At an instant t, the co-ordinates of a particle are $x = a t^{2}, y = b t^{2}$ and $z = 0.$ The magnitude of velocity of particle at an instant $t$ is

\frac{v}{\sqrt{3}}

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\frac{v}{\sqrt{2}}

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2 t \sqrt{a^{2} + b^{2}}

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t \sqrt{a^{2} + b^{2}}

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Solution

The correct option is C $2 t \sqrt{a^{2} + b^{2}}$
Given:
$x = a t^{2}, y = b t^{2}, z = 0$
Hence,
$v_{x} = \frac{d (a t)^{2}}{d t} = 2 a t v_{y} = \frac{d (b t^{2})}{d t} = 2 b t v_{z} = 0$
Therefore, magnitude of velocity is given by,
$v = \sqrt{v_{x}^{2} + v_{y}^{2} + v_{z}^{2}} = \sqrt{(2 a t)^{2} + (2 b t)^{2} + 0} = \sqrt{(2 a t)^{2} + (2 b t)^{2}} v = 2 t \sqrt{a^{2} + b^{2}}$

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Q . A t a n i n s \tan t t, t h e c o o r d i n a t e s o f a p a r t i c l e s a r e x = a t^{2}, y = b t^{2} a n d z = 0 . T h e m a g n i t u d e o f v e l o c i t y o f p a r t i c l e a t a n i n s \tan t t i s a . t \sqrt{a^{2} + b^{2}} b . \frac{v}{\sqrt{2}} c . \frac{v}{\sqrt{3}} d . 2 t \sqrt{a^{2} + b^{2}}

x - y

x = \frac{t^{3}}{3}, y = 2 t^{2}

t = 3

{cos}^{- 1} (\frac{x}{y \sqrt{z}})

(z - x - y)

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