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) $15\ \text{m/s}^2$
When $v$ is given as a function of $x$, use the chain rule:
$$a = \frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = v\frac{dv}{dx}$$
At $x = 1$: $v = 2 + 1 = 3$ m/s and $\dfrac{dv}{dx} = 4x + 1 = 5\ \text{s}^{-1}$.
$$a = 3 \times 5 = 15\ \text{m/s}^2$$
A common mistake is to report $dv/dx = 5$ as the acceleration.
Solution by Sreeraj P, M.Sc Physics