A passenger train is traveling at 29 m/s when the engineer sees a freight train

Question

A passenger train is traveling at 29 m/s when the engineer sees a freight train 360 m ahead of his train traveling in the same direction on the same track. The freight train is moving at a speed of 6.0 m/s.

(a) If the reaction time of the engineer is 0.40 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 0.80 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

distance between both for collission to be avoided= 360- 23 x0.4= 360-9.2= 350.8 m

initial velocity= 23 m/s

final velocity = 0m/s

distance = 350.8 m

so deceleration = 23^2-0^2/(2x350.8)= 0.754 m/s^2

b. then distance between them = 360- 23 x0.8= 341.6 m

final velocity of the passenger train=vf

vf^2 =vi^2 - 2 x 0.754 x 341.6= 23^2 - 2x0.754 x 341.6=

so, vf= 3.72 m/s

so passenger train approach speed = 3.72 m/s

c. time taken = 23/0.754=30.5s

distance travelled = 29 x 30.5 - 1/2 x 0.754 x 30.5^2 + 29 x 0.4= 545.4 m

time taken = 23- 3.72/0.754= 25.57 s

distance travelled = 29 x 25.57 - 1/2 x 0.754 x 25.57^2 + 29 x 0.8= 741.53- 246.49+23.2= 518.24 m

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