Q100 / 180 · NEET 2025 · Physics

In some appropriate units, time $(t)$ and position ( x ) relation of a moving particle is given by $t=x^2+x$. The acceleration of the particle is

  • A
    $-\frac{2}{(x+2)^3}$
  • Correct answer
    B
    $-\frac{2}{(2 x+1)^3}$
  • C
    $+\frac{2}{(x+1)^3}$
  • D
    $+\frac{2}{2 x+1}$

Solution

$\begin{aligned} & t=x^2+x \\ & \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{x}^2+\mathrm{x}\right] \\ & \frac{1}{\mathrm{v}}=2 \mathrm{x}+1\end{aligned}$ $v=(2 x+1)^{-1}...(i)$ $\begin{aligned} a & =\frac{v d v}{d x}=(2 x+1)^{-1}\left[-1(2 x+1)^{-2} \times 2\right] \\ a & =-1(2 x+1)^{-3} \times 2 \\ a & =-2(2 x+1)^{-3} \\ & =-\frac{2}{(2 x+1)^3}\end{aligned}$

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

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