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

Question

Dr. Abubacker · Accepted Answer

(d) Given, equation is
$y = {9x}^{2}\ \ \ldots.(i)$
Since, $x$-component of velocity remains constant, we have
$\frac{dx}{dt} = \frac{1}{3}{ms}^{- 1}\ \ \ \ldots(ii)$
From Eq. (i), we have $y$-component of velocity.
$\frac{dy}{dt} = 18x \bullet \left( \frac{dx}{dt} ight)^{2}$
$\frac{dy}{dt} = 18\left( \frac{dx}{dt} ight)^{2} = 18 	imes \left( \frac{1}{3} ight)^{2} = 2{ms}^{- 2}$
$	herefore\mathbf{a}_{y} = 2\ \widehat{\dot{\mathbf{j}}}\ {ms}^{- 2}$

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

Explanation

Reviewed by Dr. Abubacker

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