Answer:
(a) d' = 22.73 m
(b) t = 13.187 s
Solution:
As per the question:
Initial speed of both the trains, u = 0 m/s
The distance between the front ends of the train, d = 50 m
Acceleration of the train on the left,
towards right
Acceleration of the train on the right,
towards left
Relative acceleration of the train , ![a_{r} = 1.15 - (- 1.15) = 2.30 m/s^{2}](https://tex.z-dn.net/?f=a_%7Br%7D%20%3D%201.15%20-%20%28-%201.15%29%20%3D%202.30%20m%2Fs%5E%7B2%7D)
Now,
(a) Using the eqn (2) of motion, for the train on the left:
![d = ut + \frac{1}{2}a_{r}t^{2}](https://tex.z-dn.net/?f=d%20%3D%20ut%20%2B%20%5Cfrac%7B1%7D%7B2%7Da_%7Br%7Dt%5E%7B2%7D)
![50 = 0.t + \frac{1}{2}\times 2.30t^{2}](https://tex.z-dn.net/?f=50%20%3D%200.t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%202.30t%5E%7B2%7D)
![t = \sqrt{\frac{100}{2.30}} = 6.593 s](https://tex.z-dn.net/?f=t%20%3D%20%5Csqrt%7B%5Cfrac%7B100%7D%7B2.30%7D%7D%20%3D%206.593%20s)
Now, the distance covered by the train on the left before passing the front end:
![d' = ut + \frac{1}{2}a_{L}t^{2}](https://tex.z-dn.net/?f=d%27%20%3D%20ut%20%2B%20%5Cfrac%7B1%7D%7B2%7Da_%7BL%7Dt%5E%7B2%7D)
![d' = 0.t + \frac{1}{2}\times 1.15\times (6.593)^{2}](https://tex.z-dn.net/?f=d%27%20%3D%200.t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%201.15%5Ctimes%20%286.593%29%5E%7B2%7D)
d' = 43.073 m
d' = 43.073 - 25 = 22.73 m
(b) Now,
Acceleration is constant at ![a_{r} = 2.3 m/s^{2}](https://tex.z-dn.net/?f=a_%7Br%7D%20%3D%202.3%20m%2Fs%5E%7B2%7D)
Length of the trains, l = 150 m
Total distance, D = l + d = 150 + 50 = 200 m
Now, from eqn (2) of motion again:
200 = 0.t + ![\frac{1}{2}\times 2.3t^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes%202.3t%5E%7B2%7D)
t = 13.187 s