The position co-ordinates of a particle moving in a $3-D$ coordinates 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
JEE MAIN 2019, Medium
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$\mathrm{v}_{\mathrm{x}}=\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}=-\mathrm{a} \omega \sin \omega \mathrm{t}$
$\mathrm{v}_{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{a\omega} \cos \omega \mathrm{t}$
$\mathrm{v}_{\mathrm{z}}=\frac{\mathrm{d} z}{\mathrm{dt}}=\mathrm{a} \omega$
$\therefore \quad \mathrm{v}=\sqrt{\mathrm{v}_{\mathrm{x}}^{2}+\mathrm{v}_{\mathrm{y}}^{2}+\mathrm{v}_{\mathrm{z}}^{2}}=\mathrm{a} \omega \sqrt{2}$
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