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 360m 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

Here,

speed of train , v1 = 29 m/s

v2 = 6 m/s

distance , d = 360 m

relative velocity = v1 - v2 = 29 - 6

v = 23 m/s

a) distance left when deceleration starts

d = 360 - (29 - 6)* 0.40 = 350.8 m

let the deceleration is a

Using third equation of motion

v^2 - u^2 = 2 a * d

23^2 - 0^2 = 2 * a * 350.8

a = 0.754 m/s^2

the deceleration of the train must be 0.754 m/s^2

b)

distance left when deceleration starts

d = 360 - (29 - 6)* 0.80 = 341.6 m

Now, for the final velocity of approach

v^2 - 23^2 = -2 * 341.6 * 0.754

v = 3.72 m/s

the velocity of approcah is 3.72 m/s

c)

distance travelled in case 1 = 360 + 6 * (23/0.754)

distance travelled in case 1 = 543 m

----------------------

distance travelled in case 1 = 360 + 6 * ((23 - 3.72)/0.754)

distance travelled in case 1 = 513 m

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