A particle starts from the origin of coordinates at time t=0 and moves in the xy-plane with a constant acceleration $α$ in the y-direction. Its equation of motion is $y = β x^{2}$ . Its velocity component in the x-direction is

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$\frac{α}{2 β}$

$\sqrt{\frac{α}{2 β}}$

$\sqrt{\frac{2 α}{β}}$

variable

answer is D.

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Detailed Solution

Given, $y = β x^{2}$ ,

$∴ \frac{d y}{d t} = 2 β x \cdot \frac{d x}{d t} and \frac{d^{2} y}{d t^{2}} = 2 β [x \frac{d^{2} x}{d t^{2}} + {(\frac{d x}{d t})}^{2}]$

Given, $\frac{d^{2} y}{d t^{2}} = a_{y} = α$ and $\frac{d^{2} x}{d t^{2}} = a_{x} = 0 . Also, \frac{d x}{d t} = v_{x}$

Hence, $α = 2 β \cdot v_{x}^{2}$

$v_{x} = \sqrt{\frac{α}{2 β}}$

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