The position $x$ of a particle with respect to time $t$ along $x$-axis is given by $x=9 t^2$ $-t^3$ where…
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 second. What will be the position of this particle when it achieves maximum speed along the $+x$ direction?
$54 \mathrm{~m}$
$81 \mathrm{~m}$
$24 \mathrm{~m}$
$32 \mathrm{~m}$.
Solution
It is given that
$\begin{aligned}
x & =9 t^2-t^3 \\
v & =\frac{d x}{d t}=18 t-3 t^2 \\
\frac{d v}{d t} & =a=18-6 t
\end{aligned}$
for maximum speed $\frac{d v}{d t}=0$ and $\frac{d^2 v}{d t^2}$ negative
So $18-6 t=0$
$\Rightarrow \quad t=3 s$
at $\quad t=3 s, x=9(3)^2-(3)^3$
$=81-27=54 \mathrm{~m}$