JEE PYQ: Motion In Two Dimensions Question 20
Question 20 - 2019 (09 Jan Shift 2)
The position co-ordinates of a particle moving in a 3-D coordinate system is given by $x = a\cos\omega t$, $y = a\sin\omega t$ and $z = a\omega t$. The speed of the particle is:
(a) $\sqrt{2}, a\omega$ (b) $a\omega$ (c) $\sqrt{3}, a\omega$ (d) $2a\omega$
Show Answer
Answer: (a)
Solution
$v = \sqrt{v_x^2 + v_y^2 + v_z^2} = \sqrt{a^2\omega^2 + a^2\omega^2} = \sqrt{2}, a\omega$.
BETA
