An express train is running behind a goods train on the same line and in the same direction, their velocities being u1 and u2 (u1 > u2) respectively. When there is a distance x between them, each is seen from the other. At which point it is just possible to avoid a collision f1 is the greatest retardation and f2 is the greatest acceleration which can be produced in the two trains respectively?
(a) (u1 – u2)^2 = 2x(f1 + f2)
(b) (u1 + u2)^2 = 2x(f1 – f2)
(c) (u1 – u2)^2 = 2x(f1 – f2)
(d) (u1 + u2)^2 = 2x(f1 + f2)
I had been asked this question in final exam.
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The correct answer is (c) (u1 – u2)^2 = 2x(f1 – f2)
For explanation I would say: Let A be the position of the express train and B, that of the goods train when each is seen from the other, where AB = x.
Evidently, to avoid a collision the express train will apply the greatest possible retardation f1 and the goods train will apply the greatest possible acceleration f2.
Now, it is just possible to avoid a collision,if the velocities of the two trains at a point C on the line are equal, when they just touch each other at C(because after the instant the velocity of the express train will decrease due to retardation f1 and that of the goods train will increase due to acceleration f2).
Let the two trains reach the point C, at time t after each is seen by the other and their equal velocities at C bev and BC = s.
Then the equation of motion of the express train is,
v = u1 – f1t ……….(1) And x + s = u1t – (1/2)f1t^2 ……….(2)
And the equation of the motion of the goods train is,
v = u2 + f2t ……….(3) And s = u2t – (1/2)f2t^2 ……….(4)
From (1) and (3) we get,
u1 – f1t = u2 + f2t
Or (f1+ f2)t = u1 – u2
Or t = (u1 – u2)/(f1 + f2)
Again, (2)–(4) gives,
x = (u1 – u2)t – ½(t^2)(f1 + f2)
= t[(u1 – u2) – (1/2)(t)(f1 + f2)]
As, (f1 + f2)t = (u1 – u2)
So, x = (u1 – u2)/(f1 + f2)[(u1 – u2) – (1/2)(u1 – u2)]
So, (u1 – u2)^2 = 2x(f1 – f2)