Physics - Kinematics Question with Solution | TestHub

Length of each train,

$l_{\text{A}}=l_{\text{B}}=40\mathrm{0 }\text{m}$

$u{}_A=72\times \frac{5}{18}\text{m/s}$

$\text{= 20 m/s}$

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

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

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

${\text{s}}_{\text{A}}=20\times 50$

$\text{= 1000 m}$

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

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

(As motion of train B is an accelerated motion)

${\text{s}}_{\text{B}}=20\times 50+\frac{1}{2}\times 1\times {\left(50\right)}^2$

$\text{= 1000 + 1250}$

$\text{= 2250 m}$

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$\text{50 s}$.

$\therefore$Original distance between the trains

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

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

$=0\times 50+\frac{1}{2}\times 1\times {\left(50\right)}^2$

(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$\text{450 m}$