NEET 2025 · PhysicsPrevious Year Question
The relation between time t and position x of a particle moving along a straight line is given by $t = x^2 + x$. The acceleration of the particle is:
- A.$-\dfrac{2}{(2x+1)^3}$✓
- B.$\dfrac{2}{(2x+1)^3}$
- C.$-\dfrac{1}{(2x+1)^2}$
- D.$\dfrac{1}{(2x+1)^2}$
Correct Answer
(A) $-\dfrac{2}{(2x+1)^3}$
Solution & Explanation
\textbf{Answer:} (a) $-\dfrac{2}{(2x+1)^3}$ \textbf{Solution:} $\dfrac{dt}{dx}=2x+1 \Rightarrow v=\dfrac{dx}{dt}=\dfrac{1}{2x+1}$. $a=v\dfrac{dv}{dx}=\dfrac{1}{2x+1}\cdot\dfrac{-2}{(2x+1)^2}=-\dfrac{2}{(2x+1)^3}$.
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