Q 11-03-049JEE MainJEE Main 2026 (2 Apr, Shift 1)Medium
The velocity of a particle is given as $\vec v=-x\hat i+2y\hat j-z\hat k$ m/s. The magnitude of acceleration at point $(1,2,4)$ is ______ $\text{m/s}^2$.
Answer: (B) $9$
Each velocity component depends only on its own coordinate, so $a_x=\dfrac{dv_x}{dt}=\dfrac{dv_x}{dx}\,v_x$, and similarly for $y$ and $z$.
$a_x=(-1)(-x)=x,\qquad a_y=(2)(2y)=4y,\qquad a_z=(-1)(-z)=z$
At $(1,2,4)$: $\vec a=\hat i+8\hat j+4\hat k$
$|\vec a|=\sqrt{1+64+16}=\sqrt{81}=9\ \text{m/s}^2$
Solution by Sreeraj P, M.Sc Physics