Question

Two trains each of length 256 m are moving in opposite directions on two parallel railway lines. The speed of one is twice that of the other. They take 4 sec to cross each other. Find their velocities.

Answer 1

Answer:

The velocity of "train 1" is 42.667 m/s and "train 2" is 85.334 m/s.

Explanation:

Let us consider the two trains as “train 1” and “train 2”. Therefore, we get

Length of "train 1" is given as, L₁ = 256 m

Length of "train 2" is given as, L₂ =256 m

Time taken by them to cross each other, t = 4 sec

Velocity of "train 1" as “V₁” and velocity of "train 2" as “V₂”. Since we are given that the speed of the one train is twice than that of the other.

So, V₂ = 2 V₁.

Since the two train are moving in opposite directions on two parallel lines,

∴ Relative Distance = (256 + 256) m = 512 m

∴ Relative Velocity, V₁₂ = (512 / 4) m/s = 128 m/s

Also, we can deduce from the fact that the trains are moving in opposite directions, that their relative velocity is the addition of the two velocities V₁and V₂.

So, V₁₂ = V₁ + V₂

Or, 128 = V₁ + 2 V₁…… [substituting V₂ = 2 V₁]

Or, 3 V₁ = 128

Or, V₁ = 128 / 3 = 42.667 m/s

∴ V₂ = 2 V₁ = 2 * 42.67 = 85.334 m/s

Answer 2

Let the speed of train 1 is = x m/s

that of train2 is y m/s

it is clear from the question that x = 2y

(assuming that train1 moves fast)

Both of them moving in opposite direction.

$l_1= 256\:m\\\\l_2= 256 \:m$

$\boxed{Time taken= \frac{l_1 + l_2}{x + y}} \\ \\ 4= \frac{(256 + 256) }{2y + y} \\ \\ 3y = \frac{512}{4} \\ \\ y = \frac{128}{3} \\ \\ y=42.66 \: m \: per sec$
So speed of 1 st train is

$x = 2 \times 42.66 \\ \\ = 85.32 \: m \: \: per \: \: second$
Hope this short trick helps you.