Q 11-03-151JEE MainJEE Main 2019 (9 Jan, Shift 2)Easy
The position co-ordinates of a particle moving in a 3D coordinate system is given by
$$x = a\cos\omega t,\quad y = a\sin\omega t,\quad z = a\omega t$$
The speed of the particle is:
Answer: (A) $\sqrt{2}\,a\omega$
Differentiate each coordinate:
$$v_x = -a\omega\sin\omega t,\quad v_y = a\omega\cos\omega t,\quad v_z = a\omega$$
$$v = \sqrt{a^2\omega^2(\sin^2\omega t+\cos^2\omega t) + a^2\omega^2} = \sqrt{2}\,a\omega$$
Solution by Sreeraj P, M.Sc Physics