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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

Answer and Explanation:

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}


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