The position of a particle moving along the y | vedclass

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(C) The position of the particle is given by the function $y(t) = 3t^2 - t^3$.To find the time at which the particle attains the maximum positive posi

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The position of the particle is given by the function $y(t) = 3t^2 - t^3$.To find the time at which the particle attains the maximum positive position,we calculate the first derivative of $y$ with respect to $t$ and set it to zero:$\frac{dy}{dt} = 6t - 3t^2$.Setting $\frac{dy}{dt} = 0$ gives $3t(2 - t) = 0$,which results in $t = 0$ or $t = 2$.To confirm the maximum,we find the second derivative: $\frac{d^2y}{dt^2} = 6 - 6t$.At $t = 2$,$\frac{d^2y}{dt^2} = 6 - 6(2) = -6$. Since the second derivative is negative,the function attains a local maximum at $t = 2 \ s$.

The position of a particle moving along the $y-$ axis is given as $y = 3t^2 - t^3$,where $y$ is in $m$ and $t$ is in $s$. The time when the particle attains the maximum positive position will be ........ $s$.

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