Physics · Motion in one Dimension

NEET (UG) 2025 — Question 14

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

Answer: B

$\mathrm{t}=\mathrm{x}^{2}+\mathrm{x}$

$\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{x}^{2}+\mathrm{x}\right]$

$\frac{1}{\mathrm{v}}=2 \mathrm{x}+1$

\mathrm{v}=(2 \mathrm{x}+1)^{-1} …(i)

$\mathrm{a}=\frac{v \mathrm{~d} v}{\mathrm{dx}}=(2 \mathrm{x}+1)^{-1}\left[-1(2 \mathrm{x}+1)^{-2} \times 2\right]$

$a=-1(2 x+1)^{-3} \times 2$

$a=-2(2 x+1)^{-3}$

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

Answer key and solution verified before publishing.

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