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) $-\dfrac{2}{(2x + 1)^3}$
$$\frac{dt}{dx} = 2x + 1 \;\Rightarrow\; v = \frac{dx}{dt} = \frac{1}{2x + 1}$$
$$a = \frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = \left(-\frac{2}{(2x + 1)^2}\right)\cdot\frac{1}{2x + 1} = -\frac{2}{(2x + 1)^3}$$
Solution by Sreeraj P, M.Sc Physics