Question

Two trains $A$ and $B$ of length $400m$ each are moving on two parallel tracks with a uniform speed of $72km.{{h}^{-1}}$ in the same direction, with $A$ ahead of $B$. The driver of $B$ decides to overtake $A$ and accelerates by $1m.{{s}^{-2}}$. If after $50s$, the guard of $B$ just brushes past the driver of $A$, what was the original distance between them?
$A)\text{ 45}0m$
$B)\text{ }1150m$
$C)\text{ }1300m$
$D)\text{ }1250m$

Answer 1

Hint: This problem can be solved by using the concept of the relative velocity between the two trains and how it changes under acceleration. We will apply the equations of motion for uniform acceleration along with the relative velocity to get the required distance between the two trains.

Formula used:
${{v}_{AB}}={{v}_{B}}-{{v}_{A}}$
$s=ut+\dfrac{1}{2}a{{t}^{2}}$

Complete step by step answer:
To solve this problem easily, we will work with the relative velocity of train $B$ with respect to train $A$ and apply the equation of motion with uniform acceleration to get the distance between the two trains.
Now, since according to the given question, after the time period, the guard of train $B$ brushes past the driver of train $A$, hence, the total relative distance travelled by the guard will be the sum of the length of train $B$, the separation between the two trains and the length of train $A$. This is because the driver of a train sits in the front end while the guard sits at the back end.
Let the lengths of the two trains be $L=400m$.
Let the separation between the two trains be $x$.
Let the acceleration of the train $A$ be ${{a}_{A}}=0$ (Since, it is not accelerating).
Lett the acceleration of train $B$ be ${{a}_{B}}=1m.{{s}^{-2}}$
Now, the velocities of both trains are given to be equal and in the same direction.
Let the initial velocity of train $A$ be ${{u}_{A}}=72km.{{h}^{-1}}=72\times \dfrac{5}{18}=20m.{{s}^{-1}}$ $\left( \because 1km.{{h}^{-1}}=\dfrac{5}{18}m.{{s}^{-1}} \right)$
Let the initial velocity of train $B$ be ${{u}_{B}}=72km.{{h}^{-1}}=72\times \dfrac{5}{18}=20m.{{s}^{-1}}$ $\left( \because 1km.{{h}^{-1}}=\dfrac{5}{18}m.{{s}^{-1}} \right)$
The time period under consideration is $t=50s$.
Let the initial relative velocity of train $B$ with respect to train $A$ be ${{u}_{AB}}$.
Now, for two bodies $A,B$ with velocities ${{v}_{A,}}{{v}_{B}}$ in the same direction, the relative velocity ${{v}_{AB}}$ of $B$ with respect to $A$ is given by
${{v}_{AB}}={{v}_{B}}-{{v}_{A}}$ --(1)
Using (1), we get
${{u}_{AB}}={{u}_{B}}-{{u}_{A}}=0$ $\left( \because {{u}_{B}}={{u}_{A}} \right)$ --(2)
Now, we will work with this relative velocity since it is easier to work in the frame of reference of the guard on train $B$ who is moving along with the speed of train $B$.
As explained earlier, the total relative distance $s$ that he covers in the $50s$ is the sum of both the trains and the separation between the trains.
$\therefore s=L+x+L=2L+x$
Since, train $A$ is not accelerating the relative acceleration ${{a}_{AB}}$ of train $B$ with respect to train $A$ will be
${{a}_{AB}}={{a}_{B}}-{{a}_{A}}={{a}_{B}}-0={{a}_{B}}=1m{{s}^{-2}}$
Now, according to the equation of motion for constant acceleration $a$, the displacement $s$ of a body in time $t$ is given by
$s=ut+\dfrac{1}{2}a{{t}^{2}}$ --(3)
where $u$ is the initial velocity of the body.
Using (3), we get
$s={{u}_{AB}}t+\dfrac{1}{2}{{a}_{AB}}{{t}^{2}}=\left( 0 \right)\left( 50 \right)+\dfrac{1}{2}\left( 1 \right){{\left( 50 \right)}^{2}}$
$\therefore 2L+x=0+\left( \dfrac{1}{2}\times 2500 \right)$
$\therefore 2\left( 400 \right)+x=1250$
$\therefore 800+x=1250$
$\therefore x=1250-800=450m$


Therefore, the initial separation between the trains is $450m$.
Hence, the correct option is $A)\text{ 45}0m$.

Note: This question could also have been solved by considering the two trains separately and finding out the distance travelled by both of them individually in the time period and then subtracting them to get the separation between the two trains after making adjustments regarding the position of the guard and the driver and therefore, getting the initial separation between the two trains. However, this would have been a lengthier process requiring two sets of different calculations for the two trains.