Q 11-02-037NEETJEE MainMedium
A particle moves along the $x$-axis with a velocity that depends on its position as $v = 2x^2 + x$ (SI units). Its acceleration at $x = 1$ m is
Answer: (B) $15\ \text{m/s}^2$
Concept. When $v$ is given as a function of $x$, $a = v\dfrac{dv}{dx}$.
Step 1. At $x = 1$. $v = 2 + 1 = 3$ m/s; $\dfrac{dv}{dx} = 4x + 1 = 5\ \text{s}^{-1}$.
Step 2. Acceleration. $a = 3 \times 5 = 15\ \text{m/s}^2$ (not just $dv/dx$).
Answer. $15$ m/s$^2$ (Option B).
Solution by Sreeraj P, M.Sc Physics