Two trains A and B of length 400 m each are moving on two parallel...

Length of each train,

${l }_{A} = {l }_{B} = 4 0 0 \text{m}$

$u{}_{A } = 72 \times \dfrac{5}{18}\text{m/s}$

$\text{= 20 m/s}$

Distance travelled by train $\text{A}$ in $\text{50 s}$

${s}_{A} = {u}_{A} \times \text{t}$

(As for unaccelerated motion, distance = Speed x Time)

${s}_{A} = 2 0 \times 5 0$

$\text{= 1000 m}$

Distance travelled by train $\text{B}$ in $\text{50 s}$,

${s}_{B} = {u}_{B} \text{t} + \dfrac{1}{2} {a}_{B} t^{2}$

(As motion of train B is an accelerated motion)

${s}_{B} = 2 0 \times 5 0 + \dfrac{1}{2} \times 1 \times {\left(5 0\right)}^{2}$

$\text{= 1000 + 1250}$

$\text{= 2250 m}$

relative distance between the two trains ${\text{=S}}_{B} - {S}_{A}$

$\text{= 2250 - 1000}$

$\text{= 1250 m}$
we should subtract~$2l$ from it where $l$~is length of train so final answer will be 450 m

Alternate Method

Initially, both the trains are moving with same speed, the train $\text{B}$ first comes that of the train $\text{A}$ in $\text{50 s}$ with an acceleration of $\text{50 s}$.

$∴$ Original distance between the trains

= Distance covered by train $\text{B}$ with an acceleration a

$= \text{ut} + \dfrac{1}{2} {at}^{2}$

$= 0 \times 5 0 + \dfrac{1}{2} \times 1 \times {\left(5 0\right)}^{2}$

(As both train starts with same velocity, so velocity, so velocity of train when it starts accelerating can be taken as zero).

$\text{= 1250 m}$ .

we should subtract $2l$ from it where $l$ is length of train so final answer will be $\text{450 m}$