A particle moves along a parabolic path $y = 9 x^{2}$ in such a way that the x -component of velocity remains constant and has a value $\frac{1}{3} m s^{- 1} .$ The acceleration of the particle is

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$\frac{2}{3} \hat{j} m s^{- 2}$

$3 {\hat{j} m s}^{- 2}$

$\frac{1}{3} \hat{j} m s^{- 2}$

$2 \hat{j} m s^{- 2}$

answer is D.

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Detailed Solution

$Y = 9 x^{2}; V_{x} = \frac{1}{3} m / s; a_{x} = 0$

$\frac{d y}{d t} = 18 x \frac{d x}{d t} = 18 x v_{x}; \Rightarrow v_{y} = 18 x v_{x}$

$\frac{d v_{y}}{d t} = 18 [x \frac{d v_{x}}{d t} + v_{x} \cdot \frac{d x}{d t}] \Rightarrow a_{y} = 18 [x (a_{x}) + v_{x}^{2}]$

Given, $a_{x} = 0$

$a_{y} = 18 [{(\frac{1}{3})}^{2}] = 2 \hat{j} m / s^{2}$

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