The position coordinates of a particle moving | vedclass

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Question

(A) The velocity components are obtained by differentiating the position coordinates with respect to time $t$:$v_x = \frac{dx}{dt} = -a \omega \sin \o

Vedclass · Accepted Answer

$\sqrt{2} a \omega$

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(A) The velocity components are obtained by differentiating the position coordinates with respect to time $t$:$v_x = \frac{dx}{dt} = -a \omega \sin \omega t$$v_y = \frac{dy}{dt} = a \omega \cos \omega t$$v_z = \frac{dz}{dt} = a \omega$The speed $v$ is the magnitude of the velocity vector:$v = \sqrt{v_x^2 + v_y^2 + v_z^2}$$v = \sqrt{(-a \omega \sin \omega t)^2 + (a \omega \cos \omega t)^2 + (a \omega)^2}$$v = \sqrt{a^2 \omega^2 \sin^2 \omega t + a^2 \omega^2 \cos^2 \omega t + a^2 \omega^2}$$v = \sqrt{a^2 \omega^2 (\sin^2 \omega t + \cos^2 \omega t) + a^2 \omega^2}$Since $\sin^2 \omega t + \cos^2 \omega t = 1$,we get:$v = \sqrt{a^2 \omega^2 (1) + a^2 \omega^2} = \sqrt{2 a^2 \omega^2} = \sqrt{2} a \omega$

The position coordinates of a particle moving in a $3-D$ coordinate system are given by $x = a \cos \omega t$,$y = a \sin \omega t$,and $z = a \omega t$. The speed of the particle is:

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