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

Question

Dr. Abubacker · Accepted Answer

(b) $x = 9t^{2} - t^{3};v = \frac{dx}{dt} = 18t - 3t^{2}$, For maximum speed
$\frac{dv}{dt} = \frac{d}{dt}\left\lbrack 18t - 3t^{2} ightbrack = 0 \Rightarrow 18 - 6t = 0\ 	herefore t = 3\ sec$
i.e., Particle achieve maximum speed at $t = 3\ \sec$. At this instant position of this particle, $x = 9t^{2} - t^{3}$
$= 9(3)^{2} - (3)^{3} = 81 - 27 = 54\ m$

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

Explanation

Reviewed by Dr. Abubacker

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