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The coordinates of a moving particle at any time 't' are given by $x = α t^{3} and y = β t^{3}$ . The speed of the particle at time 't' is given by

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3 t^{2} \sqrt{α^{2} + β^{2}}

t^{2} \sqrt{α^{2} + β^{2}}

\sqrt{α^{2} + β^{2}}

3 t \sqrt{α^{2} + β^{2}}

answer is B.

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Detailed Solution

$x = α t^{3}$

$y = β t^{3}$

$v_{1} = 3 α t^{2}$

$v_{2} = 3 β t^{2}$

$v = \sqrt{{(3 α t^{2})}^{2} + {(3 β t^{2})}^{2}}$

$= \sqrt{9 α^{2} t^{4} + 9 β^{2} t^{4}}$

$= \sqrt{9 t^{4} (α^{2} + β^{2})}$

$= 3 t^{2} \sqrt{α^{2} + β^{2}}$

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