Question

Two trains $A$ and $B$ of length $400m$ each are moving on two parallel tracks with a uniform speed of $72km/h$ 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: 45.00

Length of each train,
$l_A=l_B=400m$
$u_A=72\times \frac{5}{18}m/s$
$=20m/s$
Distance travelled by train $A$ in $50s$
$S_A=u_A\times t$
[As for unaccelerated motion, distance $=$ speed $\times$ 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 a accelerated motion)
$S_B=20\times 50+\frac{1}{2}\times 1\times (50{)}^2$
$=1000+1250$
$=2250m$
We should subtract $2l$ from it where $l$ is length of train so final answer will be $450m$
Alternate method
Initially, both the trains are moving with same speed, the train $B$ first comes that of the train $A$ is $50s$ with an acceleration of $50s$.
$\therefore$ Original distance between the trains
$=$ Distance covered by train $B$ with an acceleration
$a=ut+\frac{1}{2}a{t}^2$
$=0\times 50+\frac{1}{2}\times 1\times (50{)}^2$
(As both train starts with 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 $450m$.