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A particle moves along the parabolic path y = ax² in such a way that the x-component of the velocity remains constant, say c. The acceleration of the particle is

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$ac \hat{k}$

$2 {ac}^{2} \hat{j}$

${ac}^{2} \hat{k}$

$a^{2} c^{2} \hat{j}$

answer is B.

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Detailed Solution

$\begin{array}{l} y = {ax}^{2} \\ \Rightarrow \frac{dy}{dt} = \dot{y} = (2 ax) \dot{x} = 2 acx \{∵ \frac{dx}{dt} = \dot{x} = c\} \end{array} Further a = \frac{d^{2} y}{{dt}^{2}} = \ddot{y} = 2 ac \dot{x} = 2 {ac}^{2}$

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