Q. If a train travelling at $72\ kmph$ is to be brought to rest in a distance of $200\ $metres, then its retardation should be
A) $20\ ms^{- 2}$
B) $10\ ms^{- 2}$
C) $2\ ms^{- 2}$
D) $1ms^{- 2}$
Explanation
(d) $u = 72\ kmph = 20\ m/s,\ v = 0$
By using $v^{2} = u^{2} - 2as \Rightarrow a = \frac{u^{2}}{2s} = \frac{(20)^{2}}{2 \times 200} = 1\ m/s^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.