Question

The position $x$ of a particle with respect to time $t$ along the $x$ -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 (in $m$ ) when it achieves maximum speed along the positive $x$ -direction?

Answer 1

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