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 m/s. if the
reaction time of the engineer is 0.4 s, what is the minimum rate at which the passenger train must lose
speed (deceleration) if the collision is to be avoided?
Answers
Explanation:
(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