36. We assume the train accelerates from rest (v00=and x00=) at a12134=+.m/suntil the midway point and then decelerates at a22=−.m / s until it comes to a stop v20=bgat the next station. The velocity at the midpoint is v1which occurs at x1= 806/2 = 403m. (a) Eq. 2-16 leads to ()2221011122 1.34 m/s403 mvvaxv¡=32.9 m/s.=(b) The time t1for the accelerating stage is (using Eq. 2-15) 2111122 403 m124.53 s21.34m/sxvtatt¡==.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.