# Two trains face each other in adjacent tracks. They are initially are at rest 40 m apart. The...

## Question:

Two trains face each other in adjacent tracks. They are initially are at rest 40 m apart. The train on the left accelerates at 1.1 m/s{eq}^{2} {/eq}. The train on the right accelerates leftward at 2.2 m/s{eq}^{2} {/eq}.

how far does the train on the left travel just before the two trains pass?

## Equation of Motion:

A body with initial velocity u is moving in a direction with uniform acceleration a covers a distance s in some time t . Then the equation which determines the distance s traveled by the body in time t is given by

{eq}s=ut+\frac{1}{2}at^2 {/eq}

• u is the initial velocity of the body
• a is the acceleration
• s is the distance covered by the body

Given:

• The distance between the two trains is {eq}d= 40 \ m {/eq}.
• The acceleration of the first train is {eq}a_1 =1.1 \ m/s^2 {/eq}.
• The acceleration of the second train is {eq}a_2 =2.2 \ m/s^2 {/eq}.
• The trains start from rest ie {eq}u_1 =u_2 = 0. {/eq}

Let the time taken by both the trains be t before they pass each other.

In time t distance covered by the first train is

{eq}\begin{align*} s_1& = u_1t +\frac{1}{2}a_1t^2 \\ & = 0 +\frac{1}{2} \times 1.1 \times t^2 \\ & = 0.55t^2 \end{align*} {/eq}

In time t distance covered by the second train is

{eq}\begin{align*} s_2& = u_2t +\frac{1}{2}a_2t^2 \\ & = 0 +\frac{1}{2} \times 2.2 \times t^2 \\ & = 1.1t^2 \end{align*} {/eq}

Now, according to the problem we have

{eq}\begin{align*} s_1 +s_2 & = d\\ 0.55t^2 + 1.1t^2 & = 40 \\ t & = 4.92 \ s \end{align*} {/eq}

Thus distance covered by the train on the left in 4.92 s is given by

{eq}\begin{align*} s_1& = 0.55t^2 \\ & = 0.55 \times 4.92^2 \ m \\ & = 13.3 \ m\Rightarrow (Answer) \end{align*} {/eq}