The position co-ordinates of a particle moving in a 3-D coordinate system is given by x = a cos ω t y = a sin ω t and z = a ω t The speed of the particle is: (a) √(2) aω (b) aω (c) √(3) aω (d) 2a ω

Answer: (a). vx = -aωsin ω t, vy = aωcos ω t, vz = aω v = √(a²ω²(sin ²ω t + cos ²ω t) + a²ω²) = √2 aω So the answer is (a) √2 aω.

The position co-ordinates of a particle moving in a 3-D coordinate… | JEE Main 2019 (9 Jan, Shift 2)

Q12JEE Main 2019 · 9 JAN · SHIFT 2Medium

Motion in a Plane (2-D Kinematics)

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:

Show answer and solution

Answer: (a)

$v_x = -a\omega\sin\omega t$ , $v_y = a\omega\cos\omega t$ , $v_z = a\omega$
$v = \sqrt{a^2\omega^2(\sin^2\omega t + \cos^2\omega t) + a^2\omega^2} = \sqrt2\,a\omega$
So the answer is (a) $\sqrt2\,a\omega$ .