Question

Two trains one of length $100\text{m}$ and another of length $125\text{m}$, are moving in mutually opposite directions along parallel lines, towards each other, each having initial speed $10\text{m/s}$. If their acceleration are $0.3{\text{m/s}}^2$ and $0.2{\text{m/s}}^2$ respectively, if there was no initial separation between them then the time they take to pass each other will be

• A. $10\text{s}$
• B. $20\text{s}$
• C. $5\text{s}$
• D. $15\text{s}$

Answer 1

The correct option is A

$10\text{s}$

Given: Length of train $A,s_1=100\text{m}$
Acceleration of train $a_A=0.3{\text{m/s}}^2$

Length of train $B,s_2=125\text{m}$
Acceleration of train $a_B=0.2{\text{m/s}}^2$

Relative initial velocity of train $Aw.r.t.$ train $B$,
$u_{AB}=u_A-u_B=10-(-10)$
$u_{AB}=20\text{m/s}$

As per the question
$a_A=0.3{\text{m/s}}^2{a}_B=0.2{\text{m/s}}^2$

Therefore,
Relative acceleration
$a_{AB}=0.3{\text{m/s}}^2-(-0.2{\text{m/s}}^2$)
$a_{AB}=0.5{\text{m/s}}^2$

As we know, from $2^{nd}$ equation of motion we get,
$s=ut+\frac{1}{2}a{t}^2\dots \left(i\right)$
As given,
$s=s_1+s_2=100\text{m}+125\text{m}=225\text{m}$

By putting the values in equation $\left(i\right)$ we get,
$225=20t+\frac{1}{2}\times 0.5\times t^2$
$0.5t^2+40t-450=0$
$t=\frac{-40\pm \sqrt{1600+4\times \left(0.5\right)\times 450}}{1}$
$=-40\pm 50$

Therefore,
$t=10\text{s}\text{(Taking}+ve\text{value}).$

Hence, option $\left(\text{B}\right)$ is the right choice.