Two trains and are approaching each other on a straight back, the former with a uniform velocity of 25 m/sec and other
15 sec when they are 225 m apar brakes are simultaneously applied to both of them. The deceleration given by the brakes
to the train B ncreases linearly with time by 0.3 w e every second While the A given a uniforme deceleration
(a) What must be the minimum deceleration of A so that the trains do not collade Ch) What is the time taken by trains to
come to stop ?
Answers
Answer:
solved
Explanation:
TrainB
V = 15 - 0.15t²
S 1= 15 t - 0.05t³
TrainA
v = 25 - at , a = deceleration of train A
s2 = 25t - at²/2
now as the trains are Approaching ; s1 + s2 = 225
25t - at²/2 + 15 t - 0.05t³ = 225
at²/2 = - 0.05t³ + 40t - 225
at² = - 0.1t³ + 80t - 450
a = - 0.1t+ 80t∧-1 - 450t∧-3
now we can minimize a
da/dt = - 0.1 - 80t∧-2 + 1350t∧-4 = 0
- 0.1t∧4 - 80t² + 1350 = 0
t∧4 + 800t² - 13500 = 0
t² = - 800 ±√640000 + 54000 ÷2 = (- 800 ±√694000) ÷2
t² = √173500 - 400 = 16.5333
t = 4.066 sec
a = -0.1t+ 80t∧-1 - 450t∧-3 = - 0.4066 + 80/4.066 - 450/4.066³
a = 19.269 - 6. 694 = 12.575 m/s²
(a) What must be the minimum deceleration of A so that the trains do not collide = a = 12.575 m/s²
Ch) What is the time taken by trains to come to stop ? t = 4.066 sec