Question

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 seconds. What will be the position of this particle when it achieves maximum speed along the $+x$ direction?

• A. 54 m
• B. 81 m
• C. 24 m
• D. 32 m

Answer 1

The Correct Option is A

$x =9 \,t ^{2}- t ^{3} \,\,\,\,\,\therefore v =18 t -3\, t ^{2}$
$\Rightarrow \frac{ d v}{ dt }=18-6\, t$
for maximum speed $\frac{ d v }{ dt }=0$, and $\frac{ d ^{2} v }{ dt ^{2}}$ negative
so $18-6 t =0 $
$\Rightarrow t =3 s$
at $t =3 s , x =9(3)^{2}-(3)^{3} $
$=81-27=54\, m$