Displacement s of a particle at time t is exp | vedclass

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(C) Given displacement $s = 2t^3 - 9t$.Velocity $v = \frac{ds}{dt} = \frac{d}{dt}(2t^3 - 9t) = 6t^2 - 9$.The velocity vanishes when $v = 0$,so $6t^2 -

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$6 \sqrt{6}$

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(C) Given displacement $s = 2t^3 - 9t$.Velocity $v = \frac{ds}{dt} = \frac{d}{dt}(2t^3 - 9t) = 6t^2 - 9$.The velocity vanishes when $v = 0$,so $6t^2 - 9 = 0 \Rightarrow t^2 = \frac{9}{6} = \frac{3}{2} \Rightarrow t = \sqrt{\frac{3}{2}}$.Acceleration $a = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 9) = 12t$.Substituting $t = \sqrt{\frac{3}{2}}$ into the acceleration equation:$a = 12 	imes \sqrt{\frac{3}{2}} = 12 	imes \frac{\sqrt{3}}{\sqrt{2}} = 6 	imes \sqrt{2} 	imes \sqrt{3} = 6\sqrt{6}$.

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.

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