Question

Two trains face each other on adjacent tracks. They are initially at rest, and their front ends are 50 m apart. The train on the left accelerates rightward at 1.15 m/s2. The train on the right accelerates leftward at 1.15 m/s2. (a) How far does the train on the left travel before the front ends of the trains pass? 25 m (b) If the trains are each 150 m in length, how long after the start are they completely past one another, assuming their accelerations are constant?

Answer 1

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, $a_{L} = 1.15 m/s^{2}$ towards right

Acceleration of the train on the right, $a_{R} = 1.15 m/s^{2}$ towards left

Relative acceleration of the train , $a_{r} = 1.15 - (- 1.15) = 2.30 m/s^{2}$

Now,

(a) Using the eqn (2) of motion, for the train on the left:

$d = ut + \frac{1}{2}a_{r}t^{2}$

$50 = 0.t + \frac{1}{2}\times 2.30t^{2}$

$t = \sqrt{\frac{100}{2.30}} = 6.593 s$

Now, the distance covered by the train on the left before passing the front end:

$d' = ut + \frac{1}{2}a_{L}t^{2}$

$d' = 0.t + \frac{1}{2}\times 1.15\times (6.593)^{2}$

d' = 43.073 m

d' = 43.073 - 25 = 22.73 m

(b) Now,

Acceleration is constant at $a_{r} = 2.3 m/s^{2}$

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 + $\frac{1}{2}\times 2.3t^{2}$

t = 13.187 s