Question bank/NEET UG/Physics

NEET UGPhysicsModerateNEET 2025

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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Sol. t=x2+xt = x^2 + x

dtdx=2x+1\frac{dt}{dx} = 2x+1

v=dxdt=1(2x+1)v = \frac{dx}{dt} = \frac{1}{(2x+1)}

dvdx=−2(2x+1)2\frac{dv}{dx} = \frac{-2}{(2x+1)^2}

a=vdvdx=[1(2x+1)][−2(2x+1)2]a = v \frac{dv}{dx} = \left[ \frac{1}{(2x+1)} \right] \left[ \frac{-2}{(2x+1)^2} \right]

=−2(2x+1)3= -\frac{2}{(2x+1)^3}

Source: geekyrakshit/indian-exam-questions / neet2025-code45

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