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(a) If the maximum acceleration that is tolerable for passengers in a subway train is1.74 m/s2and subway stations are located 816 m apart, what is the maximum speed a subway train can attain between stations? (b) What is the travel time between stations? (c) If a subway train stops for 24.0 s at each station, what is the maximum average speed of the train, from one start-up to the next?
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Added Wed, 09 Sep '15
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(a) The subway train must then be accelerating at 1.74 m/s2 for half of the distance, then decelerate at -1.74 m/s2 for the other half.

By v_f^2 = v_i^2 + 2ax

v_f^2 = 0 + 2(1.74 m/s2)(408)

v_f = 37.68 m/s

Note: the distance is halved because it takes half of the total distance to reach the max speed. Then, after this speed is reached the subway needs to start decelerating at the max acceleration because otherwise the subway would either make passengers uncomfortable or miss the stop.

(b) The travel time to go the first half is the same as the travel time to go the second half, because the average velocity (v_f+v_i)/2 is the same.

x = (v_avg)*t

408 m = (v_f + v_i)/2*t

408m = (37.68 m/s + 0)/2 * t

t = 21.66 s

The total time is equal to 2t (Because that was just the calculation for halfway). So the answer is 21.66 * 2 = 43.32 s.

(c) The average speed of the train is equal to total distance covered over total time.

At best, it takes 43.32 seconds to travel the 816 m between stations. Thus, the total distance is 816 meters. The total time is equal to the 43.32 seconds of travel plus the 24 seconds of stopping. 43.32+24 = 67.32 s. So by the average speed formula:

s = x/t = 816m/67.32s = 12.12 m/s
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