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(A) Given the position equation: $x = 9t^2 - t^3$. Velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(

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$54$

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(A) Given the position equation: $x = 9t^2 - t^3$. Velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(9t^2 - t^3) = 18t - 3t^2$. To find the time at which speed is maximum,we differentiate velocity with respect to time and set it to zero: $\frac{dv}{dt} = 18 - 6t = 0$. Solving for $t$,we get $t = 3 \ s$. Now,substitute $t = 3 \ s$ into the position equation to find the position at that instant: $x = 9(3)^2 - (3)^3 = 9(9) - 27 = 81 - 27 = 54 \ m$.

The position $x$ of a particle with respect to time $t$ along the $x$-axis is given by $x = 9t^2 - t^3$,where $x$ is in metres and $t$ is in seconds. What will be the position of this particle when it achieves maximum speed along the $x$-direction?

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