Question

The position $x$ of a particle with respect to time $t$ along the $x$ -axis is given by $x=9t^2-t^3$ where $x$ is in metres and $t$ in second. What will be the position of this particle (in $m$ ) when it achieves maximum speed along the positive $x$ -direction?

Answer 1

Answer: 54

Let us first find speed.
Speed, $v=\frac{dx}{dt}=\frac{d}{dt}(9t^2-t^3)=18t-3t^2$
For maximum speed, using $\frac{dv}{dt}=0$
$\frac{dv}{dt}=0\Rightarrow 18-6t=0\Rightarrow t=3$ ,
Now finding displacement at that time, we have :
$\Rightarrow x_{max}=81-27=54m$