A commuter train travels between two downtown stations. Because the stations are

Question

A commuter train travels between two downtown stations. Because the stations are only 1.44 km apart, the train never reaches its maximum possible cruising speed. During rush hour the engineer minimizes the travel interval t between the two stations by accelerating for a time interval t1 at a1 = 0.100 m/s2 and then immediately braking with acceleration a2 = -0.340 m/s2 for a time interval t2. Find the minimum time interval of travel t and the time interval t1. t = 1 s t1 = 2 s A commuter train travels between two downtown stations. Because the stations are only 1.44 km apart, the train never reaches its maximum possible cruising speed. During rush hour the engineer minimizes the travel interval t between the two stations by accelerating for a time interval t1 at a1 = 0.100 m/s2 and then immediately braking with acceleration a2 = -0.340 m/s2 for a time interval t2. Find the minimum time interval of travel t and the time interval t1. t = 1 s t1 = 2 s A commuter train travels between two downtown stations. Because the stations are only 1.44 km apart, the train never reaches its maximum possible cruising speed. During rush hour the engineer minimizes the travel interval t between the two stations by accelerating for a time interval t1 at a1 = 0.100 m/s2 and then immediately braking with acceleration a2 = -0.340 m/s2 for a time interval t2. Find the minimum time interval of travel t and the time interval t1. t = 1 s t1 = 2 s A commuter train travels between two downtown stations. Because the stations are only 1.44 km apart, the train never reaches its maximum possible cruising speed. During rush hour the engineer minimizes the travel interval t between the two stations by accelerating for a time interval t1 at a1 = 0.100 m/s2 and then immediately braking with acceleration a2 = -0.340 m/s2 for a time interval t2. Find the minimum time interval of travel t and the time interval t1. t = 1 s t1 = 2 s t = 1 s t1 = 2 s

Explanation / Answer

with a1 = 0.100 m /s^2 x = 1440 m = (1 / 2) a1t1^2 + v1 t2 + (1 / 2)a2 t2^2 1440 m = (1 / 2) a1t1^2 + a1 t1(-a1 t1/ a2) + (1 / 2) a2(a1 t1/ a2)^2 t2 = a1 t1/- a2 = ....... s a2 = - 0.340 m /s2 t = t1 + t2 v1 = a1t1 = - a2t2 1440 m = (1 / 2) a1 [1 -(a1 / a2)] t1^2 t1 = ........ s total time t = ......... s

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