Q. A particle starts from rest and experiences constant acceleration for 6 s. if it travels a distance $d_{1}$ in the first two second, a distance $d_{2}$ in the next two seconds and a distance $d_{3}$ in the last two second, then
Explanation
(c) Let particle start from $O$ and travels distance
$d_{1}(OA),\ d_{2}(AB),\ d_{3}(BC)$
From equation of motion, we have
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = ut + \frac{1}{2}\ at^{2}$
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$For\ OA\ :\ \ \ \ t = 2\ s,\ u = 0\ $
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ d_{1} = \frac{1}{2}\ a(2)^{2} = 2a$
$For\ OB\ :\ \ t = 4\ s,\ u = 0$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ s_{2} = \frac{1}{2}\ a(4)^{2} = 8a$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ d_{2} = 8a - 2a = 6a$
$For\ OC\ :\ t = 6\ s,\ u = 0$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ S = \frac{1}{2}\ a(6)^{2} = 18a$
Distance in last $2\ s = 18a - 8a = 10a$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ d_{1}\ :d_{2}\ :d_{3} = 2a\ :6a\ :10a$
$\ \ \ \ \ \ \ \ \ \ \ \ \ d_{1}\ :d_{2}\ :d_{3} = 1\ :3\ :5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.