A train normally travels at a uniform speed of 72 km/hon a long stretch of strai

Question

A train normally travels at a uniform speed of 72 km/hon a long stretch of straight, level track. On a particular day,the train must make a 2.0min stop at a station along this track. Ifthe train decelerates at a uniform rate of 1.0 m/s2 and, after the stop, accelerates at a rate of .50 m/s2, howmuch time is lost because of stopping at the station. Can someone show me the steps to this? A train normally travels at a uniform speed of 72 km/hon a long stretch of straight, level track. On a particular day,the train must make a 2.0min stop at a station along this track. Ifthe train decelerates at a uniform rate of 1.0 m/s2 and, after the stop, accelerates at a rate of .50 m/s2, howmuch time is lost because of stopping at the station. Can someone show me the steps to this?

Explanation / Answer

v=72km/h=72*1609m/3600s = 32.18 m/s the time to decelerates from v to zero ist1 0 = v-at1 =>t1 = v/a = 32.18m/s / 1.0m/s2 = 32.18 s the train take to speed up to v from 0 is t2 v=0+a2 t2 =>t2 =v/a2 = 32.18 m/s / 0.50m/s2 = 64.36 s the train also stop for 2.0 min = 120.0 s total time need T=32.18s + 64.36s + 120.0s = 216.54 s but we also need to consider what time it need withoutstop the distance from deceleration and acceleration is: d1=vt1-(1/2)a1t12 = 517.78m d2 = (1/2)a2t22 =0.5*0.50m/s2*(64.36s)2 = 1035.55 m d = 517.78m+1035.55m = 1553.33 m the normal time t = d/v=48.27s answer: the total time lost = 216.54 s - 48.27s = 168.27 s or 2.8 min v=0+a2 t2 =>t2 =v/a2 = 32.18 m/s / 0.50m/s2 = 64.36 s the train also stop for 2.0 min = 120.0 s total time need T=32.18s + 64.36s + 120.0s = 216.54 s but we also need to consider what time it need withoutstop the distance from deceleration and acceleration is: d1=vt1-(1/2)a1t12 = 517.78m d2 = (1/2)a2t22 =0.5*0.50m/s2*(64.36s)2 = 1035.55 m d = 517.78m+1035.55m = 1553.33 m the normal time t = d/v=48.27s answer: the total time lost = 216.54 s - 48.27s = 168.27 s or 2.8 min

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