Q146 / 180 · KCET 2022 · Physics
The displacement $x$ (in $\mathrm{m})$ of a particle of mass $m$ (in $\mathrm{kg}$ ) moving in one dimension under the action of a force, is related to time $t$ (in sec) by $t=\sqrt{x}+3$. The displacement of the particle when its velocity is zero, will be
- Correct answerAzero
- B$6 \mathrm{~m}$
- C$2 \mathrm{~m}$
- D$4 \mathrm{~m}$
Solution
Displacement of the particle related with time is given as
$$
t=\sqrt{x}+3 \Rightarrow \sqrt{x}=t-3
$$
Squaring on both sides, we get
$$
\begin{array}{ll}
\Rightarrow & x=(t-3)^2 \\
\Rightarrow & x=t^2-6 t+9
\end{array}
$$
Velocity of the particle,
$$
v=\frac{d x}{d t}=\frac{d}{d t}\left(t^2-6 t+9\right)=2 t-6
$$
when $v=0$
i.e $\quad 2 t-6=0 \Rightarrow t=36$
i.e velocity of particle is zero at $t=3 \mathrm{~s}$.
$\therefore$ Displacement of particle at $t=3 \mathrm{~s}$,
$$
\begin{gathered}
x=3^2-6 \times 3+9 \\
=0
\end{gathered}
$$
[from Eq. (i)i
Hence, the correct answer is option: 'A'.