MHT CET · Maths · Differentiation
A particle moves along a straight line according to the law \(s=16-2 t+3 t^{3}\), where \(s\) metres is the distance of the particle from a fixed point at the end of \(t\) second. The acceleration of the particle at the end of \(2 \mathrm{~s}\) is
- A \(3.6 \mathrm{~m} / \mathrm{s}^{2}\)
- B \(36 \mathrm{~m} / \mathrm{s}^{2}\)
- C \(36 \mathrm{~km} / \mathrm{s}^{2}\)
- D \(360 \mathrm{~m} / \mathrm{s}^{2}\)
Answer & Solution
Correct Answer
(B) \(36 \mathrm{~m} / \mathrm{s}^{2}\)
Step-by-step Solution
Detailed explanation
Given, \(s=16-2 t+3 t^{3}\)
\(
\begin{array}{lc}
\Rightarrow \frac{d s}{d t}=-2+9 t^{2} \\
\Rightarrow \frac{d^{2} s}{d t^{2}}=18 t
\end{array}
\)
Now, the acceleration of the particle at the end of \(t=2 \mathrm{~s}\) is
\(f=\frac{d^{2} s}{d t^{2}}=18 \times 2\)
\(=36 \mathrm{~m} / \mathrm{s}^{2}\)
\(
\begin{array}{lc}
\Rightarrow \frac{d s}{d t}=-2+9 t^{2} \\
\Rightarrow \frac{d^{2} s}{d t^{2}}=18 t
\end{array}
\)
Now, the acceleration of the particle at the end of \(t=2 \mathrm{~s}\) is
\(f=\frac{d^{2} s}{d t^{2}}=18 \times 2\)
\(=36 \mathrm{~m} / \mathrm{s}^{2}\)
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