Q 11-02-003NEETNEET 2025Top questionMedium
In 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
Answer: (B) $-\dfrac{2}{(2x + 1)^3}$
Concept. Get $v = \dfrac{dx}{dt}$ by inverting $\dfrac{dt}{dx}$, then $a = v\dfrac{dv}{dx}$.
Step 1. Velocity.
$$\frac{dt}{dx} = 2x + 1 \;\Rightarrow\; v = \frac{1}{2x + 1}$$
Step 2. Acceleration.
$$a = v\frac{dv}{dx} = \frac{1}{2x + 1} \cdot \left(-\frac{2}{(2x + 1)^2}\right) = -\frac{2}{(2x + 1)^3}$$
Answer. $-\dfrac{2}{(2x + 1)^3}$ (Option B).
Solution by Sreeraj P, M.Sc Physics