A Particle moves uniformly with speed v along the parabolic path y=kx2. Taking k as a positive constant, the magnitude of acceleration of the particle at x=0 is given by:

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A Particle moves uniformly with speed $v$ along the parabolic path $y = {kx}^{2}$ . Taking $k$ as a positive constant, the magnitude of acceleration of the particle at $x = 0$ is given by:

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$\frac{k v^{2}}{2}$

$k v^{2}$

$\frac{k v^{2}}{4}$

$2 k v^{2}$

answer is A.

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Detailed Solution

At $x = 0, v_{y} = 0; v_{x} = v$ and $a_{x} = 0;$ [ $∵$ particle moves uniformly]

$v_{y} = 2 k x v_{x}; a_{y} = 2 k [v_{x}^{2} + x a_{x}]; ∴ a_{y} = 2 k v^{2} \vec{a} = 2 k v^{2} \hat{j}$

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