In some appropriate units, time (t) and position (x) relation of a moving particle is given by t = x² + x. The acceleration of the particle is (a) +(2)/((x + 1)³) (b) +(2)/(2x + 1) (c) -(2)/((x + 2)³) (d) -(2)/((2x + 1)³)

Answer: (d). dt/dx = 2x + 1 ⇒ v = (1)/(2x + 1) dv/dx = -(2)/((2x + 1)²) a = vdv/dx = -(2)/((2x + 1)³) So the answer is (d).

In some appropriate units, time (t) and position (x) relation of a… | NEET 2025

Q4NEET 2025Medium

Velocity, Acceleration & Variable Motion

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

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Answer: (d)

$\dfrac{dt}{dx} = 2x + 1 \Rightarrow v = \dfrac{1}{2x + 1}$
$\dfrac{dv}{dx} = -\dfrac{2}{(2x + 1)^2}$
$a = v\dfrac{dv}{dx} = -\dfrac{2}{(2x + 1)^3}$
So the answer is (d).

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