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:

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NEET 2025 Official Paper (Held On: 04 May, 2025)
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  1. \(-\frac{2}{(x+2)^3} \)
  2. \(-\frac{2}{(2x+1)^3}\)
  3. \(+\frac{2}{(x+1)^3}\)
  4. \(+\frac{2}{2x+1}\)

Answer (Detailed Solution Below)

Option 2 : \(-\frac{2}{(2x+1)^3}\)
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Correct option is: (2) −2 / (2x + 1)3

t = x2 + x

We have

dt/dx = 2x + 1

⇒ v = dx/dt = 1 / (2x + 1)

⇒ dv/dx = −2 / (2x + 1)2

⇒ a = v × dv/dx = [1 / (2x + 1)] × [−2 / (2x + 1)2]

= −2 / (2x + 1)3

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