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 $A$ and the original distance between them is $x$ , then calculate value of $\frac{x}{10}$ ?

Answer 1

Answer: 45

Length of each train,
$l_A=l_B=400m$
$u_A=72\times \frac{5}{18}m{s}^{-1}$
$=20m{s}^{-1}$
Distance travelled by train $A$ in $50s$
$s_A=u_A\times t$
(As for uniform motion, distance = Speed x Time)
$s_A=20\times 50$
$=1000m$
Distance travelled by train $B$ in $50s$ ,
$s_B=u_{B}t+\frac{1}{2}{a}_B{t}^2$
(As motion of train $B$ is an accelerated motion)
$s_B=20\times 50+\frac{1}{2}\times 1\times {\left(50\right)}^2$
$=1000+1250$
$=2250m$
relative distance between the two trains $=S_B-S_A$
$=2250-1000$
$\text{= 1250 m}$
We should subtract $2l$ from it where $l$ is length of train.
Original distance between them is $1250-\left(2\times 400\right)=450m$ .
The value of $\frac{450}{10}=45$ .
Alternate Method
Initially, both the trains are moving with same speed, the train $\text{B}$ first comes that of the train $A$ in $50s$ with an acceleration of $50s$ .
$\therefore$ Original distance between the trains
$=$ Distance covered by train $B$ with an acceleration $a$
$=ut+\frac{1}{2}a{t}^2$
$=\left(0\times 50\right)+\frac{1}{2}\times 1\times {\left(50\right)}^2$
(As both trains start with the same velocity, so velocity, so velocity of train when it starts accelerating can be taken as zero).
$=1250m$ .
we should subtract $2l$ from it where $l$ is length of train so final answer will be $450m$