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PhysicsNEET 2025Displacement, velocity, acceleration; graphs

Motion in a Straight Line

In some appropriate units, the time (tt) and position (xx) relation of a moving particle is given by t=x2+xt = x^2 + x. The acceleration of the particle is:

  1. A.−2(x+2)−3-2(x + 2)^{-3}
  2. B.−2(2x+1)−3-2(2x + 1)^{-3}✓
  3. C.(x+1)−3(x + 1)^{-3}
  4. D.2(2x+1)22(2x + 1)^{2}

Solution

t=x2+x⇒dtdx=2x+1⇒v=12x+1t = x^2 + x \Rightarrow \frac{dt}{dx} = 2x + 1 \Rightarrow v = \frac{1}{2x+1}. Then a=dvdt=dvdxv=−2(2x+1)2⋅12x+1=−2(2x+1)−3a = \frac{dv}{dt} = \frac{dv}{dx}v = -\frac{2}{(2x+1)^2}\cdot\frac{1}{2x+1} = -2(2x+1)^{-3}.

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