Motion In One Dimension Question 1
Question 1 - 2024 (01 Feb Shift 1)
A particle is moving in one dimension (along $\mathrm{x}$ axis) under the action of a variable force. It’s initial position was $16 \mathrm{~m}$ right of origin. The variation of its position $(\mathrm{x})$ with time $(\mathrm{t})$ is given as $x=-3 t^{3}+18 t^{2}+16 t$, where $x$ is in $m$ and $t$ is in $s$. The velocity of the particle when its acceleration becomes zero is $\mathrm{m} / \mathrm{s}$.
Show Answer
Answer: (52)
Solution:
$x=3 t^{3}+18 t^{2}+16 t$
$\mathbf{v}=-9 t^{2}+36+16$
$a=-18 t+36$
$a=0$ at $t=2 s$
$v=-9(2)^{2}+36 \times 2+16$
$v=52 \mathrm{~m} / \mathrm{s}$
