs at x 1 = 806/2 = 403m. (a) Eq. 2-16 leads to () 22 2 10 1 1 1 2 2 1.34 m/s 403 m vv a x v ¡ = 32.9 m/s. = (b) The time t 1 for the accelerating stage is (using Eq. 2-15) 2 1 1 1 1 2 2 403 m 1 24.53 s 21 . 3 4 m / s xv t a t t ¡ == . Since the time interval for the decelerating stage turns out to be the same, we double this result and obtain t = 49.1 s for the travel time between stations.
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## This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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