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

Answer: (A)

Given : $x=9 t^{2}-t^{3}\,\,\,$...(i)
Speed $v=\frac{d x}{d t}=\frac{d}{d t}\left(9 t^{2}-t^{3}\right)=18 t-3 t^{2}$
For maximum speed, $\frac{d v}{d t}=0$
$ \Rightarrow 18-6 t=0$
$\therefore t=3\, s $
$\therefore x_{\max }=81 \, m -27 \, m =54\, m$
(From $x=9 t^{2}-t^{3}$ ).