MHT CET · Maths · Differentiation
The displacement ' \(\mathrm{S}\) ' of a moving particle at a time \(t\) is given by \(S=5+\frac{48}{t}+t^3\). Then its acceleration when the velocity is zero, is
- A \(12\)
- B \(20\)
- C \(16\)
- D \(24\)
Answer & Solution
Correct Answer
(D) \(24\)
Step-by-step Solution
Detailed explanation
Given, \(S=5+\frac{48}{t}+t^3\)
Velocity \((V)=\frac{d S}{d t}=0-\frac{48}{t^2}+3 t^2\)
\(\begin{aligned}
\therefore \quad & V=\frac{-48}{t^2}+3 t^2 ...(i)\\
& \text { But } V=0 ...[Given]\\
& \Rightarrow \frac{-48}{t^2}+3 t^2=0 \\
& \Rightarrow t=2
\end{aligned}\)
Now, \(\mathrm{A}=\frac{\mathrm{dV}}{\mathrm{dt}}\)
\(\begin{aligned}
& =\frac{d}{d t}\left(\frac{-48}{t^2}+3 t^2\right) \\
& =\frac{96}{t^3}+6 t
\end{aligned}\)
Acceleration at \(\mathrm{t}=2\) is \(\frac{96}{8}+12=24\)
Velocity \((V)=\frac{d S}{d t}=0-\frac{48}{t^2}+3 t^2\)
\(\begin{aligned}
\therefore \quad & V=\frac{-48}{t^2}+3 t^2 ...(i)\\
& \text { But } V=0 ...[Given]\\
& \Rightarrow \frac{-48}{t^2}+3 t^2=0 \\
& \Rightarrow t=2
\end{aligned}\)
Now, \(\mathrm{A}=\frac{\mathrm{dV}}{\mathrm{dt}}\)
\(\begin{aligned}
& =\frac{d}{d t}\left(\frac{-48}{t^2}+3 t^2\right) \\
& =\frac{96}{t^3}+6 t
\end{aligned}\)
Acceleration at \(\mathrm{t}=2\) is \(\frac{96}{8}+12=24\)
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