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The position $x$ of a particle with respect to time $t$ along $x-$axis is given by $x = 9{t^2} - {t^3}$ where $x$ is in metres and $t$ in seconds. What will be the position of this particle when it achieves maximum speed along the $x$ direction ?..........$m$
$54$
$81$
$24$
$32$
Solution
$\begin{array}{l}
Give:\,x = 9{t^2} – {t^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( i \right)\\
Speed\,v = \frac{{dx}}{{dt}} = \frac{d}{{dt}}(9{t^2} – {t^3}) = 18t – 3{t^2}.\\
For\,{\rm{maximum}}\,speed,\,\frac{{dv}}{{dt}} = 0\,\, \Rightarrow \,18 – 6t = 0\\
\therefore \,\,t = 3\,\,\,\,s.\\
\therefore {x_{\max }} = 81\,\,m – 27\,\,m = 54\,m.\,\left( {From\,x\, = \,9{t^2} – {t^3}} \right)
\end{array}$
