AP EAMCET · Maths · Application of Derivatives
Displacement \(s\) of a particle at time \(t\) is expressed as \(s=2 t^3-9 t\). Find the acceleration at the time when the velocity vanishes.
- A 6
- B \(6 \sqrt{3}\)
- C \(6 \sqrt{6}\)
- D \(3 \sqrt{6}\)
Answer & Solution
Correct Answer
(C) \(6 \sqrt{6}\)
Step-by-step Solution
Detailed explanation
\(s=2 t^3-9 t\) Velocity \(=\frac{d s}{d t}=6 t^2-9=0 \Rightarrow t=\frac{3}{\sqrt{6}}\) Acceleration \(=\frac{d^2 s}{d t^2}=\left.12 t \Rightarrow \frac{d^2 s}{d t^2}\right|_{\left(t=\frac{3}{\sqrt{6}}\right)}=6 \sqrt{6}\)
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