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Physics · PYQ NEET UG 2025
हिन्दीIn 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
Correct answer · Option 2
\( -\frac{2}{(2x+1)^3} \)
Explanation
Differentiating \( t = x^2 + x \) w.r.t time \( t \): \( 1 = (2x+1)v \implies v = \frac{dx}{dt} = \frac{1}{2x+1} \). Acceleration \( a = \frac{dv}{dt} = \frac{d}{dx}(\frac{1}{2x+1})\frac{dx}{dt} = \frac{-2}{(2x+1)^2} \cdot \frac{1}{2x+1} = \frac{-2}{(2x+1)^3} \).
