The position co-ordinates of a particle moving in a 3-D co-ordinates system is given by
$x = 2 \cos 10 t m$
$y = 2 \sin 10 t m$
and $z = 20 t m$
The speed of the particle is ( in $m s^{- 1}$ )

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$20 \sqrt{3}$

$20 \sqrt{2}$

answer is A.

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Detailed Solution

$V_{x} = \frac{d x}{d t} = - 20 \sin 10 t$

$V_{y} = \frac{d y}{d t} = 20 \cos 10 t$

$V_{z} = \frac{d z}{d t} = 20$

$∴$ speed $v = \sqrt{v_{x}^{2} + v_{y}^{2} + v_{z}^{2}}$

$v = \sqrt{{(- 20 \sin 10 t)}^{2} + {(20 \cos 10 t)}^{2} + 20^{2}}$

$v = 20 \sqrt{\sin^{2} 10 t + \cos^{2} 10 t + 1}$

$v = 20 \sqrt{2} m s^{- 1}$

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