KCET · Physics · Motion In One Dimension
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
- A zero
- B \(6 \mathrm{~m}\)
- C \(2 \mathrm{~m}\)
- D \(4 \mathrm{~m}\)
Answer & Solution
Correct Answer
(A) zero
Step-by-step Solution
Detailed explanation
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
\(\Rightarrow x=(t-3)^2\)
\(\Rightarrow x=t^2-6 t+9\)
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 \(t=3 \mathrm{~s}\),
\(\begin{gathered}x=3^2-6 \times 3+9 =0\end{gathered}\quad\)[from Eq. (i)i
\(t=\sqrt{x}+3 \Rightarrow \sqrt{x}=t-3\)
Squaring on both sides, we get
\(\Rightarrow x=(t-3)^2\)
\(\Rightarrow x=t^2-6 t+9\)
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 \(t=3 \mathrm{~s}\),
\(\begin{gathered}x=3^2-6 \times 3+9 =0\end{gathered}\quad\)[from Eq. (i)i
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