Question

A particle starts from rest and experiences constant acceleration for $6 \,s$. If it travels a distance $d_1$ in the first two seconds, a distance $ d_{2} $ in the next two seconds and a distance $ d_{3}$ in the last two seconds, then

• A. $d_{1}: d_{2}: d_{3}=1: 1: 1$
• B. $d_{1}: d_{2}: d_{3}=1: 2: 3$
• C. $d_{1}: d_{2}: d_{3}=1: 3: 5$
• D. $d_{1}: d_{2}: d_{3}=1: 5: 9$

Answer 1

Answer: (C)

Let pardcle start from 0 and travels distance
$d_{1}(O A), d_{2}(A B), d_{3}(B C)$. .
From equation of motion, we have
$S=u t+\frac{1}{2} g t^{2}$
where $u$ is initial velocity, $t$ - time and $g$ acceleration due to gravity.
For $O A: t=2 s, \,\,u=0$
$d_{1}=\frac{1}{2} a(2)^{2}=2 a$
For $O B: r=4 s, \,\,u=0$
$\therefore d_{2}=\frac{1}{2} a(4)^{2}=8 a$
For $O C: u=0, t=6 s$
$\therefore S=\frac{1}{2} a(6)^{2}=18 a$
Distance in last $2 s=18 a-8 a=10 a$
$\therefore d_{1}: d_{2}: d_{3}=2 a: 6 a: 10 a$
$d_{1}: d_{2}: d_{3}=1: 3: 5$