Question

A train is moving at a constant speed $V$ when its driver observes another train in front of him on the same track and moving in the same direction with a constant speed $v$. If the distance between the trains is $x$, then what should be the minimum retardation of the train so as to avoid collision?
A. $\dfrac{{{{\left( {V + v} \right)}^2}}}{x}$
B. $\dfrac{{{{\left( {V - v} \right)}^2}}}{x}$
C. $\dfrac{{{{\left( {V + v} \right)}^2}}}{{2x}}$
D. $\dfrac{{{{\left( {V - v} \right)}^2}}}{{2x}}$

Answer 1

Use the third kinematic equation. This equation gives the relation between the final velocity, initial velocity, acceleration and displacement. Calculate the relative velocity of the first train with respect to the second train and final relative velocity of the first train with respect to the second train. Using a third kinematic equation, calculate the retardation of the train.

Formula used:
The expression for third kinematic equation is
${v^2} = {u^2} + 2as$ …… (1)
Here, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $s$ is the displacement.

Complete step by step answer:
We have given that the speed of the first train on the track is $V$ and the speed of the second train moving on the same track is $v$.We have also given that the distance between the two trains is $x$. We have asked to calculate the retardation of the trains in order to avoid collision between the trains.

Let us first calculate the initial relative velocity of the first train with respect to the second train.The initial relative velocity of the first train with respect to the second train is given by
${u_{rel}} = V - v$
The final relative velocity of the first train with respect to the second train is zero because after travelling some distance the velocities of the two trains becomes the same so the distance between them remains the same and also the collision between the two trains is avoided.
${v_{rel}} = 0$

Let us calculate the retardation of the train using equation (1).Rewrite equation (1) for the relative final velocity between the two trains.
$v_{rel}^2 = u_{rel}^2 + 2ax$
Substitute $0$ for ${v_{rel}}$ and $V - v$ for ${u_{rel}}$ in the above equation.
$0 = {\left( {V - v} \right)^2} + 2ax$
$\therefore a = - \dfrac{{{{\left( {V - v} \right)}^2}}}{{2x}}$
Therefore, the retardation of the train is $\dfrac{{{{\left( {V - v} \right)}^2}}}{{2x}}$.The negative sign indicates the negative acceleration or retardation of the train.

Hence, the correct option is D.

Note: One can also solve the same question by another method. One can also calculate the initial relative velocity of the second train with respect to the first train and final relative velocity of the second train with respect to the first train to calculate the retardation of the train. The only difference one will find is that the acceleration will not have the negative sign we consider as the retardation.