Question

Two straight railway lines meet at right angles. A train starts from the junction along one line and at the same time instant, another train starts towards the junction from a station on the other line and they move at the same uniform velocity.When will they be nearest to each other?

(a) When they are equal distance from the junction

(b) When they are in unequal distance from the junction

(c) When they form a right angle at the junction

(d) Data not sufficient

I had been asked this question during an interview.

Enquiry is from Calculus Application in division Application of Calculus of Mathematics – Class 12

Answer 1

Right answer is (a) When they are equal distance from the junction

Explanation: Let OX and OY be two straight railway lines and they meet at O at right angles.

One train starts from the junction and moves with uniform velocity u km/hr along the line OY.

And at the same instant, another train starts towards the junction O from station A on the line OX with same uniform velocity u km/hr.

Let C and B be the position of the two trains on lines OY and OX respectively after t hours from the start.

Then OC = AB = ut km. Join BC and let OA = a km and BC = x km.

Then OB = a – ut.

Now, from the right angled triangle BOC we get,

BC^2 = OB^2 + OC^2

Or x^2 = (a – ut)^2 + (ut)^2

Thus, d(x^2)/dt = 2(a – ut)(-u) + u^2(2t)

And d^2(x^2)/dt^2 = 2u^2 + 2u^2 = 4u^2

For maximum or minimum value of x^2(i.e., x) we must have,

d(x^2)/dt = 0

Or 2(a – ut)(-u) + u^2(2t) = 0

Or 2ut = a   [Since u ≠ 0]

Or t = a/2u

Again at t = a/2u we have, d^2(x^2)/dt^2 = 4u^2 > 0

Therefore, x^2(i.e., x) is minimum at t = a/2u

Now when t= a/2u, then OC = ut = u(a/2u) = a/2 and OB =a – ut = a – u(a/2u) = a/2 that is at t = a/2u we have, OC = OB.

Therefore, the trains are nearest to each other when they are equally distant from the junction.