Mathematics in Physics

JEE Main level, single correct, medium

37 of 121

Question 37 of 121

The position co-ordinates of a particle moving in a 3D3D coordinate system is given by

x=acos⁡⁡ωtx=a\cos ⁡\omega t

y=asin⁡⁡ωty=a\sin ⁡\omega t

and z=aωtz=a\omega t

The speed of the particle is:

Show answer and solution

Answer: (A) 2aω\sqrt{2}a\omega

x=acos⁡⁡ωtx=a\cos ⁡\omega t

y=asin⁡⁡ωty=a\sin ⁡\omega t

z=aωtz=a\omega t

vx=−aωsin⁡⁡ωt{v}_{x}=-a\omega \sin ⁡\omega t

vy=aωcos⁡⁡ωt{v}_{y}=a\omega \cos ⁡\omega t

v2=aω{v}_{2}=a\omega

v=vx2+vy2+vz2v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}+{v}_{z}^{2}}

=(−aωsin⁡⁡ωt)2+(aωcos⁡⁡ωt)2+(aω)2=\sqrt{{\left(-a\omega \sin ⁡\omega t\right)}^{2}+{\left(a\omega \cos ⁡\omega t\right)}^{2}+{\left(a\omega \right)}^{2}}

=aω2=a\omega \sqrt{2}

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