NEET 2025, Physics Q15: Motion In One Dimension

NEET2025Physics

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

Solution

Differentiate $t = x^{2} + x$ w.r.t. $x$ : $\frac{d t}{d x} = 2 x + 1$ , so $v = \frac{d x}{d t} = (2 x + 1)^{- 1}$ .

Acceleration: $a = v \frac{d v}{d x} = (2 x + 1)^{- 1} \cdot [- 1 \cdot (2 x + 1)^{- 2} \cdot 2] = - \frac{2}{( 2 x + 1 ) ^{3}}$ .

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NEET 2025, Physics Q15: Motion In One Dimension

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