The displacement s of a particle,in meters,at | vedclass

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(B) Given,the displacement of a particle at time $t$ is $s = \frac{t^3}{3} - 6t$. Velocity $v$ is the rate of change of displacement with respect to t

Vedclass · Accepted Answer

$2\sqrt{6} 	ext{ m/s}^2$

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(B) Given,the displacement of a particle at time $t$ is $s = \frac{t^3}{3} - 6t$. Velocity $v$ is the rate of change of displacement with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(\frac{t^3}{3} - 6t) = t^2 - 6$. When the velocity vanishes,$v = 0$,so $t^2 - 6 = 0$,which gives $t = \sqrt{6} 	ext{ s}$ (considering $t > 0$). Acceleration $a$ is the rate of change of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(t^2 - 6) = 2t$. Substituting $t = \sqrt{6} 	ext{ s}$ into the acceleration equation,we get $a = 2(\sqrt{6}) = 2\sqrt{6} 	ext{ m/s}^2$.

The displacement $s$ of a particle,in meters,at any time $t$ in seconds is expressed as $s = \frac{t^3}{3} - 6t$. Find the acceleration at the time when the velocity vanishes.

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