ahead of his train traveling in the same direction on the same track. The freigh

Question

ahead of his train traveling in the same direction on the same track. The freight train is moving at a speed of 6.4 m/s.
(a) If the reaction time of the engineer is 0.37 s, what is the minimum (constant) rate at which the passenger train must lose speed if a collision is to be avoided?

(b) If the engineer's reaction time is 0.86 s and the train loses speed at the minimum rate described in Part (a), at what rate is the passenger train approaching the freight train when the two collide?

(c) For both reaction times, how far will the passenger train have traveled in the time between the sighting of the freight train and the collision?

Explanation / Answer

by the time the brakes are hit, the passenger train is within 321.5m of the freight train, so we write equations of motion for the two trains: xp(t)=position of passenger train = 31 t - 1/2 a t^2 xf(t)=position of freight train = 321.5 + 6.4t if they collided, xf = xp and we would have 31t - 1/2 a t^2 = 321.5 + 6.4t or 1/2 a t^2 -24.6 t +321.5 =0 now, we can solve the quadratic for the time of collision: t = {24.6 +/- Sqrt[24.6^2 -4 a(321.5)]}/a if there is a real solution for t, then the terms in the square root must be greater than zero; if the discriminant (the part in the square root) is less than zero, then there is no real solution, and there is no collision so, we want to find that value of a such that 24.6^2 -4(321.5)a < 0 this gives us the minimum value of acceleration: a=24.6^2/4*321.5 = 0.47m/s/s a smaller acceleration will yield a collision

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