Motion In One Dimension Question 8
Question 8 - 2024 (29 Jan Shift 2)
A particle is moving in a straight line. The variation of position ’ $x$ ’ as a function of time ’ $t$ ’ is given as $x=\left(t^{3}-6 t^{2}+20 t+15\right) \mathrm{m}$. The velocity of the body when its acceleration becomes zero is :
(1) $4 \mathrm{~m} / \mathrm{s}$
(2) $8 \mathrm{~m} / \mathrm{s}$
(3) $10 \mathrm{~m} / \mathrm{s}$
(4) $6 \mathrm{~m} / \mathrm{s}$
Show Answer
Answer: (2)
Solution:
$x=t^{3}-6 t^{2}+20 t+15$
$\frac{d x}{d t}=v=3 t^{2}-12 t+20$
$\frac{d v}{d t}=a=6 t-12$
When $a=0$
$6 t-12=0 ; t=2 \mathrm{sec}$
At $t=2 \mathrm{sec}$
$v=3(2)^{2}-12(2)+20$
$v=8 \mathrm{~m} / \mathrm{s}$
