Q. The engine of a train can impart a maximum acceleration of $1ms^{- 2}$and the brakes can give a maximum retardation of $3ms^{- 2}$. The least time during which a train can go from one place to the other place at a distance of 1.2 km is nearly
A) 108 s
B) 191 s
C) 56.6 s
D) Time is fixed
Explanation
(c) $1 = \frac{v}{t_{1}},3 = \frac{v}{t_{2}},1200 = \frac{1}{2}(t_{1} + t_{2})v$,
$1200 = \frac{1}{2}\left( v + \frac{v}{3} \right)v = \frac{1}{2}\frac{4v^{2}}{3} = \frac{2v^{2}}{3}$
or $v^{2} = 1800,$
$\therefore\ 1200 = \frac{1}{2}t \times \sqrt{1800}$
$t = \frac{2400}{\sqrt{1800}}$s =$\frac{2400}{42.43}$s =56.6 s
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.