a. Dp + Df = 345 m.
27*(T-0.36) + 5.9T = 345.
27T-9.72 + 5.9T = 345.
32.9T = 354.72.
T = 10.78 s. = Time they meet.
V = Vo + a*T = 0.
a = -Vo/T = -27/10.78 = -2.50 m/s^2. =
Min. rate.
A passenger train is traveling at 27 m/s when the engineer sees a freight train 345 m ahead of his train traveling in the same direction on the same track. The freight train is moving at a speed of 5.9 m/s.
(a) If the reaction time of the engineer is 0.36 s, what is the minimum (constant) rate at which the passenger train must lose speed if a collision is to be avoided? (m/s/s)
(b) If the engineer's reaction time is 0.73 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? (m/s)
(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? (km)
1 answer