Q. A man runs at a speed of $4ms^{- 1}$to overtake a standing bus. When he is 6 m behind the door at $t = 0$, the bus moves forward and continues with a constant acceleration of $1.2ms^{- 2}$. The man reaches the door in time $t$. Then,
A) $4t = 6 + 0.6t^{2}$
B) $1.2t^{2} = 4t$
C) $4t^{2} = 1.2t$
D) $6 + 4t = 0.2t^{2}$
Explanation
(a) As the clear from figure
[image:3_image189.emf]
$4t = 6 + \frac{1}{2} \times 1.2 \times t^{2}$
or $4t = 6 + 0.6t^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.