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 54 m

Given, $x=9t^2-t^3$,
Speed $v=\frac{dx}{dt}=\frac{d}{dt}(9t^2-t^3)=18t-3t^2$.
For maximum speed $\frac{dv}{dt}=0\Rightarrow 18-6t=0$
$\therefore t=3\text{s}$.
$\therefore x_{max}=81-27=54\text{m}$
from $(x=9t^2-t^3)$
Hence, the correct answer is option (a).