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 second. What will be the position of this particle when it achieves maximum speed along the positive $x$-direction?

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

Answer 1

The correct option is C 54 m

We have,

$\begin{array}{c}x=9t^2-t^3\\ v=\frac{dx}{dt}=18t-3t^2\end{array}$

As $v$ is max,

So,

$\begin{array}{c}\frac{dv}{dt}=0\\ \frac{dv}{dt}=18-6t=0\\ t=3\\ xatt=3\sec \\ =t^2\left(9-t\right)\\ =9\left(9-3\right)\\ 54m\end{array}$

So,

Option $C$ is correct answer.