If the displacement (s in metre) of a moving | vedclass

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(C) The displacement of the particle is given by $s = t^3 - 6t^2 + 18t + 9$.The velocity $v$ is the first derivative of displacement with respect to t

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$6$

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(C) The displacement of the particle is given by $s = t^3 - 6t^2 + 18t + 9$.The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = 3t^2 - 12t + 18$.To find the minimum velocity,we find the derivative of velocity with respect to time and set it to zero: $\frac{dv}{dt} = 6t - 12$.Setting $\frac{dv}{dt} = 0$,we get $6t - 12 = 0$,which implies $t = 2 \ s$.Now,substitute $t = 2 \ s$ into the velocity equation to find the minimum velocity:$v_{min} = 3(2)^2 - 12(2) + 18 = 3(4) - 24 + 18 = 12 - 24 + 18 = 6 \ m \ s^{-1}$.

If the displacement ($s$ in metre) of a moving particle in terms of time ($t$ in second) is $s = t^3 - 6t^2 + 18t + 9$,then the minimum velocity attained by the particle is (in $m \ s^{-1}$)

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