Q. A man is 45 m behind the bus, when the bus starts accelerating from rest with acceleration $2.5\ {ms}^{- 2}.$ With what minimum velocity should the man start running to catch the bus?
A) $12\ {ms}^{- 1}$
B) $14\ {ms}^{- 1}$
C) $15\ {ms}^{- 1}$
D) $16\ {ms}^{- 1}$
Explanation
(c) In order that the man catches the bus, let his minimum velocity be $v,$ then from equation of motion.
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$v^{2} = u^{2} + 2as$
We have, $u = 0,\ a = 2.5\ {ms}^{- 2},\ s = 45\ m$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ v^{2} = 2 \times 2.5 \times 45$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \sqrt{225} = 15\ {ms}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.