JEE Main 2026 · 2 Apr · Shift 1 · Physics Q28

PhysicsSingle correct+4 / -1Question 28 of 75

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 ________ m/s².

\sqrt6

\sqrt{33}

Answer and worked solution

Official answer: B

Acceleration. $a_x=\frac{dv_x}{dt}=-v_x=x$ , $a_y=2v_y=4y$ , $a_z=-v_z=z$ .

At (1, 2, 4). $\vec a=(1,8,4)$ , so $|\vec a|=\sqrt{1+64+16}=9$ m/s².

(B)

Topics:Motion in a plane

JEE Main 2026 · 2 Apr · Shift 1, Physics question 28