The coordinates of a moving particle at any time $t$ are given by $x = c t^{2}$ and $y = b t^{2}$ . Then the speed of the particle is given by

2 t (c + b)

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2 t \sqrt{b c}

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

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

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Solution

The correct option is D $2 t \sqrt{c^{2} + b^{2}}$
$\frac{d x}{d t} = 2 c t$ and $\frac{d y}{d t} = 2 b t$
$∴ v = \sqrt{{(\frac{d x^{2}}{d t}) + {(\frac{d y}{d t})}^{2}}}$
$= \sqrt{{{(2 c t)}^{2} + {(2 b t)}^{2}}}$

$= 2 t \sqrt{(c^{2} + b^{2})}$

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