Q. A passenger train is running on a railways track with a speed $v_{1}$. While driving, the driver discovers that another goods train is travelling in front of the passenger train with a speed $v_{2}\left( v_{1} > v_{2} \right).$ Its retardation after applying brakes is $a$. The least distance the passenger train must travel to avoid collision with goods train is
A) $\frac{v_{1}^{2} - v_{2}^{2}}{2a}$
B) $\frac{v_{2}^{\ } - v_{1}^{\ }}{a}$
C) $\frac{v_{2}^{\ } + v_{1}^{\ }}{2a}$
D) $\frac{v_{2}^{2} + v_{1}^{2}}{2a}$
Explanation
(a) Here, initial velocity of passenger train $u = v_{1};$ final velocity $v = v_{2},a = - a,$ distance $s = ?$
As $v^{2} = u^{2} + 2as,$ so$\ \ v_{2}^{2} = v_{1}^{2} + 2( - a)s$
or $s = (v_{1}^{2} - v_{2}^{2})/2a$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.