Q107 / 160 · AP EAMCET 2024 (22 May Shift 1) · Mathematics
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.
- A6
- B$6 \sqrt{3}$
- Correct answerC$6 \sqrt{6}$
- D$3 \sqrt{6}$
Solution
$s=2 t^3-9 t$
Velocity $=\frac{d s}{d t}=6 t^2-9=0 \Rightarrow t=\frac{3}{\sqrt{6}}$ Acceleration $=\frac{d^2 s}{d t^2}=\left.12 t \Rightarrow \frac{d^2 s}{d t^2}\right|_{\left(t=\frac{3}{\sqrt{6}}\right)}=6 \sqrt{6}$
Velocity $=\frac{d s}{d t}=6 t^2-9=0 \Rightarrow t=\frac{3}{\sqrt{6}}$ Acceleration $=\frac{d^2 s}{d t^2}=\left.12 t \Rightarrow \frac{d^2 s}{d t^2}\right|_{\left(t=\frac{3}{\sqrt{6}}\right)}=6 \sqrt{6}$
Hence, the correct answer is option: 'C'.