PhysicsNEET 2025Displacement, velocity, acceleration; graphs

Motion in a Straight Line

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

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

Solution

$t = x^{2} + x \Rightarrow \frac{d t}{d x} = 2 x + 1 \Rightarrow v = \frac{1}{2 x + 1}$ . Then $a = \frac{d v}{d t} = \frac{d v}{d x} v = - \frac{2}{( 2 x + 1 ) ^{2}} \cdot \frac{1}{2 x + 1} = - 2 (2 x + 1)^{- 3}$ .

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Motion in a Straight Line

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