Q103 / 200 · NEET 2007 · Physics

The position $x$ of a particle with respect to time $t$ along $x$-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 when it achieves maximum speed along the $+x$ direction?

  • Correct answer
    A
    $54 \mathrm{~m}$
  • B
    $81 \mathrm{~m}$
  • C
    $24 \mathrm{~m}$
  • D
    $32 \mathrm{~m}$.

Solution

It is given that $\begin{aligned} x & =9 t^2-t^3 \\ v & =\frac{d x}{d t}=18 t-3 t^2 \\ \frac{d v}{d t} & =a=18-6 t \end{aligned}$ for maximum speed $\frac{d v}{d t}=0$ and $\frac{d^2 v}{d t^2}$ negative So $18-6 t=0$ $\Rightarrow \quad t=3 s$ at $\quad t=3 s, x=9(3)^2-(3)^3$ $=81-27=54 \mathrm{~m}$

Hence, the correct answer is option: 'A'.

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