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?

Answer 1

(a) 
The closing speed is 29 - 6 = 23 m/s. In 0.4 s the distance between the trains has reduced from 360 m to 360 - 0.4*23 = 350.8 m The distance between the trains in time t is given by s = 350.8 - 23*t + 0.5*a*t². When the trains meet, the closing velocity has dropped to 0, and this happened in time t, so the acceleration is a = 23/t, or t = 23/a. Putting this into the distance equation above gives 

s = 350.8 - 23*23/a + 0.5*a*(23/a)² = 350.8 - 23²/a + 0.5*23²/a = 350.8 - 0.5*23²/a 

set s = 0 to get 0 = 350.8 - 0.5*23²/a, a = 0.5*23²/350.8 = 0.75 m/s² 

(b). The distance at start of acceleration is now 360 - 0.8*23 = 341.6 m. The distance traveled in time t is 23*t - 0.5*a*t² so t to crash is set by 341.6 = 23*t - 0.5*0.75*t². The velocity at crash time is vc = 23 - a*t; t = (23 -vc)/0.75. using this t in the formula above: 

341.6 = 23*(23 -vc)/0.75 - 0.5*1.33*[(23 -vc)/0.75]² 
341.6 = 23²/0.75* - 23*vc/0.75 - 0.5*(23 -vc)²/0.75 
341.6 = 23²/0.75* - 23*vc/0.75 - 0.5*23²/0.75 + 23*vc/0.75 - 0.5*vc²/0.75 
341.6 = 0.5*23²/0.75 - 0.5*vc²/0.75 
512.4 = 23² - vc² 
vc = 4.1 m/s 

(c) 

for case (a) the time for the trains to meet after the reaction time is t = 23/a = 23/0.75 = 30.7 s; The distance traveled by the fast train is 29*0.4 + 29*t - 0.5*a*t² = 29*0.4 + 29*30.7 - 0.5*0.75*30.7² = 537 m 

for case (b) the time after reaction to crash is (23 - 4.1)/a = 25.2 s and the distance is 29*.8 +29*25.2 - 0.5*0.75*25.2² = 516 m




Hope this helps....