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NEET 2025

Exam: NEET
Year: 2025
Subject: Physics
Chapter: Motion in One Dimension

NEET 2025 · Physics · Motion in One Dimension

In some appropriate units, time (t) and position ( x ) relation of...

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

Options

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

Correct answer

B. $-\frac{2}{(2 x+1)^3}$

Step-by-step 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}

$

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