Question

Two trains $A$ and $B$ of length $400m$ each are moving on two parallel tracks with a uniform speed of $72kmh^{- 1}$ in the same direction, with $A$ ahead of $B$ . The driver of $B$ decides to overtake $A$ and accelerates by $1ms^{- 2}$ . 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

Length of each train,
$l_{A}=l_{B}=400m$
$u_{A}=72\times \frac{5}{1 8}ms^{- 1}$
$=20ms^{- 1}$
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)
$s_{B}=20\times 50+\frac{1}{2}\times 1\times \left(5 0\right)^{2}$
$=1000+1250$
$=2250m$
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 $a$
$=ut+\frac{1}{2}at^{2}$
$=\left(0 \times 50\right)+\frac{1}{2}\times 1\times \left(5 0\right)^{2}$
(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 $2 l$ from it where $l$ is length of train so final answer will be $450m$