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If the greatest admissible acceleration of a train is 3 feet/sec square, calculate the least time taken from one station to another at a distance of 10m.
2 AnswersMaximum permissible acceleration = 0.9144 m/s.
If initial velocity=0;
t=((2x10)/0.9144)1/2
=4.67 sec.
i assume that the train starts from the first station and crosses the second station with the same acceleration.
so the initial velocity is zero.
the acceleration is 0.9m/s2.
time taken is (2*10/0.9)(1/2) = 4.71 s.
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