Q. Two trains one of $100\ m$ and another of length 125m, are moving in mutually opposite directions along parallel lines, meet each other, each with speed 10m/s. If their acceleration are $0.3m/s^{2}$and $0.2m/s^{2}$respectively, then the time taken to pass each other will be
A) $5\ s$
B) $10\ s$
C) $15\ s$
D) $20\ s$
Explanation
(b) Relative velocity of one train $w.r.t.$ other
$= 10 + 10 = 20m/s$
Relative acceleration=$0.3 + 0.2 = 0.5m/s^{2}$
If train crosses each other then from $s = ut + \frac{1}{2}at^{2}$
$As,\ s = s_{1} + s_{2} = 100 + 125 = 225$
$\Rightarrow 225 = 20t + \frac{1}{2} \times 0.5 \times 0.5 \times t^{2} \Rightarrow 0.5t^{2} + 40t - 450 = 0$
$\Rightarrow t = \frac{- 40 \pm \sqrt{1600 + 4.(005) \times 450}}{1} = - 40 \pm 50$
$\therefore t = 10sec$ (Taking +ve value)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.