The position co-ordinates of a particle moving in a 3-D coordinates system is given by $$ \begin{aligned} & x=a \cos \omega t \\ & y=a \sin \omega t \\ & z=a \omega t \end{aligned} $$ The speed of the particle is:

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Kinematics

The position co-ordinates of a particle moving in a 3-D coordinates system is given by

x = a cos ω t y = a sin ω t z = aω t

The speed of the particle is:

(A) $2 aω$ ✓ Correct
(B) $aω$
(C) $3 aω$
(D) $2 aω$

Answer: (A)

Solution

The speed is just the magnitude of the velocity, so differentiate each coordinate. With

x = a cos ω t

y = a sin ω t

and

z = aω t

we get

v_{x} = - aω sin ω t

v_{y} = aω cos ω t

and

v_{z} = aω

. Then

v = v_{x}^{2} + v_{y}^{2} + v_{z}^{2} = a^{2} ω^{2} (sin^{2} ω t + cos^{2} ω t) + a^{2} ω^{2}

. The trig identity collapses the first part to

a^{2} ω^{2}

, leaving

v = 2 a^{2} ω^{2} = 2 aω

. Physically the particle is spiralling up a helix at constant speed.

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

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